Fundementals Of Nuclear Physics

核能发电基本常识

认识科学做一个人聪明的外行人 http://www.jxcad.com.cn/?u=74032

核能发电基本常识

认识科学做一个人聪明的外行人 http://www.jxcad.com.cn/?u=74032

Atoms and Nuclei Each topic is subdivided into a number of episodes. An episode represents a coherent section of teaching – perhaps one or two lessons. Each episode contains links to a number of activities.

Quantum physics Episode 500: Preparation for quantum physics topic Episode 501: Spectra and energy levels Episode 502: The photoelectric effect

Lasers Episode 503: Preparation for lasers topic Episode 504: How lasers work

Wave particle duality Episode 505: Preparation for wave-particle duality topic Episode 506: Particles as waves Episode 507: Standing waves

Radioactivity Episode 508: Preparation for radioactivity topic Episode 509: Radioactive background and detectors Episode 510: Properties of radiations Episode 511: Absorption experiments Episode 512: Nuclear equations Episode 513: Preparation for exponential decay topic Episode 514: Patterns of decay Episode 515: The radioactive decay formula Episode 516: Exponential and logarithmic equations

Accelerators and detectors Episode 517: Preparation for accelerators and detectors topic Episode 518: Particle accelerators Episode 519: Particle detectors

Rutherford’s experiment Episode 520: Preparation for Rutherford scattering Episode 521: Rutherford’s experiment Episode 522: The Size of the Nucleus

Nuclear stability Episode 523: Preparation for nuclear stability topic. Episode 524: Stable nuclides Episode 525: Binding energy

Nuclear fission Episode 526: Preparation for nuclear fission topic Episode 527: Nuclear transmutation Episode 528: Controlling fission

X-ray and neutron diffraction Episode 529: Preparation for X-ray and neutron diffraction topic Episode 530: X-ray diffraction Episode 531: Neutron diffraction

Particles and antiparticles Episode 532: Preparation for particle physics topic Episode 533: The particle zoo Episode 534: Antiparticles and the lepton family Episode 535: Particle reactions Episode 536: Vector bosons and Feynman diagrams

Quarks Episode 537: Preparation for deep scattering and quarks topic Episode 538: Electron scattering Episode 539: Deep inelastic scattering Episode 540: Quarks and the standard model

资料来源： http://www.iop.org/activity/education/Projects/Teaching%20Advanced%20Physics/Atomic%20and%20 Nuclei/index.html

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Episode 500: Preparation for the quantum physics topic It is most likely that this will be a topic that is completely unfamiliar to all students from pre-16 level courses. It is also one where the impact on everyday life is apparently very limited and where students think that they have no direct experience of its consequences. Nothing could be further from the truth: e.g. all modern electronics relies on quantum physics. However these very facts make it a most fascinating subject and one whose very novelty should attract the interest of your students.

Episode 501: Spectra and energy levels Episode 502: The photoelectric effect

Main aims Students will: 1. Know that atoms absorb and emit light as quanta (photons). 2. Explain how this is used to explain emission spectra. 3. Know how to calculate photon energies. 4. Describe the photoelectric effect, and explain it in terms of photons and electrons. 5. Describe an experiment to determine Planck’s constant using the photoelectric effect.

Prior knowledge Students should be familiar with the nuclear model of the atom. They should know that metals contain ‘free’ (conduction) electrons.

Where this leads These episodes (and work on lasers) lead on to general ideas about wave-particle duality.

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Episode 501: Spectra and energy levels Summary Demonstration: Looking at emission spectra. (20 minutes) Discussion: The meaning of quantisation. (20 minutes) Demonstration: Illustrating quantisation. (10 minutes) Discussion: Energy levels in a hydrogen atom. (10 minutes) Worked example + Student Questions: Calculating frequencies. (20 minutes) Discussion: Distinguishing quantisation and continuity. (5 minutes) Worked example: Photon flux. (10 minutes) Student calculations: Photon flux. (20 minutes) Student experiment: Relating photon energy to frequency. (30 minutes)

Demonstration: Looking at emission spectra Show a white light and a set of standard discharge lamps: sodium, neon, hydrogen and helium. Allow students to look at the spectrum of each gas. They can do this using a direct vision spectroscope or a bench spectroscope, or simply by holding a diffraction grating up to their eye. What is the difference? (The white light shows a continuous spectrum; the gas discharge lamps show line spectra.)

Emission and absorption spectra

(Diagram: resourcefulphysics.org)

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The spectrum of a gas gives a kind of 'finger print' of an atom. You could relate this to the simple flame tests that students will have used at pre-16 level. Astronomers examine the light of distant stars and galaxies to discover their composition (and a lot else).

Discussion: The meaning of quantisation Relate the appearance of the spectra to the energy levels within the atoms of the gas. Students will already have a picture of the atom with negatively charged electrons in orbit round a central positively charged nucleus. Explain that, in the classical model, an orbiting electron would radiate energy and spiral in towards the nucleus, resulting in the catastrophic collapse of the atom. This must be replaced by the Bohr atomic structure – orbits are quantised. The electron’s energy levels are discrete. An electron can only move directly between such levels, emitting or absorbing individual photons as it does so. The ground state is the condition of lowest energy – most electrons are in this state. Think about a bookcase with adjustable shelves. The bookshelves are quantised – only certain positions are allowed. Different arrangement of the shelves represents different energy level structures for different atoms. The books represent the electrons, added to the lowest shelf first etc

Demonstration: Illustrating quantisation Throw a handful of polystyrene balls round the lab and see where they settle. The different levels on which they end up – the floor, on a desk, on a shelf – gives a very simple idea of energy levels. Some useful clipart can be found below TAP 501-1: The emission of light from an atom Resourceful Physics >Teachers>OHT>Emission of Light An energy input raises the electrons to higher energy levels. This energy input can be by either electrical, heat, radiation or particle collision. When the electrons fall back to a lower level there is an energy output. This occurs by the emission of a quantum of radiation.

Discussion: Energy levels in a hydrogen atom Show a scale diagram of energy levels. It is most important that this diagram is to scale to emphasise the large energy drops between certain levels. The students may well ask the question, “Why do the states have negative energy?” This is because the zero of energy is considered to be that of a free electron 'just outside' the atom. All

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energy states 'below' this – i.e. within the atom are therefore negative. Energy must be put into the atom to raise the electron to the 'surface' of the atom and allow it to escape.

Worked example + student questions: Calculating frequencies Calculate the frequency and wavelength of the quantum of radiation (photon) emitted due to a transition between two energy levels. (Use two levels from the diagram for the hydrogen atom.) E2 – E1 = hf Point out that this equation links a particle property (energy) with a wave property (frequency). Ask your students to calculate the photon energy and frequency for one or two other transitions. Can they identify the colour or region of the spectrum of this light? Emphasise the need to work in SI units. The wavelength is expressed in metres, the frequency in hertz, and the energy difference in joules. You may wish to show how to convert between joules and electronvolts.

Discussion: Distinguishing quantisation and continuity The difference between the quantum theory and the classical theory is similar to the difference between using bottles of water (quantum) or water from a tap (classical). The bottles represent the quantum idea and the continuous flow from the tap represents the classical theory. The quantisation of energy is also rather like the kangaroo motion of a car when you first learn to drive – it jumps from one energy state to another, there is no smooth acceleration. It is all a question of scale. We do not 'see' quantum effects generally in everyday life because of the very small value of Planck's constant. Think about a person and an ant walking across a gravelled path. The size of the individual pieces of gravel may seem small to us but they are giant boulders to the ant. We know that the photons emitted by a light bulb, for example, travel at the speed of light (3 × 108 m s-1) so why don’t we feel them as they hit us? (Although all energy is quantised we are not aware of this in everyday life because of the very small value of Planck’s constant.) Students may worry about the exact nature of photons. It may help if you give them this quotation from Einstein: ‘All the fifty years of conscious brooding have brought me no closer to the answer to the question, “What are light quanta?”. Of course, today every rascal thinks he knows the answer, but he is deluding himself.’ (Illustrations: resourcefulphysics.org)

Worked example: Photon flux Calculate the number of quanta of radiation being emitted by a light source.

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Consider a green 100 W light. For green light the wavelength is about 6 × 10-7 m and so: Energy of a photon = E = hf =hc / λ= 3.3 × 10-19 J The number of quanta emitted per second by the light N = 100 × λ / hc = 3 × 1020 s-1.

Student calculations: Photon flux TAP 501-2: Photons streaming from a lamp TAP 501-3: Quanta

Student experiment: Relating photon energy to frequency TAP 501-4: Relating photon energy to frequency. Students can use LEDs of different colours to investigate the relationship between frequency and photon energy for light.

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TAP 501-1: The emission of light from an atom

Electron in orbit round a nucleus in an atom

Energy input. The electron is excited and rises to a higher energy level (shell).

The electron falls back to its original energy level and energy is emitted in the form of radiation. The bigger the drop the greater the energy emitted and the shorter wavelength the radiation has (blue light).

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TAP 501-2: Photons streaming from a lamp What to do Complete the questions below on the sheet. Provide clear statements of what you are estimating; show what calculations you are performing and how these give the answers you quote. Try to show a clear line of thinking through each stage.

Steps in the calculation 1.

Estimate the power of a reading lamp in watts.

2.

Estimate the average wavelength of a visible photon.

3.

Calculate the energy transferred by each photon.

4.

Calculate the number of photons emitted by the lamp in each second.

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Practical advice This question, or a substitute for it, needs to come early on in the discussion of photons to avert questions concerning our inability to be aware of single photons. However, single photon detectors are now used in astronomy etc.

Alternative approaches This may be prefaced or supplemented by such a calculation performed in class. It is well done by linking to other such questions that yield large numbers.

Social and human context Every time we meet a pervasive quantity like power it is useful to compare it to our place in the Universe (75 W or so as a useful power output over any length of time) and to compare developed and developing countries in this respect.

Answers and worked solutions 1.

P = 40 W

2.

λ = 5 × 10–7 m

3.

Calculate the frequency of the photons corresponding to this wavelength:

f = =

c

λ 3 × 10 8 m s −1 5 × 10 −7 m

= 6 × 1014 Hz. Now calculate the energy of each photon:

E = hf

= 6 × 10 −34 J s −1 × 6 × 1014 Hz = 4 × 10 −19 J. 4.

Energy per second = 40 J s–1 Energy per photon = 4 × 10–19 J. energy per second photons per second = energy per photon =

40 J s -1 4 × 10 −19 J

= 1× 10 20.

External reference 9

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This activity is taken from Advancing Physics chapter 7, question 20E

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TAP 501- 3: Quanta Speed of electromagnetic radiation in free space (c) = 3.00 x 108 m s-1 Planck’s constant (h) = 6.63 x 10-34 J s 1. Write down the equation for the quantum energy of a photon in terms of its frequency.

2. Calculate the energies of a quantum of electromagnetic radiation of the following wavelengths: (a)

gamma rays

wavelength

10-3 nm

(b)

X rays

wavelength

0.1 nm

(c)

violet light

wavelength

420 nm

(d)

yellow light

wavelength

600 nm

(e)

red light

wavelength

700 nm

(f)

microwaves

wavelength

2.00 cm

(g)

radio waves

wavelength

254 m

3. Calculate the wavelengths of quanta of electromagnetic radiation with the following energies: (a)

6.63 x 10-19 J

(b)

9.47 x 10-25 J

(c)

1.33 x 10-18 J

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(d)

3.98 x 10-20 J

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Practical advice Pupils may need to be reminded that a wavelength of 10-3 nm is 1 x 10-12 m and that some students could need help in using their calculators.

Answers and worked solutions 1

E = hf

2 (a)

f = c/λ

E = hf so E = h c/λ

E = (6.63 x 10-34 x 3 x 108) / (1 x 10-12) = 1.99 x 10-13 J (b)

E = 1.99 x 10-15 J

(c)

E = 4.74 x 10-19 J

(d)

E = 3.01 x 10-19 J

(e)

E = 2.84 x 10-19 J

(f)

E = 9.95 x 10-24 J

(g)

E = 7.83 x 10-19 J

3 (a)

λ = hc/E

(b)

0.21 m

(c)

1.5 x 10-7 m

(d)

λ = (6.63 x 10-34 x 3 x 108) / 6.63 x 10-19 = 3 x 10-7 m (150 nm)

-6

5 x 10 m

13

(300 nm)

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TAP 501- 4: Relating energy to frequency Photons have a characteristic energy Light of a particular colour is a stream of photons of a specific frequency. Light appears granular when seen at the finest scale. A bright light delivers lots of energy every second. If light is granular then the amount of energy must be related to the number of granules arriving each second. The intensity of the light will also depend on the energy delivered by each granule. This activity relates the energy of each photon to the frequency of that photon.

You will need 9

multiple LED array

9

peering tube

9

power supply, 5 V (smooth and regulated)

9

multimeter

9

five 4 mm leads

Measuring energy The energy released by each electron as it travels through the LED is transferred to a photon. To measure the energy released by each electron measure the potential difference across the LED when it just glows. Then we multiply this figure by the charge on the electron (1.6 × 10–19 C). The quantity that characterises the photon is the frequency so we then seek to find a connection between this frequency and the energy. 1.

Set up the circuit and check that each LED can be lit by altering the pd 100 Ω

select LEDs one at a time by flying lead

select LED by flying lead

+5 V

to 5 V power supply

Knob on pot Alter pd so that LED just lights

V 0V

470 nm

502 nm

to voltmeter

2.

Calculate the frequency of the LEDs.

14

563 nm

585 nm

620 nm

650 nm

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3.

Measure the pd just sufficient to strike each LED. At this pd the energy supplied by the electrons is all transferred to photons. Use the peering tube to cut out room lighting.

4.

Look for a pattern connecting energy to frequency (plot a graph!). You should be prepared to re-measure any points that do not fit and to check your results with those from other measurements.

5.

See if you can quantify the relationship. By how much does the energy of the photon change for each hertz?

Energy and frequency 1.

The energy associated with a photon is related to its frequency.

2.

This relationship introduces the important quantity, h, the Planck constant.

Practical advice We suggest setting up several competing research groups, and actively encouraging students to form a consensus about the relationship between frequency and energy. An appropriate degree of collaboration gets the correct answer; inappropriate degrees yield a work of fiction or no consensus. Thus can physics progress. Students will know about the existence of an LED from previous work on electricity and will know that it conducts in one direction only. Thus electrons, simply introduced as what moves when electricity is conducted, can be presented as meeting an electrical barrier when the LED is reverse biased and falling down that barrier when forward biased. This simple model of the action of an LED is enough for this purpose. Connecting this electrical model to an energetic model requires the notion of potential difference to be reviewed as being likely to be the sensible way of determining the height of the barrier and the energy as being the potential difference times the charge on the electron. Analogies with the energy released in falling down a hill can reinforce this idea. So as to make sure that none of the electrical energy is dissipated we need to insist that we require the smallest pd across the photodiode. This energy, plotted against the frequency of emitted light (taken from the manufacturer's specifications), can then be used. Experience shows that the measurements made by the students may not be so accurate, and that encouragement to settle on a simple pattern, together with the consensual approach suggested above and the ability to make several measurements before deciding on the accurate answer, will be necessary. A class using a graph-plotting package may make this review and interaction more likely. Students should easily establish E = h f. A consensus on the value of h should give a value that is far from embarrassing. Students who are red/green or other forms of colour blindness will get different results. Often red/green colour blind students need a higher striking pd to see some light from the LED or may not be able to see particular wavelengths of light.

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Sample results: LED

Wavelength

Frequency

colour

/ nm

14

/ 10 Hz

Blue

470

Green

Striking pd

Energy

h

/V

/J

–34

6.38

2.38

0.381

6.0

563

5.33

1.69

0.270

5.1

Yellow

585

5.12

1.63

0.261

5.1

Orange

620

4.83

1.48

0.237

4.9

Red

650

4.62

1.47

0.235

5.1

/ 10

Js

Technician’s information The array of LEDs is used to make the connection between the frequency at which a photon is emitted and the energy carried by that photon. Measurements are made of the minimum pd required to just turn an LED on and of the wavelength of light from it. The wavelength may be better taken directly from the manufacturer's specification. Each LED emits photons of one characteristic frequency, specified in many catalogues by the peak wavelength emitted by the LED. You need a wide range of wavelengths, fairly evenly spaced, so as to get a reasonable graph of pd against frequency or wavelength. You may find a wider range of LEDs available at lower cost than when this design was first produced, if so then exceeding the range suggested is fine; reducing that range does not yield a reliable graph. To measure the energy carried by each photon students will need to be able to measure the pd applied across each LED in turn. Students then plot energy / frequency. So it may be useful to mark the peak frequencies or wavelengths emitted by the LEDs on the apparatus.

How to make it The requirement is to be able to apply a variable pd, 0 - 3 V, across a series of LEDs, one at a time in turn. 5 LEDs are sufficient. Increasing the range of frequencies is more important than adding more LEDs. We suggest that the LEDs be mounted in plastic ducting, a 150 mm length of 40 mm by 25 mm proved sufficient to mount all the components neatly. Here is a circuit diagram:

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100 Ω

select LEDs one at a time by flying lead

select LED by flying lead

+5 V

to 5 V power supply

Knob on pot Alter pd so that LED just lights

V 0V

470 nm

502 nm

563 nm

585 nm

620 nm

650 nm

to voltmeter

Variations are, of course possible. You may choose not to have sockets (which enable the current through the LEDs to be measured), or perhaps to have only one return wire so that the LEDs come on one after another as the pd is increased. A suitable protective resistor (roughly equal in value to the resistance of the shortest wavelength LED) in series with the wiper arm of the potentiometer will prevent applying too much pd across the LEDs.

Important features •

Use as wide a range of wavelengths for the LEDs as are currently available.

•

The LEDs should all be mounted in clear plastic, not mounted in coloured plastic.

•

The LEDs should all have approximately the same power output.

•

Provide a narrow opaque tube, about the length of a pencil, through which to peer at each LED, excluding extraneous light, so as to detect when it just comes on, and obtain a good value for the pd needed. Black paper or card is usually sufficient for the tube.

•

The arrangement of the potential divider together with the protective resistor to protect the LEDs allows the apparatus to be driven from a 5 V supply whilst protecting the LEDs. Other arrangements might also work.

Alternative approaches To establish the connection between the energy associated with each click as a Geiger counter detects a gamma ray photon we suggest measuring the energy required to release one photon of light in a light-emitting diode. This has the advantage that we can cheaply try out several frequencies and rapidly obtain a picture of how energy varies with frequency. It is, of course, not the only technique for establishing a link between frequency and energy. You may substitute others. The photoelectric effect has been used for this for many years, as has appeal to the evidence of spectra.

Social and human context

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Studying engineering barriers of different heights to release different amounts of energy takes us into the structures of semiconductor materials and engineering band gaps, which, whilst fascinating, does not contribute to the central understanding sought here. It is, however, central to quantum engineering. This can be used to start a discussion linking the potential barrier to an energetic understanding of the situation.

External reference This activity is taken from Advancing Physics chapter 7, activity 10E

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Episode 502: The photoelectric effect This episode introduces an important phenomenon. Light releases electrons from metal surfaces.

Summary Demonstration: The basic phenomenon. (15 minutes) Discussion: Summarising the phenomenon. (10 minutes) Discussion: An analogy. (5 minutes) Student questions: Using the photoelectric equation. The Millikan experiment: to verify Einstein’s photo-electric relationship (30 minutes) Student experiment: Measuring Planck’s constant. (30 minutes)

Demonstration: The basic phenomenon Introduce the topic by demonstrating the electroscope and zinc plate experiment. TAP 502-1: Simple photoelectric effect demonstration Point out to the students that the photoelectric effect is apparently instantaneous. However, the light must be energetic enough, which for zinc is in the ultraviolet region of the spectrum. If light were waves, we would expect the free electrons to steadily absorb energy until they escape from the surface. This would be the case in the classical theory, in which light is considered as Negatively waves. We could wait all day and still the red charged zinc light would not liberate electrons from the zinc plate Ultra violet plate. light So what is going on? We picture the light as quanta of radiation (photons). A single electron captures the energy of a single photon. The emission of an electron is instantaneous as long as the energy of each incoming quantum is big enough. If an individual photon has insufficient energy, the electron will not be able to escape from the metal.

Discussion: Summarising the phenomenon Summarise the important points about the photoelectric effect.

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Gold leaf falls immediately the zinc plate is illuminated with ultra violet light (Diagram: resourcefulphysics.org)

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There is a threshold frequency (i.e. energy), below which no electrons are released. The electrons are released at a rate proportional to the intensity of the light (i.e. more photons per second means more electrons released per second). The energy of the emitted electrons is independent of the intensity of the incident radiation. They have a maximum KE.

Discussion: An analogy Try this analogy, which involves ping-pong balls, a bullet and a coconut shy. A small boy tries to dislodge a coconut by throwing a ping-pong ball at it – no luck, the ping-pong ball has too little energy! He then tries a whole bowl of ping-pong balls but the coconut still stays put! Along comes a physicist with a pistol (and an understanding of the photoelectric effect), who fires one bullet at the coconut – it is instantaneously knocked off its support. Ask how this is an analogy for the zinc plate experiment. (The analogy simulates the effect of infrared and ultra violet radiation on a metal surface. The ping-pong balls represent low energy infrared, while the bullet takes the place of high-energy ultra violet.) Now you can define the work function. Use the potential well model to show an electron at the bottom of the well. It has to absorb the energy in one go to escape from the well and be liberated from the surface of the material.

Units The electronvolt is introduced because it is a convenient small unit. You might need to point out that it can be used for any (small) amount of energy, and is not confined to situations involving electrically accelerated electrons.

Potential well It is useful to compare the electron with a person in the bottom of a well with totally smooth sides. The person can only get out of the well by one jump, they can't jump half way up and then jump again. In the same way an electron at the bottom of a potential well must be given enough energy to escape in one 'jump'. It is this energy that is the work function for the material. Now you can present the equation for photoelectric emission: high energy violet quantum

Electron leaves metal Electron energy

Quantum energy

Potential well

Work function

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(Diagram: resourcefulphysics.org)

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Energy of photon E = hf Picture a photon being absorbed by one of the electrons which is least tightly bound in the metal. The energy of the photon does two things. Some of it is needed to overcome the work function φ. The rest remains as KE of the electron.

hf = φ + (1/2) mv2 A voltage can be applied to bind the electrons more tightly to the metal. The stopping potential Vs is just enough to prevent any from escaping:

hf = φ + eVs

Student questions: Using the photoelectric equation Set the students some problems using these equations. TAP.502-2: Photoelectric effect questions TAP 502-4: Student Question. The Millikan experiment The Millikan experiment question may best come after TAP 502-3: Student experiment: Measuring threshold frequency.

Student experiment: Measuring Planck’s constant TAP502-3: Measuring threshold frequency Students can measure Planck’s constant using a photocell. (Some useful clipart is given here below)

to oscilloscope

21

coloured filter

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TAP 502-1: Simple photoelectric effect demonstration Resources needed 9

Gold leaf electroscope or coulomb meter

9

Zinc plate attachment (sand-papered clean to remove oxidation)

9

Laser (class 2)

9

Mains lamp (a desk lamp is ideal)

9

Ultra violet lamp with clear quartz envelope

Safety A class 2 laser requires a warning: Do not stare down the beam. A short-wave UV lamp must be shielded so that the UV emerges through a hole. The hole is always directed away from eyes. The presence of UV can be demonstrated by showing fluorescence of paper.

Technique Attach the zinc plate to the top of the electroscope. (A coulomb meter can be used instead of the electroscope.) Charge the plate negatively.

o

No effect

No effect

With U.V. leaf falls immediately (Diagrams: resourcefulphysics.org)

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Shine red laser light onto the cleaned zinc plate – no effect. Use a mains light bulb emitting white light – no effect. Use an ultra violet lamp – the leaf falls immediately.

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TAP 502-2: Photoelectric effect questions hf = φ + (1/2) mv2 and hf = φ + eVs e = 1.60 x 10-19 C, h = 6.63 x 10-34 J s, mass of electron = 9.11 x 10-31 kg 1

The work function for lithium is 4.6 x 10-19 J.

(a)

Calculate the lowest frequency of light that will cause photoelectric emission.

(b)

What is the maximum energy of the electrons emitted when light of 7.3 x 1014 Hz is used?

2

Complete the table.

Metal Sodium

Work Function

Work Function

Frequency

Maximum KE of

/eV

/J

used /Hz

Ejected electrons /J

2.28

6 x 10

14

3.68 x 10-19

Potassium

0.32 x 10-19 1 x 1015

Lithium

2.9

Aluminium

4.1

0.35 x 10-19

Zinc

4.3

1.12 x 10-19

Copper

7.36 x 10-19

1 x 1015

3

The stopping potential when a frequency of 1.61 x 1015 Hz is shone on a metal is 3 V.

(a)

What is energy transferred by each photon?

(b)

Calculate the work function of the metal.

(c)

What is the maximum speed of the ejected electrons?

4

Selenium has a work function of 5.11 eV. What frequency of light would just eject electrons? (The threshold frequency is when the max KE of the ejected electrons is zero)

5

A frequency of 2.4 x 1015 Hz is used on magnesium with work function of 3.7 eV.

(a)

What is energy transferred by each photon?

(b)

Calculate the maximum KE of the ejected electrons.

(c)

The maximum speed of the electrons.

(d)

The stopping potential for the electrons.

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Answers and worked solutions 1(a)

hf = φ hf = 4.60 x 10-19 f = 4.60 x 10-19 / 6.63 x 10-34 = 6.94 x 1014 Hz

(b)

hf = φ + (1/2) mv2. (6.63 x 10-34 x 7.30 x 1014) = 4.60 x 10-19 + (1/2) mv2 4.84 x 10-19 - 4.60 x 10-19 = (1/2) mv2 = 0.24 x 10-19 J

2

Metal

Work Function

Work Function

Frequency

Maximum KE of

/ eV

/J

used / Hz

ejected electrons / J

-19

6 x 10

14

0.35 x 10-19

Sodium

2.28

3.65 x 10

Potassium

2.30

3.68 x 10-19

6 x 1014

0.32 x 10-19

Lithium

2.90

4.64 x 10-19

1. x 1015

1.99 x 10-19

Aluminium

4.10

6.56 x 10-19

1.04 x 1015

0.35 x 10-19

Zinc

4.30

6.88 x 10-19

1.2 x 1015

1.12 x 10-19

Copper

4.60

7.36 x 10-19

1 x 1015

0

For copper 1 x 1015 Hz is below the threshold frequency so no electrons are ejected. 3 (a)

1.07 x 10-18 J

(b)

hf = φ + eVs, so φ = hf - eVs, so φ = 1.07 x 10-18 – (1.6 x 10-19 x 3) = 5.9 x 10-19 J

(c)

eVs = (1/2) mv2 so (1.60 x 10-19 x 3) = 0.5 x 9.11 x 10-31 x v2 so v2 =1.04 x 1012 and v = 1.02 x 106 m s-1

4

1.2 x 1015 Hz

5 (a)

1.6 x 10-18 J

(b)

(1/2) mv2 = 1.x 10-18 J (c)

(d)

v2 = 1.1 x 1012 so v = 1.1 x 106 m s-1

eVs = (1/2) mv2 so eVs = 1.00 x 10-18 and Vs = 0.63 V

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TAP502- 3: Measuring threshold frequency Use a white light source and a set of coloured filters to find the threshold frequency, and hence the work function, of the photosensitive material in a photocell. This may well be a standard piece of kit in your school or college. Use a spreadsheet to plot and analyse a graph of your results. This experiment uses a photocell to investigate the photoelectric effect. Light of various frequencies is incident on the cell and photoelectrons are emitted and then form an electric current. A white light source is shone through various coloured filters to produce a series of different frequencies of light falling on the photocell. The current of photoelectrons produced in the cell maybe amplified internally and is measurable on the ammeter. Otherwise an amplifying picoammeter is needed.

The potential divider provides an adjustable voltage. The incident frequency, threshold frequency, and stopping potential are related by the following equation: hf =eV + hfo

Procedure For each coloured filter, adjust the potential divider until the stopping potential has been reached. Record the stopping potential and the wavelength of light transmitted by the filter. Select the middle of the range as the transmitted frequency. NB: the filter might have written on it the range of wavelengths it transmits measured in

Enter your results of frequency and stopping potential on a spreadsheet and plot a graph of frequency f versus stopping potential V. Use your graph to determine the value of the threshold frequency f and hence calculate the work function φ. Express your result in J and in eV. Estimate the uncertainties in your measurements of V and in the values of f that you have used. Use these estimates to add error bars to your graph and hence estimate the uncertainty in your values of f0 and φ. Decide how you could use your graph to determine the value of Planck's constant if you knew only the value of e and not h.

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Data Planck constant h = 6.60 x 10-34 J s electron charge e = 1.67 x 10-19 C speed of light c = 3.00 x 108 m s-1 1eV = 1.67 x 10-19 J

Practical advice This section revisits ideas about charge, energy and potential difference and introduces the idea of a stopping potential in order to measure the kinetic energy of photoelectrons. Students need to realise that, if a charged particle is accelerated by a pd, energy is transferred to it, whereas if it moves the other way, it loses kinetic energy, and that the two situations are the exact reverse of each other. Students should appreciate that the work function represents the minimum amount of work that an electron must do in order to get free and that the expression for kinetic energy represents the maximum possible kinetic energy of the photoelectron. This energy is only attainable if the energy transfer from photon to electron is 100% efficient and there is no energy dissipation, e.g. due to heating. This is quite a demanding activity, as it involves a relatively complicated and unfamiliar set-up. Make sure students appreciate the use of the potential divider, (as in Episode 118 The Potential Divider). In analysing their results, students need to plot a graph and determine the y-intercept. Students will have met graphs of the type y = mx + c before but still might not be very confident in using them, this might need some discussion. We recommend using a spreadsheet graphing package here. Students should also take account of experimental uncertainties; the most significant is likely to be in the frequency, as the filters available do not have a very definite cut-off wavelength. There might also be some uncertainty in deciding exactly the pd at which the photocell current drops to zero.

Einstein's ideas Albert Einstein explained the photoelectric effect in a paper published in 1905. It was the second of five ground-breaking papers he wrote that year. In the first paper, Einstein explained the mysterious Brownian motion of particles contained in pollen grains as due to the random impact of much smaller particles. This work led to the acceptance of the molecular or atomic nature of matter, which until then had been quite speculative. Einstein's third paper that year is now his most famous. Here Einstein introduced his Special Theory of Relativity which, in a later paper, led to probably the most famous equation in science: E=mc2, which describes the equivalence of mass and energy. But it was Einstein's second paper, that contained his work on the photoelectric effect, that at the time was the most revolutionary of the three, and it was for this work that Einstein was eventually awarded the Nobel Prize, in 1922. (The Nobel Committee works somewhat more slowly than the speed of light!) In this paper Einstein broke away from the idea that light (electromagnetic radiation) is continuous in nature and introduced us to the idea of the quantum (plural quanta) or photon as a `packet' of light. (The term quantum is used for any packet of energy, while a photon is a quantum associated with electromagnetic radiation.) The wave model of light had been fairly conclusively established a century earlier, mainly due to the work of Thomas Young, who demonstrated and explained interference patterns. But the wave model cannot explain the photoelectric effect;

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Einstein realised this and took the bold step of putting forward a completely different model in order to explain the following experimental results: •

for any given metal, with radiation below a certain threshold frequency no electrons are released even if the radiation is very intense;

•

provided the frequency is above the threshold, some electrons are released instantaneously, even if the radiation is very weak;

•

the more intense the radiation, the more electrons are released;

•

the kinetic energy of the individual photoelectrons depends only on the frequency of the radiation and not on its intensity.

Einstein was the first to use the equation E = hf to explain the photoelectric effect. It is known as the Planck equation, and h is called Planck's constant, because Max Planck had already proposed that when electromagnetic radiation was absorbed or emitted, energy was transferred in packets. That work earned Planck the 1918 Nobel Prize.

External references This activity is taken from Salters Horners Advanced Physics, section DIG, activity 30 and the Einstein notes above from Salters Horners Advanced Physics, section DIG additional sheet 11

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TAP 502- 4: The Millikan experiment to verify Einstein’s photo-electric relationship. The photoelectric effect was – well known by the end of the 19th century. Its explanation was one of Einstein’s first applications of his photon model for light. He devised an equation relating the energy of the photoelectrons to the frequency of the light and the work function of the metal used. In 1916, American physicist Robert Millikan completed some experiments that tested Einstein's equation. Though Millikan had been firmly against quantum theory, the results convinced him that Einstein was right.

In Millikan's photoelectric apparatus, monochromatic light was incident on each of the newly-cleaned metal plates in turn. The resulting photoelectrons were collected by electrode C and flowed, via the variable resistor and microammeter, back to the plate assembly. By adjusting the potentiometer, Millikan found the potential difference that was just large enough to stop the electrons moving round the circuit. For each metal, he used radiation of several different frequencies, noting the `stopping voltage' for each. This graph shows some of his results for sodium.

(a)

Use the graph to find the threshold frequency for sodium.

(b)

Calculate the work function for sodium using your answer to (a). h = 6.63 x 10-34 J s.

(c)

Use your answer to (b) to find the maximum kinetic energy of photoelectrons emitted from sodium when the frequency of the incident light is 10 x 1015 Hz. Give your answer in joules and in electronvolts. e = 1.6 x 10-19 C

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External reference This activity is taken from Salters Horners Advanced Physics, section DIG, additional sheet 13

Answers and worked solutions

External reference Answers and worked solutions are taken from Salters Horners Advanced Physics, section DIG, additional sheet 14

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Episode 503: Preparation for lasers topic It is likely that all pupils will have experienced laser technology in their homes or at school but may well not familiar with the range of uses of lasers, their operation and safety. Section TAP 504-2 gives details of the safety measures required. Lasers form rather more of a 'tool' and so demonstration experiments are relatively few. However some ideas and an explanation of the workings and other details of a laser are given.

ruby rod (Diagram: resourcefulphysics.org)

Episode 504: How lasers work

Main aims As regards post-16 examinations, the formal requirements laid down are modest, or non-existent, so that the level of treatment here is not detailed. Students will: 1. Outline the principles of operation of a laser. 2. State some uses of high and low energy lasers.

Prior knowledge Students should know about the wave nature of light, including interference. They should also know how light is emitted by electron transitions within atoms.

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Episode 504: How lasers work This episode considers uses of lasers, and the underlying theory of how they work.

Safety: Ensure that you are familiar with safety regulations and advice before embarking on any demonstrations (see TAP 504-2).

Summary Demonstration: Seeing a laser beam. (10 minutes) Discussion: Uses of lasers. (15 minutes) Discussion: Safety with lasers. (10 minutes) Discussion: How lasers work. (20 minutes) Worked examples: Power density. (10 minutes) Student calculations: (10 minutes)

Demonstration: Seeing a laser beam A laser beam can be made visible by blowing smoke or making dust in its path. Its path through a tank of water can be shown by adding a little milk. Show laser light passing through a smoke filled box or across the lab and compare this with a projector beam or a focussed beam of light from a tungsten filament light bulb. Show the principle of optical fibre communication by directing a laser beam down a flexible plastic tube containing water to which a little milk has been added. Show a comparison between the interference pattern produced by a tungsten filament lamp (with a 'monochromatic' filter) and that produced by a laser.

Discussion: Uses of lasers Talk about where lasers are used – ask for suggestions from the class. As far as possible this should be an illustrated discussion with a CD player, a laser pointer, a set of bar codes, a bar code reader and the school’s laser with a hologram available for demonstration. TAP 504-1: Uses of lasers Show the list of uses. Invite students to consider the uses shown in the list. Can they say why lasers are good for these? The reasons might be: •

A laser beam can be intense.

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•

A laser beam is almost monochromatic.

•

A laser beam diverges very little.

•

Laser light is coherent.

Discussion: Safety with lasers Lasers must be used with care. Use the text as the basis of a discussion of the precautions which must be taken. TAP 504-2: Lasers and safety

Discussion: How lasers work If students are familiar with energy level diagrams for atoms, and of the mechanisms of absorption and emission of photons, you can present the science behind laser action. Point out the difference between: (a) excitation – an input of energy raises an electron to a higher energy level (b) emission - the electron falls back to a lower energy level emitting radiation and (c) stimulated emission – the electron is stimulated to fall back to a lower energy level by the interaction of a photon of the same energy Define population inversion: Usually the lower energy levels contain more electrons than the higher ones (a). In order for lasing action to take place there must be a population inversion. This means that more electrons exist in higher energy levels than is normal (b).

For the lasing action to work the electrons must stay in the excited (metastable) state for a reasonable length of time. If they 'fell' to lower levels too soon there would not be time for the stimulating photon to cause stimulated emission to take place.

(a)

(b) (Diagram: resourcefulphysics.org)

‘Laser’ stands for Light Amplification by Stimulated Emission of Radiation. The diagrams in TAP 503 shows the ruby laser and the snowball effect of photons passing down a laser tube, and the diagram above shows the three level laser action. The electrons are first ‘pumped’ up to the higher energy level using photons. They then drop down and accumulate in a relatively stable energy level, where they are stimulated to all drop back together to the ground state by a photon whose energy is exactly the energy difference to the ground state.

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Discuss coherent and non-coherent light. Coherent light is light in which the photons are all in 'step' – in other words the change of phase within the beam occurs for all the photons at the same time. There are no abrupt phase changes within the beam. Light produced by lasers is both coherent and monochromatic (of one 'colour').

Incoherent sources emit light with frequent and random changes of phase between the photons. (Tungsten filament lamps and 'ordinary' fluorescent tubes emit incoherent light)

Worked examples: Power density The laser beam also shows very little divergence and so the power density (power per unit area) diminishes only slowly with distance. It can be very high. For example consider a light bulb capable of emitting a 100 W of actual light energy. At a distance of 2 m the power density is 100 / 4πr2 = 2 W m-2. The beam from a helium-neon gas laser diverges very little. The beam is about 2 mm in diameter 'close' to the laser spreading out to a diameter of about 1.6 km when shone from the Earth onto the Moon! At a distance of 2 m from a 1 mW laser the power density in the beam would be 0.001 / (π × 0.0012) = 320 W m-2! This is why you must never look directly at a laser beam or its specular reflection.

Student calculations Ask the class to calculate the power densities for a 100 W lamp and a 1 mW laser at the Moon. (Distance to Moon = 400 000 km; diameter of laser beam at Moon = 1.6 km)

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TAP 504-1: Uses of lasers •

repairing damaged retinas

•

bar code reader

•

communication via modulated laser light in optical fibres

•

neurosurgery - cutting and sealing nerves sterilization – key hole surgery (microsurgery)

•

laser video and audio discs (CD and DVD)

•

cutting metal and cloth

•

laser pointer

•

laser printer

•

holography - three-dimensional images

•

surveying - checking ground levels

•

production of very high temperatures in fusion reactors

•

laser light shows

•

making holes in the teats of babies' bottles

•

cutting microelectronic circuits

•

physiotherapy - using laser energy to raise the temperature of localised areas of tissue e.g. removing tonsure tumours

•

laser lances for unblocking heart valves; removal of tattoos or birth marks

•

laser guidance systems for weapons

•

laser defence systems ('Star Wars')

•

a modification of the Michelson-Morley experiment to check the existence of the ether

•

distance measurement

•

laser altimeters

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TAP 504-2: Lasers and safety The following is an extract from the CLEAPSS Science Publications Handbook Section 12.12:

12.12 Lasers A low-power, continuous-wave, helium-neon laser is useful for teaching wave optics because it produces a beam of light that is: •

highly monochromatic (very narrow spread of wavelengths) and coherent (the same phase) both across the beam and with time;

•

of high intensity;

•

of small divergence (typically 1 mm in diameter when it leaves the laser, 5 mm in diameter 4 m away).

Because of its intensity, care must be taken to prevent a beam from a laser falling on the eye directly or by reflection, as it could damage the retina. In fact, it is now known that eyes could be damaged by the beam from the low power Class II lasers used in schools only if a deliberate attempt were made to stare at the laser along the beam or to concentrate the beam from a IIIa laser with an instrument; the normal avoidance mechanism of the body would prevent damage in other cases. Nevertheless, it is good practice to take precautions to avoid a beam falling on the eye. Lasers should be positioned so that beams cannot fall on the eyes of those present, i.e., are directed away from spectators. Ancillary optical equipment must be arranged so that reflected beams cannot reach eyes. Ten years ago, the lasers affordable by schools would work for only a few years and then only if run periodically. Those currently available have longer lives and do not require this.

DES Administrative memorandum 7/70 This advises on the use of lasers in schools and FE colleges. It confines use at school level to teacher demonstration. Safety rules include the avoidance of direct viewing; screening pupils who should be at least 1 m away; keeping background illumination as high as possible (so that eye pupils are as small as possible); displaying warning notices; keeping lasers secure from unauthorised use or theft; use of goggles by the demonstrator. Because of the more recent realisation that the lasers used in schools and colleges will not harm an eye by a beam accidentally falling on it, these rules are not strictly necessary but following them is good training. However, goggles are expensive, impractical and do not reduce hazard as they make it harder for the demonstrator to see the beam, stray reflections etc.

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Episode 505: Preparation for wave-particle duality topic In the first quarter of the twentieth century physicists began to realise that particles did not always behave like particles – they could behave like waves. They called this wave-particle duality. This theory suggests that there is no basic distinction between a particle and a wave. The differences that we observe arise simply from the particular experiment that we are doing at the time. As with quantum theory, this is a section of the course that candidates will find completely new. They are unlikely to have already met the wave nature of particles or the wave nature of electrons bound within atoms.

Episode 506: Particles as waves Episode 507: Standing waves

Main aims Students will: 1. Understand that electron diffraction is evidence for wave-like behaviour. 2. Use the de Broglie equation. 3. Identify situations in which a wave model is appropriate, and in which a particle model is appropriate, for explaining phenomena involving light and electrons. 4. Use a standing wave model for electrons in an atom.

Prior knowledge Students should have an understanding of wave phenomena, including diffraction and interference. They should know how to calculate momentum. This work follows on from a study of the photoelectric effect.

Where this leads These episodes merely skim the surface of quantum physics. For students who wish to learn a little more, here is some suggested reading: The new quantum universe; Tony Hey and Patrick Walters; CUP Quantum physics: An introduction; J Manners; IoPP You can extend the idea of electrons-as-waves further, to the realm of the atom.

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Episode 506: Particles as waves This episode introduces an important phenomenon: wave - particle duality. In studying the photoelectric effect, students have learned that light, which we think of as waves, can sometimes behave as particles. Here they learn that electrons, which we think of as particles, can sometimes behave as waves.

Summary Demonstration: Diffraction of electrons. (30 minutes) Discussion: de Broglie equation. (15 minutes) Worked examples: Using the equation. (15 minutes) Student questions: Using the equation. (30 minutes) Discussion: Summing up. (10 minutes) Student questions: Practice calculations. (30 minutes)

Demonstration: Diffraction of electrons The diffraction of electrons was first shown by Davisson and Germer in the USA and G P Thomson in the UK, in 1927 and it can now be observed easily in schools with the correct apparatus. Show electron diffraction. It will help if students have previously seen an electron-beam tube in use (e.g. the fine beam tube, or e/m tube). Evacuated tube Diffraction rings

Thin graphite screen Electron gun

………

Electrons show particle properties Electrons show wave properties (Diagram: resourcefulphysics.org)

TAP 506-1: Diffraction of electrons

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Before giving an explanation, ask them to contemplate what they are seeing. It is not obvious that this is diffraction/interference, since students may not have seen diffraction through a polycrystalline material. (Note that the rigorous theory of crystal diffraction is not trivial - waves scattered off successive planes of atoms in the graphite give constructive interference if the path difference is a multiple of a wavelength, according to the Bragg equation. The scattered waves then appear to form a wave that appears to “reflect” off the planes of atoms, with the angle of incidence being equal to the angle of reflection. In the following we adopt a simplification to a 2D case - see TAP 506-1) Qualitatively it can help to show a laser beam diffracted by two ‘crossed’ diffraction gratings. Rotate the grating, and the pattern rotates. If you could rotate it fast enough, so that all orientations are present, you would see the array of spots trace out rings. TAP506-2: Diffraction of light From their knowledge of diffraction, what can they say about the wavelength of the electrons? (It must be comparable to the separation of the carbon atoms in the graphite.) How does wavelength change as the accelerating voltage is increased? (The rings get bigger; wavelength must be getting smaller as the electrons move faster.)

Discussion: de Broglie equation In 1923 Louis de Broglie proposed that a particle of momentum p would have a wavelength λ given by the equation: wavelength of particle λ = h/p where h is the Planck constant, or λ = h/mv for a particle of momentum mv. The formula allows us to calculate the wavelength associated with a moving particle.

Worked examples: Using the equation 1.

Find the wavelength of an electron of mass 9.00 × 10-31 kg moving at 3.00 × 107 m s-1. λ = h/p = [6.63 × 10-34] / [9.00 × 10-31 × 3.00 × 107] = 6.63 × 10-34 / 2.70×10-23 = 2.46×10-11 m = 0.025 nm

This is comparable to atomic spacing, and explains why electrons can be diffracted by graphite. 2.

Find the wavelength of a cricket ball of mass 0.15 kg moving at 30 m s-1. λ = h/p = [6.63 × 10-34] / [0.15 × 30] = 1.47×10-34 m = 1.5 10-34 J s (to 2 s.f.)

This is a very small number, and explains why a cricket ball is not diffracted as it passes near to the stumps.

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3.

It is also desirable to be able to calculate the wavelength associated with an electron when the accelerating voltage is known. There are 3 steps in the calculation.

Calculate the wavelength of an electron accelerated through a potential difference of 10 kV. Step 1: Kinetic energy Ek = eV = 1.6 × 10-19 × 10000 = 1.6 × 10-15 J Step 2: EK = ½ mv2 = ½m (mv) 2 = p2 / 2m, so momentum

p = √2mEk = √2 × 9.1 × 10-31 × 1.6 × 10-15 = 5.4 × 10-23 kg m s-1 Step 3: Wavelength λ = h / p = 6.63 × 10-34 / 5.4 × 10-23 = 1.2 × 10-11 m = 0.012 nm.

Student questions: Using the equations A useful set of questions can be found at Resourceful Physics on the web This can also be accessed from http://resourcefulphysics.org/download/361/waves_and_particles.doc and is available to subscribers.

Discussion: Summing up You may come across a number of ways of trying to resolve the wave-particle dilemma. For example, some authors talk of ‘wavicles’. This is not very helpful. Summarise by saying that particles and waves are phenomena that we observe in our macroscopic world. We cannot assume that they are appropriate at other scales. Sometimes light behaves as waves (diffraction, interference effects), sometimes as particles (absorption and emission by atoms, photoelectric effect). Sometimes electrons (and other matter) behave as particles (beta radiation etc), and sometimes as waves (electron diffraction). It’s a matter of learning which description gives the right answer in a given situation. The two situations are mutually exclusive. The wave model is use for ‘radiation’ (i.e. anything transporting energy and momentum, e.g. a beam of light, a beam of electrons) getting from emission to absorption. The particle (or quantum) model is used to describe the actual processes of emission or absorption.

Student question: Interpreting electron diffraction patterns TAP 506-3: Interpreting electron diffraction patterns TAP 506-4: Electron diffraction question

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TAP 506- 1: Diffraction of electrons Start by setting up the electron diffraction demonstration tube according to the manufacturer’s instructions. Check the connections before switching on as it is easy to burn out the graphite grid unless manufacturer’s instructions are correctly followed.

Wire carefully, no bare conductor above 40 V School EHT supplied are limited to a maximum output current of less than 5 mA so they are safe, but can still make the user jump. The lead connecting the supply to the anode should have a female connector so that no metal is exposed. You will need two supply voltages:

9

a low-voltage supply to provide the current to heat the cathode;

9

a high-voltage supply to provide the potential difference between the cathode and the anode, to accelerate the electrons.

These are both usually supplied by the EHT unit.

Qualitative experiments Check that you can identify the following: •

the electron gun, which includes the cathode and the anode;

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•

the graphite target;

•

the fluorescent screen, which will glow when the electrons strike it.

When the cathode has heated up (you will see it glowing), increase the accelerating voltage V. An invisible beam of electrons emerges from the electron gun and passes through the graphite film. To show that there is an electron beam present, bring up a bar magnet close to the tube and observe the pattern shifting. Magnetic fields deflect moving electrons. So this qualitative demo shows both the wave aspects (the diffraction pattern) and the particle aspects (deflection by a magnetic field) in the same equipment. Look for a diffraction pattern on the screen at the end of the tube. What shape is the pattern? Where can you see constructive interference? And destructive interference? Now predict: If you increase the accelerating voltage, how will the energy and speed of the electrons change? How will this affect the diffraction pattern? Test your ideas.

Quantitative experiments Choose a feature of the diffraction pattern that is easy to measure. It might be the diameter of the first bright ring, or of the first dark ring. We will call this chosen quantity d. Investigate how d depends on V. Make several measurements. Determine the mathematical relationship between d and V. If the relationship is of the form d proportional to Vn, you can find the value of n by plotting a log-log graph. The gradient of the graph is equal to the value of n.

What other information would you need if you were to use this experiment to determine the atomic spacing in graphite?

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Practical advice It is difficult to present a convincing discussion of wave–particle duality in a few lines. We have adopted the approach that it is an observed phenomenon, witness the diffraction of electrons. The de Broglie equation wavelength = h/p simply allows us to translate between the wave and particle pictures. Increasing the accelerating voltage increase the electrons’ energy and momentum; greater momentum means shorter wavelength, which in turn means that the electrons are diffracted through a smaller angle. Hence the diameters of the diffraction rings get smaller. Students could make measurements of the rings and relate these to the accelerating voltage. A log-l graph will reveal the relationship between them. It is reasonable to say that ring diameter is proportional to wavelength; a log-log graph of diameter against voltage will thus be a straight line with gradient (-1/2).

Apparatus 9

electron diffraction tube

9

power supply (6.3 V) for cathode

9

e.h.t. supply (0 – 5000 V dc) with voltmeter

9

connecting leads

Great care must be taken to set the tube up correctly. The graphite can be damaged by incorrect connections. Notice that the positive e.h.t supply terminal is used without the protective resistor in some set ups. Take care. As the discussion in this section is almost entirely in terms of electron diffraction, students might get the idea that only electrons exhibit wave–particle duality whereas in fact it applies to all particles. You might like to mention that neutron diffraction is also used as a tool for probing material structure. Large bio-molecules have now been diffracted. The chosen method is to introduce wave particle duality via the experiment however a little mathematics may be of use. This mathematics is given so an experimental or theoretical route can be chosen or the mathematics later used to back up the experiment.

E=hf and since fλ = c then E =hc/λ Taking E = mc2 then mc2 = hc/λ and so mc = h/λ momentum = mass x velocity = p =mc And this gives the de Broglie relation p = h/λ Do notice that the switch from mc to mv has not been justified other than it works. (Note, this derivation really only refers to photons and not to electrons). For the diffraction tube where V is the anode cathode pd and v the electron speed then

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½ mv2 = eV so v = (2eV/m) 1/2 giving mv = m (2eV/m) 1/2 = (2meV) 1/2

p = h/λ = (2meV) 1/2 Tube screen Graphite Target

d

θ

D

Simplifying the Bragg formula into a two dimensional treatment for a transmission grating, for diffraction θ = λ/b (approximately) where b is the separation between the planes of carbon atoms. D, if required, can be obtained from manufacturers details. θ = d/D from geometry so d/D = λ/b or λ proportional to d as diffracting gap size b and target screen distance D are constants. So using the momentum and accelerating pd equation above then

p = h/λ = (2meV) 1/2 or 1/λ is proportional to V1/2 or

λ is proportional to V-1/2

External references This activity is taken from Salters Horners Advanced Physics, section PRO, activity 18, with extra material added in the last two sections, adapted from Revised Nuffield Advanced Science Physics teachers guide, book 2, section L2.

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TAP 506-2: Diffraction of light Apparatus 9

torch bulb in holder with power supply

9

diffraction gratings, 2 each of various spacing, e.g. 100 lines mm-1, 300 lines mm-1

9

red, blue and green filters (one of each colour)

9

2 glass slides (e.g. microscope slides)

9

lycopodium dust

Safety Lycopodium is dried pollen. Some people are allergic to pollen so, before use, it is wise to check for hay-fever sufferer. (See Safeguards in the School Lab 4.4). Lycopodium powder in air will explode so extinguish all flames before using it.

Technique Observe what happens when light is diffracted by: •

a grating of parallel lines;

•

a regular grid of crossed lines;

•

a random array of fine dust particles.

Explore the effects of changing the separation of the lines and the wavelength of the light. Set up a small (6 V) lamp. You are going to observe how light from the lamp is diffracted in different situations.

Write down or sketch what you observe. Hold a diffraction grating and a colour filter close to your eye. Look through it at the lamp. Rotate the grating. (A diffraction grating consists of many parallel, equally spaced lines, as much as several thousands in each centimetre.) Repeat with a finer or a coarser grating (using the same colour filter as before). Now use two diffraction gratings. Place one on top of the other, so that their lines are at 90 degrees to each other. You have made a grid of lines through which the light is diffracted. Look through the grid at the lamp. Try holding the gratings so that their lines cross at a different angle – say, 60° or 45°. Try combining gratings with different line spacing. Finally, sprinkle fine dust (e.g. lycopodium powder) on to a glass slide. Blow off any excess dust. Look through the slide at the lamp. (Now you are looking at light diffracted by a random array of tiny particles.)

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Repeat the above, using a different colour filter. Sum up your observations by saying how the diffraction patterns depend on: •

the separation of the diffracting objects

•

the geometry of the diffracting objects

•

the wavelength of the light.

Practical advice This activity is intended to remind students of diffraction effects they may have previously observed. Points to bring out are: •

the size of the pattern increases with increasing wavelength (red light is diffracted more than blue)

•

the size of the pattern depends inversely on the spacing of the diffracting objects (the finer the grating, the more widely spaced the diffraction images)

•

the geometry of the diffraction pattern depends on the geometry of the diffracting objects

•

a random arrangement of small obstacles gives rise to a diffraction pattern that is a set of concentric rings.

External reference This activity is taken from Salters Horners Advanced Physics, section PRO, activity 17

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TAP 506-3: Interpreting electron diffraction patterns For this question to make sense, students need to be aware that an obstacle (e.g. a nucleus) is equivalent to a ‘hole’ of the same size. So far they have only met diffraction by ‘gaps’. Use the electron diffraction patterns shown in Figures (a) and (b) to determine the diameters of carbon and oxygen nuclei.

The diffraction patterns represented by the two graphs in Figures (a) and (b) were obtained using electrons accelerated to 420 MV.

By following the steps below, you will get a flavour of how such diffraction data are interpreted. You will also see some of the equations that must be used when electrons are moving at relativistic speeds (that is, close to the speed of light).

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Calculate the kinetic energy Ek of an electron accelerated through 420 MV. To calculate the momentum of the electron, you cannot use p=mv and EK = ½ mv2, because these equations don’t apply to relativistic particles. Instead, it turns out that the electron’s momentum p is related to its kinetic energy by Ek ~ pc where c is the speed of light. Calculate the momentum of the electron. The de Broglie relationship p = hλ applies to relativistic and non-relativistic electrons because Relativity Theory is used in its derivation. Calculate the electron’s de Broglie wavelength. The angle of the first minimum θmin on the graph is a clearly identifiable point, and this is used to calculate the diameter of the nuclei. From the graphs in Figure (a) and (b), determine the angle of the first minimum for carbon nuclei, and for oxygen nuclei. The smaller the nucleus, the more it diffracts the electrons. From the angles you found, which are smaller, carbon nuclei or oxygen nuclei? Is this what you would expect from their atomic numbers? The diameter d of the nucleus is related to the electron wavelength and the diffraction angle θmin by

Use this relationship to obtain estimates of the diameters of carbon and oxygen nuclei. Explain why your answers are only estimates.

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External references This activity is taken from Salters Horners Advanced Physics, section PHM, activity 19 which was an adaptation of Revised Nuffield Advanced Physics section L question 37(L).

Answers and worked solutions Ek = 420 x 106 x 1.60 x 10-19 = 6.72 x 10-11 J Ek ~ pc so p = 6.72 x 10-11 / 3 x 108 = 2.24 x 10-19 N s

λ = h/p = 6.63 X 10 -34 / 2.24 x 10-19. = 2.96 x 10-15 m (a)

carbon nuclei, 50°

(b)

oxygen nuclei. 42°

d = 1.22λ θ min where d is the diameter of the nucleus Remember, the angle should be in radians or use sin θ. (a)

carbon d = (1.22 x 2.96 x 10-15) / 0.766 = 4.7 x 10-15 m

(b)

oxygen d = (1.22 x 2.96 x 10-15) / 0.669 = 5.3 x 10-15 m

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TAP 506- 4: Electron diffraction questions 1

In an electron diffraction experiment using graphite the larger ring formed by rows of carbon atoms 1.23 x 10-10 m apart was formed at an angle of 0.167 radian.

(a)

What is the wavelength? [θ in radians = λ / b where b is the diffracting object size]

(b)

Write an expression for the kinetic energy of an electron (½ mv2) in terms of its charge, and accelerating voltage V

(c)

Obtain an expression for momentum, p, in terms of e, V and m.

The accelerating voltage was 5000 V (d)

Work out h in mv = h/ λ

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Answers (a)

0.205 x 10-10 m

(b)

KE = 0.5mv2 = Ve

(c)

p = mv = (2meV) 1/2

(d)

h = 7.78 x 10-34 J s (the accepted value is 6.6 x 10-34 J s)

External reference This activity is taken from Revised Nuffield Advanced Physics Unit L question 32.

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Episode 507: Electron standing waves You could extend the idea of electrons-as-waves further, to the realm of the atom.

Summary Demonstration: Melde’s experiment. (20 minutes) Discussion: Electron waves in atoms. (10 minutes) Demonstration: Standing waves on a loop. (10 minutes) Student Question Electron standing waves (10 mins)

Demonstration:

L

Melde’s experiment TAP 507-1: Standing waves – for electrons?

(Diagram: resourcefulphysics.org)

In this section we are going to introduce the idea of standing waves within an atom. It is therefore useful first to demonstrate standing waves on a stretched elastic cord. This is known as Melde’s experiment.

M

(A very simple alternative to the vibration generator is an electric toothbrush.) Show that there are only certain frequencies at which standing wave loops occur.

Discussion: Electron waves in atoms The waves on the string are 'trapped' between the two fixed points at the ends of the string and cannot escape. If the electron has wave properties and it is also confined within an atom we could imagine a sort of standing wave pattern for these waves rather like the standing waves on a stretched string. The electrons are 'trapped' within the atom rather like the waves being 'trapped' on a stretched string. The boundaries of these electron waves would be the potential well formed 'within' the atom. This idea was introduced because the simple Rutherford model of the atom had one serious disadvantage concerning the stability of the orbits. Bohr showed that in such a model the electrons would spiral into the nucleus in about 10-10 s, due to electrostatic attraction. He therefore proposed that the electrons could only exist in certain states, equivalent to the loops on the vibrating string.

nucleus

electron

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If your students have met the idea of angular momentum, you could tell them that Bohr proposed that the angular momentum of the electrons in an atom is quantised, in line with Planck's quantum theory of radiation. He stated that the allowed values of the angular momentum of an electron would be integral multiples of h/2π. This implied a series of discrete orbits for the electron. We can imagine the electron as existing as a wave that fits round a given orbit an integral number of times.

Demonstration: Standing waves on a loop The wire loop is a two dimensional analogy of electron waves in an atom. As it vibrates at the correct frequency, an integral number of waves fit round the orbit. These waves represent the electron waves in an atom.

Student question: Electron standing waves de Broglie waves can be imagined as forming standing waves which fit into an atom TAP 507-2: Electron standing waves

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TAP 507-1: Standing waves - for electrons? Electrons have a frequency too Electrons can be modelled as having a frequency. In another context you have seen how superposition of waves in the laboratory produces standing waves. Here it is useful to put these two together, describing electrons with standing waves, with their wavelength described by de Broglie’s relationship: λ = h / mv .

Here you look at some consequences of this step. You will need:

9

vibration generator

9

signal generator

9

rubber cord

9

G-clamp

9

wooden blocks

Vibrating a rubber cord

Fre quency Adjust 5 3

7 8

2

1

10Hz

100H z

1kHz

10kHz

9

100 kHz

Frequency ran ge 100 0 100

10

1

1000

10

100

Frequency

10 Hz

Wave

Outp uts

A

power

Make sure that you can get clear patterns with this apparatus. Note that only whole numbers of half wavelengths fit onto the cord. Electrons trapped in an atom are also constrained. Describing them as waves, where the amplitude tells you the chance of find them in any one place, constrains you to draw the waves in a similar way.

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Making links Putting relationships together with these results allows you to predict that electrons will have certain allowed energy levels only. simplest pattern

next simplest pattern

L

L

λ 2

from diagrams

from de Broglie

from kinetic energy

λ= h mv

Ek =

1 mv2 2

λ 2

λ = 2L

λ=L

h mv = λ

Ek =

mv =

h 2L

Ek =

h2 8mL2

mv =

h L

p2 2m

p = mv

Ek =

h2 2mL2

2 ? λ and 4 ? Ek

See if you can continue the series for the next two standing wave patterns that will fit onto the cord. You have 1. Reminded yourself about standing waves. 2. Seen some of the consequences of using standing waves to model electrons in atoms.

External reference This activity is taken from Advancing Physics chapter 17, 140E

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TAP 507-2: Electron standing waves If an electron is confined in a definite space, the de Broglie waves can be imagined as forming standing waves that fit into that space. h = 6.6 x 10-34 J s, charge on electron = 1.6 x 10-19 C, mass of electron is 9.1 x 10-31 kg

de Broglie standing wave r = 1.0 × 10–10 m

r = 0.1 × 10–10 m

diameter 2.0 × 10–10 m diameter 0.2 × 10–10 m

Suppose the standing wave fits with one half-wavelength across the diameter of the atom. 1.

Write down the wavelength of the standing wave if the atom is imagined to have radius

r = 1.0 × 10–10 m.

2.

Write down the wavelength of the standing wave if the atom is imagined ten times smaller, with radius r = 0.1 × 10–10 m.

3.

Other standing waves could fit inside the same diameter. Would their wavelengths be longer or shorter than the waves shown here?

Electron momentum The momentum of an electron with de Broglie waves of wavelength λ is m v = h / λ. If the wavelength is the largest possible, the momentum must be the smallest possible. 4.

Calculate the smallest possible momentum of the electron, if the atom is imagined to have radius r = 1.0 × 10–10 m.

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5.

Calculate, or write down directly from the answer to question 4, the smallest possible momentum of the electron if the atom is imagined to have radius r = 0.1 × 10–10 m.

6.

Write a few lines explaining why an electron confined in a smaller space has a larger minimum momentum.

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Answers and worked solutions 1.

The waves have half a wavelength fitting into the diameter, so λ = 4.0 × 10–10 m.

2.

The radius is 10 times smaller so the wavelength is 10 times shorter: λ = 0.4 × 10–10 m.

3.

More half-wavelength loops have to be fitted into the same length, so the wavelengths will all be smaller.

4.

mv =

h

λ

6.6 × 10 −34 J s = 4.0 × 10 −10 m = 1.65 × 10 − 24 kg m s −1 . 5.

As for question 4 with the wavelength 10 times smaller, so the momentum is 10 times larger: mv = 16.5 × 10–24 kg m s–1.

6.

The smaller the space the shorter the wavelength. But the shorter the wavelength the greater the momentum, since mv = h / λ.

External reference This activity is taken from Advancing Physics chapter 17, and uses part of question 160s.

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Episode 508 See Episode 513

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Episode 509: Radioactive background and detectors This episode introduces the ubiquitous nature of radioactivity, and considers its detection. It draws on students’ previous knowledge, and emphasises the importance of technical terminology.

Summary Demonstration: Detecting background radiation (10 minutes) Discussion: Sources of background radiation (15 minutes) Demonstration: Radioactive dust (10 minutes) Discussion + survey: Sources of radiation – should we worry? (15 minutes) Demonstration + discussion: Am-241 source, plus use of correct vocabulary. (10 minutes) Demonstration: Spark detector (10 minutes)

Demonstration: Detecting background radiation Use a Geiger counter to reveal the background laboratory. What is the ‘signal’ like? (It is discrete, Does it vary from place to place in the room? (No; it this is an opportunity to discuss the need to make longer-term measurements.) Does it vary from time roughly constant.) Count for 30 s to get a total count N; repeat several random variation. Calculate the average value of N.

radiation in the erratic / random.) may appear to; multiple or to time? (No, it’s to scaler/ratemeter

(Note: a good rule of thumb is that the standard so roughly two-thirds of values of N will be within ±

times to show deviation is √Nave, √Nave.)

Which is better: 10 counts of 30 s, averaged to get the activity, or one count of 300 s? (They amount to the same thing. The statistics of this is probably beyond most A-level students.) Calculate the background count rate from the data (typical value is 0.5 counts per second or 30 counts per minute, but this varies a lot geographically.)

Discussion Sources of background radiation Look at charts showing sources of background radiation. Consider how these might vary geographically, with time, occupation etc.

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Pie chart for background radiations Note that pie charts in text books etc showing relative contributions are often calibrated in units of equivalent dose of 0.4% fallout

0.4% air travel

0.2% occupational 12% internal such as food etc.

<0.1% nuclear waste

10% cosmic rays

14% gamma rays from the ground

12% medical, X rays etc.

50% radioactive gases in the home

(Diagram: resourcefulphysics.org)

radiation called sieverts (symbol Sv); sievert is a unit which takes account of the effects of different types of radiation on the human body. 1 Sv = 1 J kg-1 = 1 m2 s-2 TAP 509-1: Doses TAP 509-2: Whole body dose equivalents

Demonstration: Radioactive dust Airborne radioactive substances are attracted to traditional computer and TV screens that use a high voltage. Similarly a ‘charged’ balloon will also accumulate radioactive dust and have an activity larger than the average background. The fresh dust in vacuum cleaner bags has a noticeably higher activity too. Set up a Geiger counter to measure the activity of vacuum cleaner dust; don’t forget to measure background rate also. TAP 509-3: Radiation in dust

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Discussion + survey: Sources of radiation – should we worry? So “radiation is all around us”. Indeed, most substances, and things, are radioactive. Students are radioactive! Typically 7000 Bq. So “it’s dangerous to sleep with somebody”! However, most of the resulting radiation is absorbed within the ‘owners’ body. Introduce activity of a sample as a quantity, measured in becquerels (Bq). Mass of typical student = 70 kg, so specific activity = 7000/70 = 100 Bq kg-1. The Radiation Protection Division of the Health Protection Agency (formerly the National Radiological Protection Board, NRPB) defines a radioactive substance as having specific activity ≥ 400 Bq kg-1. Should we worry about this? Ask students to complete this survey, and then re-visit at the end of the topic. For each statement, indicate whether they think it is true or false, or they don’t know. S1

Radioactive substances make everything near to them radioactive.

S2

Once something has become radioactive, there is nothing you can do about it.

S3

Some radioactive substances are more dangerous than others.

S4

Radioactive means giving off radio-waves.

S5 Saying that a radioactive substance has a half life of three days means any produced in six days.

now will all be gone

Demonstration + discussion: Am-241 source, plus use of correct vocabulary Radioactive sources Follow the local rules for using radioactive sources, in particular do not handle radioactive sources without a tool or place them in close proximity to your body. Place an Am-241 source close to the GM tube and measure the count rate, which will be impressive, compared to background. (Some end window GM tubes will not detect alpha emission from AM-241 but only weak gamma).

(Diagram: resourcefulphysics.org)

Geiger Muller tube

Radioactive source

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From here on, start to use appropriate technical vocabulary, drawing on students’ earlier experience. For example: Substances are radioactive, they emit ‘radiation’ when they decay. Why are some substances radioactive? (They contain unstable nuclei inside their respective atoms.) The unstable nucleus is called the mother and when it (she?) decays a daughter nucleus is produced (it’s not quite like human procreation!). Eventually the activity of a radioactive substance must cease. However, point out that the decay of the americium doesn’t seem to be getting any less. There are a very large number of nuclei in there! Am-241 is used in smoke alarms, so it won’t ‘run out’. They are supplied with 33.3 kBq sources. The half-life is 458 years. What particular property does nuclear radiation have – what does it do to the matter through which it passes? (It is ionizing radiation. It creates ions when it interacts with atoms.) What is an ion? (A neutral atom which has lost or gained (at least one) electron.) Simplified diagram of a smoke alarm

Alpha source

Current due to ionisation flows from + to -

Current flow stopped by smoke (Diagram: resourcefulphysics.org)

To knock an electron from an atom, the ionizing radiation transfers energy to the atom – this is how nuclear radiation is detected. It is not difficult to detect the presence of a single ion – electron pair, so it’s easy to detect the decay of single radioactive nucleus. Chemists can detect microgrammes or nanogrammes of chemical substances, physicists can detect individual atomic events.

Demonstration: Spark detector Show a spark detector responding to the proximity of an alpha source. (NB At first, do not refer to the source as an alpha source.) Move the source away a few centimetres; you do not need much distance in air to absorb this radiation. Ask students to recall which type of radiation is easily absorbed by air. (Alpha.)

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0V

+ 5000 V

gauze (Diagram: resourcefulphysics.org)

thin wire

sparking here

TAP 509-4: Rays make ions Comment that a GM tube is not dissimilar to a spark counter, but rolled up to have cylindrical symmetry. At this point, you could discuss the sophisticated design of a GM tube and associated counter. The end window is usually made of mica and has a plastic cover, with holes, to protect the mica. anode

low pressure neon or argon gas

+450 V

0V thin end window radioactive particle

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TAP 509-1: Doses

Radiation dose measurements and units

dose in gray

activity in becquerel = disintegrations per second

radiation source

energy in different types of radiation

dose in gray = energy deposited per kg

dose equivalent in sievert = dose in gray × quality factor

Radiation quality factors radiation

factor

dose equivalent of 1 gray

alpha

20

20 Sv

beta

1

1 Sv

gamma

1

1 Sv

x-rays

1

1 Sv

10

10 Sv

neutrons

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Practical advice This gives a simple view of the connection between gray and sievert as units of radiation dose measurements. It may be best to change the size of the diagram and use it as an OHT.

External reference This activity is taken from Advancing Physics chapter 18, display material 20O

67

food and drink 370 µSv

medical procedures 500 µSv

air travel 8 µSv nuclear power 3 µSv nuclear weapons fallout 10 µSv

Artificial sources 530 µSv 22% of total work 9 µSv

radon gas from ground and buildings 800 µSv

gamma radiation from ground and buildings 380 µSv

cosmic and solar radiation 310 µSv

Natural sources 1860 µSv 78% of total

Typical whole body dose equivalent per year from various sources (Europe)

INDEX

TAP 509-2: Whole body dose equivalents

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Practical advice This shows the breakdown of whole body doses, from different sources. It may be best to change the size of the diagram and use it as an OHT.

External reference This activity is taken from Advancing Physics chapter 18, display material 30O.

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TAP 509-3: Radiation in dust Ionisation in the home Radon gas is present in the atmosphere of our homes; its daughter products can contribute to the background count as constituents of household dust. In this activity you are asked to collect dust and assess its level of radioactivity. You will need

9

two sheets of kitchen paper towel

9

radiation sensor, alpha sensitive (NB not a standard GM tube)

9

stop watch

9

rubber band

What to do 1. Collect some household dust on a sheet of kitchen paper towel. One way to do this is to cover your finger with the paper and then to wipe your finger over the surface of a dusty television or computer monitor screen. Do this for the whole screen, more than one screen if possible – you need a substantial layer of dust. Put the paper carefully to one side. Do not dislodge the dirt. (Incidentally, one reason for suggesting the dust from a television screen is because it is charged. It therefore attracts the ionised daughter products from, amongst other things, radon decays.) 2. Take a new clean piece of identical kitchen paper. Fix it around the thin window end of the sensor using a rubber band. Take a background measurement for a substantial time, several thousand seconds or even overnight if possible. Automating the capture of the data will prove useful. 3. Without changing any other conditions replace the clean paper with the dusty paper. Wrap it around the GM tube with the dust fingerprint side next to the window so that alpha particles are not absorbed by the paper. Take care not to get the dust onto the sensor. 4.

Measure the radioactive count from the dust for the same length of time as before.

5. Look critically at the results. Are they significantly different? Remember that variation in a radiation experiment is equal to the square root of the measurement itself. Do the two results differ by substantially more than this?

You have Measured some of the ambient radiation from a household.

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Practical advice This experiment needs care but it does yield interesting results. It requires long timings and not a little luck. Household dust has radioactive products from a number of decays. In trials a count for 600 s produced the following results: background count (no paper): about 230 counts; background count (with paper): about 240 counts; dust count: about 295 counts. There was a substantial amount of dust on the paper from three computer monitors.

Alternative approaches The use of TASTRAK plastic is a possible substitute here. The plastic known as CR-39 was developed in 1933 and in 1978 it was found to be an excellent detector of charged particles, which could be revealed by etching the plastic. TASTRAK is a version of CR-39 developed specifically to detect alpha particle tracks. It can be used to demonstrate the detection of radon. The suppliers, TASL, offer kits of TASTRAK plastic for class use which may then be returned to the Track Analysis Group for processing free of charge. This arrangement is STRICTLY for UK schools, colleges and universities only. Track Analysis Systems operate a free etching service. It is not recommended that you attempt the etching process yourself. When returning the exposed slides, remember to declare whether you require them to be in microscope or slide projector form. You can find more information about TASTRAK from http://www.tasl.co.uk/schools.htm or Track Analysis Systems Ltd, H H Wills Physics Laboratory, Tyndall Avenue, Bristol. BS8 1TL, U.K. Tel: +44 (0)117 926 0353, Fax: +44 (0)117 925 172

Be safe Students should wash their hands before eating after collecting the dust.

External references This activity is taken from Advancing Physics chapter 18, activity 20H.

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TAP 509- 4: Rays make ions The particles and rays that come from radioactive (unstable) nuclei are known generically as ionising radiation. Ionising radiations do just what their name says – they ionise atoms by removing one or more of the electrons outside the nucleus. It is this property that helps us to detect the radiations in many of the instruments that may already be familiar to you. In this activity you will learn to use a simple detector of ionising radiation – the spark detector.

You will need 9

spark counter

9

EHT power supply, 0 – 5 kV, dc

9

leads

9

almost pure alpha source (The only ‘pure’ alpha source available in schools is Pu-239. Am-241 emits gammas too. However, since the spark counter only responds to the alphas, any alpha-emitting source will do here)

9

forceps or tweezers for handling the radioactive source (or source holder)

Radioactive sources Follow the local rules for using radioactive sources, in particular do not handle radioactive sources without a tool or place them in close proximity to your body.

Ionisation and its effects First, some ideas about ionisation and its effects. Ionisation is the name given to the process in which an atom or molecule gains or loses one or more electrons. In the detection of radioactivity we are normally concerned with the loss of an electron. The mechanism is that radiation is emitted from an unstable nucleus; the particle or ray carries energy away with it. It is this loss of energy that allows the radioactive atom to become more stable. As the particle moves away through the air or a more dense substance it comes close to atoms, sometimes sufficiently close to interact. The particle can simply bounce off the atom – an elastic collision – or it can interact inelastically and transfer some its energy to the atom. This gain of energy by the atom leads to the removal of one of its electrons. You can think of the particle or ray as having knocked the electron free of its atom. Energy taken from the incoming particle or ray is needed for this to occur. Can you think of examples in which ionisation is used to detect radiation?

Getting going 1.

Now, before setting it up, look closely at the spark detector. You should be able to see a grid or array of very fine wires running parallel and close to the surface of a metal plate or above a wire. There may also be an arrangement for holding a radioactive source a fixed distance from the wire array. Your power supply will maintain a high potential difference between the plate and the wires. There will be a large electric field in the space between the wires and the plate.

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Wire carefully, EHT supply I use Although school EHT supplies are current-limited and safe, using the extra limiting resister (50 MΩ) reduces the shock current to a trivial level. 2. Now connect up the detector to the power supply. Ask for help if you are not certain how to do this. To be safe the wires (the part you are most likely to touch) should be at earth (0 V) potential. Adjust the EHT until sparks just pass, then reduce it slightly. 3. Insert a source of alpha particles into the holder and ensure that the source is pointing array and about 10 mm from it. Take the usual handling precautions with the source. –

0V

e.h.t.

towards the wire

+

+ 5 kV

5 kV 0V

4.

Does an increase in the potential difference make a significant change in the sparking rate?

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The alpha particles ionise the air molecules as they travel from the source. The sparking is a result of what happens when the ionisation occurs in the strong electric field produced by the spark detector. Look at the diagram. The field is directed between the wire and the gauze, or between the wires and the plate. So, once the electron and its ion (the original atom) are separated, the electron moves towards the wire and the ion moves towards the plate. The ion has a large mass and so gains speed slowly. The electron, however, has a low mass so its acceleration is large. (You might want to estimate how large it is: to do this use the distance from wire to plate and the voltage setting to make a reasonable estimate of the electric field strength, hence the force and the acceleration.) If an electron can gain enough energy from the field it will be able to collide with another gas atom and cause a further ionisation. So one electron has now produced two: the original one plus the one from this second ionisation. Both of these can go on to accelerate and ionise again. The number of electrons in the space builds up rapidly and eventually the air contains enough electrons in one region to allow a spark to jump from wire to plate. This is known as a cascade process.

You have seen that 1. Radioactive particles are often detected using ionisation. Examples include the spark Geiger–Müller tube, the cloud and bubble chambers, and photographic detection methods.

detector, the

2. Such methods show us where the particles have caused ionisation. They are not direct particles themselves.

detections of the

3. Some methods rely on a cascade process in which ions are multiplied by successive strong electric field.

ionisations in a

Also try TAP 519: Particle Detectors

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Practical advice Students may need close supervision in using this apparatus. They will also need guidance in the correct earth setting to avoid electric shock. The instructions provided in the activity may need to be altered to suit the style of spark detector you have.

Radioactive sources Follow the local rules for using radioactive sources, in particular do not handle radioactive sources without a tool or place them in close proximity to your body.

Wire carefully, EHT supply in use

Social and human context Every time your teeth are x-rayed, ionisation processes are used to expose the film. The ubiquitous Geiger–Müller tube involves a controlled cascade between the central wire and the outer cylinder inside the tube. We rely heavily on the ionising effects of radiation for its detection.

External reference This activity is taken from Advancing Physics chapter 18 activity 50E

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Episode 510: Properties of radiations The focus of this episode is the properties of ionizing radiations. It is a good idea to introduce these through a consideration of safety.

Summary Discussion: Ionising radiation and health. (10 minutes) Demonstration: Deflection of beta radiation. (10 minutes) Student activity: Completing a summary table. (10 minutes) Student experiment: Inverse square law for gamma radiation. (30 minutes) Discussion: Safety revisited. (5 minutes)

Discussion: Ionising radiation and health Why are radioactive substances hazardous? It is the ionising property of the radiation that makes it dangerous to living things. Creating ions can stimulate unwanted chemical reactions. If the radiation has enough energy it can split molecules. Disrupting the function of cells may give rise to cancer. Absorption of radiation exposes us to the risk of developing cancer. Thus it is prudent to avoid all unnecessary exposure to ionising radiation. All deliberate exposure must have a benefit that outweighs the risk. Radioactive contamination is when you get a radioactive substance on, or inside, your body (by swallowing it or breathing it in or via a flesh wound). The contaminating material then irradiates you. How can you handle sources safely in the lab? Point out that you will be safe if you follow your local rules which will incorporate the following: •

always handle sources with tongs

•

point the sources away from your body (and not at any anybody else)

•

fix the source in a holder which is not adjacent to where your body will be when you take measurements

•

replace sources in lead-lined containers as soon as possible

•

wash hands when finished

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Radioactive sources Follow the local rules for using radioactive sources, in particular do not handle radioactive sources without a tool or place them in close proximity to your body.

Demonstration: Deflection of beta radiation Show the deflection of β by a magnetic field. (Make sure you have a small compass to determine which are the N and S poles of the magnet.) Is the deflection consistent with the LH rule? (Yes; need to recall that electron flow is the opposite of conventional current.) Why is this demo is no good with the α source? (α particles are absorbed too quickly by the air.)

alpha

magnetic field

gamma

beta

NB The diagram above is common in textbooks, but is ONLY illustrative. For the curvature shown of beta particles, the curvature of the alpha tracks would be immeasurably small.

Student activity: Completing a summary table Display the table, with only the headings and first column completed. Ask for contributions, or set as a task; compile results. TAP 510-1: α, β and γ radiation

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INDEX

name

symbol

nature

elec charge

"stopped by"

ionising "power"

what is it?

alpha

α

particle

+2

mm air; paper

very good

He nucleus

beta

β

particle

-1

mm Al

medium

very fast electron

gamma

γ

wave

0

cm Pb *

relatively poor

electromagnetic radiation

Can you see any patterns in the table? (Most ionising - the largest electrical charge - is the least penetrating.) Can you explain this? (The most ionising lose energy the quickest.) How can the electrical charge determined? Deflection in a magnetic field.) * NB Gamma radiation is never completely absorbed (unlike alpha and beta) it just gets weaker and weaker until it cannot be distinguished form the background.

Student experiment: Inverse square law for gamma radiation Note: since you are unlikely to have sufficient gamma sources for several groups to work simultaneously, this experiment can be part of a circus with others in the next episode. Alternatively, it could be a demonstration. Gamma radiation obeys an inverse square law in air since absorption is negligible. (Radiation spreads out over an increasing sphere. Area of a sphere = 4 π r2, so as r gets larger, intensity will decrease as 1/r2. The effect of absorption by the air will be relatively small. TAP 5102: Range of gamma radiation (Some students could do an analogue experiment with light, with an LDR or solar cell as a detector.) When detecting γ radiation with a Geiger tube you may like to aim the source into the side of the tube rather than the window at the end. The metal wall gives rise to greater ‘secondary electron emission’ than the window and this increase the detection efficiency. Radioactive source

Correct readings for background.

How can we get a straight line graph? We to scaler/ratemeter expect I ∝ r-2, so a plot of I versus r–2 should be direct proportion (i.e. a straight line through the origin). It is much easier to see if a graph is a straight line, rather than a particular curve. Lift the graph and look along the line – it’s easy to spot a trend away from linear. However, two points are worth noting: (i) Sealed γ sources do not radiate in all directions, so do not expect perfect 1/r2 behaviour, and (ii) you do not know exactly

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where in the Geiger tube the detection is taking place, so plotting I-1/2 against r gives an intercept, the systematic error in the measurement.

Discussion: Safety revisited Return briefly to the subject of safe working. Background radiation is, say, 30 counts per minute. How far from a gamma source do you have to be for the radiation level to be twice this? Would this be a safe working distance? (Probably.) How much has your lifetime dose of radiation been increased by an experiment like the above? (Perhaps one hour at double the background radiation level – a tiny increase. It will be safe enough to carry out a few more experiments.)

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TAP 510-1: α, β and γ radiation Using your knowledge from pre-16 level work, or using information from textbooks, draw up a table to summarise the properties of α, β and γ radiation. The table below provides a framework for your summary. Write items from the following list into appropriate cells of the table. Note that some of the items are not needed and some may be used more than once.

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Practical Advice Some text books may be needed.

External reference This activity is taken from Salters Horners Advanced Physics, section DIG, activity 25

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TAP 510-2: Range of gamma radiation Penetrating electromagnetic radiation Gamma radiation is emitted by a nucleus in order to drop from an excited state to a more stable one. Gamma rays are high-frequency members of the electromagnetic spectrum. In this activity you will assess the penetrating power of gamma rays and also the way in which their intensity changes with distance.

You will need 9

gamma source and holder

9

set of absorbers: various thicknesses of lead

9

a Geiger–Müller tube and associated power supply

9

suitable means for mounting the GM tube a fixed measurable distance from the source

9

forceps or tweezers for handling the radioactive source

9

stop clock

9

metre rule

9

micrometer screw gauge or vernier calliper for making thickness measurements

9

set square

Radioactive sources Follow the local rules for using radioactive sources, in particular do not handle radioactive sources without a tool or place them in close proximity to your body.

Making measurements 1.

Set up the GM tube and counter to take a background reading. This background value will need to be subtracted from each subsequent reading you take when using the gamma source.

2.

Fix the gamma source 10–15 cm from the front window of the GM tube (do not remove any plastic cap that is protecting it). The separation of the source and counter and the orientation of the source must be fixed – not too hard to see why so take care. (Because we are investigating absorption, we want a fixed geometry so that the results are not confused by the inverse square law effect investigated later.)

3.

Take a reading of the number of gamma photons over a sensible time period and convert this to a rate of arrival per second.

4.

Insert one of your lead sheets between the tube and source close to the GM tube, finding the new count rate for each absorber thickness. Be sensible – a short preliminary

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experiment using a very thin absorber and a very thick one in order to get a feeling for the range of count rates that you will encounter will save you time later. You can also plot as you go (count rate against thickness of lead), looking for patterns and anomalies.

Analysis 5.

Try to find the thickness of lead for which half the incident gamma radiation is absorbed. Is the pattern exponential? Can you check?

Comparisons with beta particles (To be done if your class has carried out the activity dealing with the range of beta particles. TAP episode 511-2) 6.

Do the gamma rays penetrate further into the lead than beta particles? Which are the better ionisers?

7.

Plot a graph of corrected count rate against thickness. Draw a smooth curve through the points.

8.

If you have carried out the 'more sophisticated analysis' for beta particles you might like to repeat the logarithmic plot here too.

The inverse square law – a fresh start Theory predicts that electromagnetic radiation should vary with distance according to an inverse square rule. With a little care, you can test this. 1.

Remove the lead sheets – they are not needed again – and move the gamma source closer to the tube. This starting distance must not be so close that your GM tube is overwhelmed by counts and misses some out. Some trial and error may be needed.

2.

Measure the distance from the thin window of the GM tube to the front of the source. A possible way to do this is to use a set square to transfer the measurement to a metre rule taped to the bench.

3.

Obtain the average count rate for this distance.

4.

Repeat the experiment over a wide range of distances, taking care with the alignment of the sources.

5.

Does the inverse square law describe your results?

A more careful analysis You measured a distance from the window of the GM tube to the front of the source. This leads to two problems: the source itself is on a foil and lies a few millimetres behind the wire mesh of its holder and the average detection point is somewhere inside the GM tube. Theory predicts that the intensity of the gamma rays varies with the inverse square of the distance:

y∝

1 x2

which can be rewritten as

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INDEX

y =

k x2

However, the two problems mean that the true value of x is unknown and that it should be written as x + c where c is a correction that converts your (inaccurate) measurement of distance x into the correct value. So the equation becomes y =

1

(x + c )2

and a plot of y against x2 will not be a straight line. The trick is to take square roots of both sides: 1

y

2

=

1

(x + c )

leading to 1 y

1 2

= x+c

So if the gamma rays obey an inverse square law, a plot of 1 / y1/2 against x ought to be a straight line. The intercept on the x-axis will give you an estimate of c, the error in the distance determination. Plot your data in this way and decide if they verify the inverse square law.

You have seen that 1.

Gamma radiation is very penetrating. It will go through metres of lead and concrete. Gammas are poor ionisers. Half thicknesses can be measured, to characterise absorbers.

2.

Gamma rays, like all electromagnetic radiation, obey the inverse square law. Double your distance from the source and you reduce the intensity by four times.

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Practical advice Students will need to take care with the measurements, particularly with alignments. You may choose to make up apparatus to allow this to be easily controlled. The more careful analysis is only for mathematically competent students. It can be omitted without loss.

Be safe Follow local rules for using radioactive sources.

Social and human context Nuclear reactors and other nuclear machines are surrounded by huge amounts of concrete or lead to prevent the radiation emitted during the reactions causing a hazard to us. (When would you choose concrete and when lead? When there is room for lots of concrete, it is used because it is much cheaper than lead.)

External reference This activity is taken from Advancing Physics chapter 18, 110E

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Episode 511: Absorption experiments This gives students the opportunity to work with radioactive sources. Radioactive source

Absorbing material

Absorbing material holder

Summary Student experiments: Absorption of radiation + report back. (40 minutes) Demonstration: Absorption of radiation by living matter Student experiment: (optional)

Student experiments Groups could work in parallel and report back to a plenary session. Remind them to correct for the background count (taken at least twice – at the start and end of the main experiments and the two results averaged). Range of alpha radiation TAP 511-1: Use a spark counter, Range of beta radiation TAP 511-2: The range of beta particles in aluminium and lead Range of gamma radiation An optical analogue for the absorption of γs by lead is the absorption of light by successive microscope slides.

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TAP 511-3: Absorption in a liquid Absorption of γs is an example of exponential decrease – check the data for a constant ‘half thickness’, thus suggesting the type of physics involved. (Each mm of absorber is reducing the intensity by the same fraction.) TAP 511-4: Absorbing radiations

Demonstration: Absorption of radiation by living matter To simulate the absorption of radiation by living matter use slices of different vegetables as absorbers, or a slice of bacon to represent human flesh. TAP 511-5: Absorption in biological materials

Student experiments: Optional The first requires a sealed radium-226 source. Because Ra-226 is the parent to a chain of radioactive daughters, granddaughters and so on, you get a mixture of αs, βs and γs emitted. Challenge students to use absorbers to establish that all three radiation types are being emitted. (The maximum energies are: α = 7.7 MeV, β = 3.3 MeV, γ = 2.4 MeV). The second is an extension of the βabsorption experiment. You could speculate that some β particles might be ‘back-scattered’ (like Rutherford’s α particle scattering that first demonstrated the existence of the nucleus). A quick try shows that some β particles are indeed back-scattered.

Radioactive sources Follow the local rules for using radioactive sources, in particular do not handle radioactive sources without a tool or place them in close proximity to your body. Deliberately placing a radioactive source in contact with the skin would increase your dose of ionising radiation unnecessarily and increase the risks to your health. This is a criminal offence.

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TAP 511- 1: Use a spark counter The alpha particle is one type of emission that is possible from the nuclei of some atoms. This activity allows you to investigate how far these alphas can travel in air and other materials.

You will need 9

spark counter and leads

9

EHT power supply, 0–5 kV, dc

9

metre rule

9

set of absorbers including very thin tissue paper

9

plane mirror

9

alpha source and a suitable holder (since a spark counter does not respond to betas or gammas, the source does not have to be a pure alpha source.)

9

plastic tweezers for handling the absorbers

Starting out If you have not used a spark counter before, then make sure you know how it works.

Wire carefully, EHT supply in use Although school EHT supplies are current-limited ad safe, using the extra limiting resister (e.g. 50 MΩ) reduces the shock current to a trivial level.

Radioactive sources Follow the local rules for using radioactive sources, in particular do not handle radioactive sources without a tool or place them in close proximity to your body.

Support the alpha source in a holder about 10 cm from the active region of the spark counter, so that no sparking occurs. You need to be able to vary and measure this distance. If you are using a ruler, remember that you can use a plane mirror to align the pupil of your eye, the plate and the wires in the same plane. To set up the spark counter at its correct working voltage, find the setting which gives a spark with no source and then reduce the voltage slightly. Remember that it uses high voltages that could give you an electric shock.

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Retort stand with clamp

Radioactive source in holder

absorber when needed

0V

+ 5000 V

_

+ EHT

Gradually move the source towards the detector. Take care here, do not touch the source, detector wires or plate. At what distance do the sparks start? Does the sparking begin suddenly or does it start gradually? Try to explain what you see in terms of the initial energies of the alpha particles as they leave the nuclei. If you can, change the alpha source to one in which the alpha particles are emitted from a different nucleus. Is the distance at which sparking starts the same? Again, try to explain this. Leave the source at a distance such that sparking is occurring freely. Insert thin materials between the detector and the source (use tweezers to avoid getting close to the source). Draw up a table of how effective the absorber materials are at stopping the alpha particles. You might like to see if you can think of some property of the absorber which determines its effectiveness, to check for patterns.

You have seen that 1.

The alpha particles emitted from one isotope all leave with the same energy. They tend to lose this energy by collision (ionisation) with the air molecules at about the same rate and so the sparking stops suddenly as the source is moved away from the counter.

2.

Alpha particles are good ionisers. So, conversely, they are poor at penetrating substances. Even the thinnest of materials will absorb them

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Practical advice This short experiment needs care by the students or teachers carrying it out. You may prefer to demonstrate the result, at the same time emphasising the meanings of ionisation and the cascade process that is going on in the electric field produced by the spark counter. There are useful estimation possibilities here for brighter students: the strength of the electric field, the relative acceleration of the electron and the oxygen / nitrogen ion, and so on.

Be safe The student text contains safety warnings, but it would be as well to remind the students that there are both high potentials and radioactive materials here. Some models of spark counter have a lead designed for connection to a pulse counter and the terminal plug could float at a high voltage, although it is designed not to. It is as well to tape the plug up when not in use. Since the use with a counter enables quantitative studies, it would be unwise to remove it. Ensure the most accessible electrode of the spark counter is earthed.

Wire carefully, EHT supply in use Although school EHT supplies are current-limited ad safe, using the extra limiting resister (e.g. 50 MΩ) reduces the shock current to a trivial level.

Radioactive sources Follow the local rules for using radioactive sources, in particular do not handle radioactive sources without a tool or place them in close proximity to your body.

External reference This activity is taken from Advancing Physics chapter 18, 70E

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TAP 511- 2: The range of beta particles in aluminium and lead The background Beta particles are emitted by the unstable nuclei of some radioactive atoms. They consist of electrons given out when a neutron in the nucleus converts to a proton plus an electron. The electron is too energetic to remain inside the nucleus and is ejected. The beta particles are particles that can ionise materials through which they pass and they will continue to move through these materials until they have completely used up all the energy they had when they left the nucleus. In this experiment you will look at the thickness of material needed to absorb the electron, in other words to take away all its energy.

You will need 9

pure beta source and suitable holder

9

radiation sensor, beta sensitive, e.g. end window GM tube with ratemeter or scaler (counter)

9

set of absorbers

9

suitable method for holding the absorbers

9

forceps or tweezers for manipulating the absorbers and the source

9

suitable means for measuring the thickness of the absorbers, e.g. micrometer screw gauge, vernier calliper, etc

Radioactive sources Follow the local rules for using radioactive sources, in particular do not handle radioactive sources without a tool or place them in close proximity to your body.

Measurements… This activity requires the use of a radiation sensor, counter and power supply. Ensure that you know how to use them before you begin. 1.

Set up the sensor and counter to take a background reading. This background value will need to be subtracted from each subsequent reading you take when using the beta source.

2.

Fix the beta source probably 10–20 cm from the front window of the sensor (do not remove any plastic cap that is protecting it). The separation of source and counter and the orientation of the source must be fixed – not too hard to see why so take care.

3.

Take a reading of the number of beta particles over a sensible time period and convert this to a rate of arrival per second. If this number is greater than 1000, increase the separation

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of source and sensor and take another reading. (A ratemeter will give counts per second directly.) 4.

Beginning with the aluminium foils and sheets, insert absorbers of varying thickness into the space between the source and the sensor, finding the new count rate for each absorber thickness. Be sensible – a short preliminary experiment using a very thin absorber and a very thick one in order to get a feeling for the range of count rates that you will encounter will save you time later. You can also plot as you go (count rate against thickness of aluminium), looking for patterns and anomalies.

… and analysis 5.

How thick does the aluminium have to be in order to reduce the count rate by one-half? Use your graph to benefit from the smoothing present in all the data you have gathered.

6.

If possible, use the graph to estimate what extra thickness reduces the count rate to one-quarter of what it was at the start. Is it the same as that required to reduce it by half? (It should NOT be!)

7.

You should look for a pattern.

8.

Repeat the experiment with lead absorbers. Does the lead obey the same rule as the aluminium?

Comparisons You can repeat the measurements and analysis for different materials, if provided.

You have seen that 1.

Different materials are able to absorb beta particles by different amounts. The ability to absorb depends on factors such as the extent to which the material is ionised by the betas and the density of the material (the number of atoms a beta meets per unit length of its travel).

Note on absorption of beta versus gamma radiation This activity is taken from Advancing Physics; which has treated beta radiation as being absorbed in the same way as gamma radiation; however, this is not strictly correct. Betas gradually lose energy as they pass through an absorber so that they become easier to stop as they approach the end of the path.

Practical advice Students require some skills before they can tackle this experiment: 1.

Knowledge of safe handling techniques for the radioactive sources.

2.

The ability to set up a radiation sensor and counter or ratemeter correctly. If the power supply has a variable output, students will need to be told the correct p.d. for the tube.

3.

A knowledge of the importance of the background count.

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You may need to tell your students an appropriate time over which they should count the arrival of betas in order to give sensible statistics and values. A trial will calibrate your beta sources with your sensors. Slower students might stop after point 7.

Be safe Students need to be confident and safe in their handling of the source in this activity. You might like to consider having special holders made up for the sensor and source, which allow the alignment and separation to be easily maintained, or use a radioactivity bench which you may already have.

Technician's note The range of thicknesses of supplied foils in any one material will need to be sufficient to ensure that students can actually reduce the count rate by a half. Ideally, they should be able to see a reduction well beyond this ratio in order to answer the question in point 9 of the activity. Note that some sensors have a plastic cap that is loaded with atoms of high atomic number to improve the gamma sensitivity. This cap should be removed for beta counting.

Radioactive sources Follow the local rules for using radioactive sources, in particular do not handle radioactive sources without a tool or place them in close proximity to your body.

Social and human context This property of absorption is important to us. A number of devices use it, a good example being the film badges worn by personnel involved in work with radioactive materials. The varying thicknesses of absorber in these badges enable the assessment of absorbed energies once the film has been developed.

External reference This activity is taken from Advancing Physics chapter 18, 90E

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TAP 511- 3: Absorption in a liquid An analogue for the absorption of gamma radiation.

Absorption and energy transfer Absorption occurs when light travels through a medium – whether air or a liquid or a transparent solid. Transfer of energy from light to the medium occurs with absorption. The more absorption, the more energy is transferred. This experiment, which can be tackled at a number of levels, will help you to understand the way in which the energy is removed from the light beam. This is a particularly important piece of physics in the manufacture and the technology of glass fibre optics.

Collect these 9

absorption tank

9

power supply 5 V dc

9

LED and a suitable phototransistor

9

digital multimeter

9

copper (II) sulfate solution, about 5% by volume

What to do 1. Connect up the circuit as shown.

10k

mA

2. Be especially careful with the polarity of the power supplies. You could easily damage the phototransistor by connecting the supply the wrong way round. Ask your teacher to check if you are unsure. 3. Light emitted from the LED passes through the copper sulfate and is detected by the phototransistor circuit. The output shown on the meter is a measure of the intensity (brightness) of the light emerging from the liquid after absorption has taken place.

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water phototransistor

LED

copper sulphate

4. Slide the tank between the light source and detector at 90° to the direction in which the light is shone. By sliding the tank in this way you can vary the thickness of copper (II) sulfate solution through which the light travels and therefore assess the loss of intensity. 5. Plot a graph of meter reading (a measure of intensity, remember) against solution thickness. 6. You can take this analysis as far as you wish: Is there an obvious rule which describes the variation shown in your graph? Have you met such a rule in science before? (Hint: how does the radioactive count-rate change when a radioactive isotope decays?) Why does this rule apply? 7. Find out how thick the liquid has to be to absorb half the light. 8. Could you simulate the absorption experiment here on a computer or a calculator? (Hint: imagine that there are 100 units of light energy coming out of the light and that every 0.1 mm 1% of the energy remaining is removed. How much is left after the first 0.1 mm, answer: 99 units, after 0.2 mm 99% of 99 units [0.9801 units], after 0.3 mm 99% of 99% of 99 units [0.9703 units]. Use a spreadsheet to carry out this tedious calculation for 100 mm, 200 mm, 300 mm, and so on.)

The relationship between amount of light absorbed and thickness of absorbing medium Light is absorbed when it passes through a medium – however transparent it appears to be. A constant fraction of the incident light is removed per unit distance travelled. Mathematically, this is known as an exponential relationship. Radioactive decay is another phenomenon involving an exponential relationship – leading to the idea of half-life of an isotope, the time taken for half the number of radioactive atoms in a sample to decay. In a similar way we can think of a 'half-distance', in light absorption, over which half the light is absorbed.

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INDEX

Practical advice The light-absorbing properties of materials are not required by the specification, but this may be of interest to some students. The equipment needed here is easy to build. It requires a phototransistor and a light-emitting diode. The circuit is shown in the diagram(s). One power supply will suffice for both halves of the circuit. The LED and transistor need to face each other across a gap of perhaps 10-12 cm. In trials the devices were rigidly mounted as shown in the diagram, but they could be free standing. This, however, needs careful alignment. In the gap sits a rectangular tank of dimensions 10 cm x 30 cm. These dimensions are not critical. A vertical barrier separates the two halves of the tank. One half is filled with water, the other half with the 5% (w/v) copper (II) sulfate solution. The arrangement is known as a Bjerrum wedge and is used for colorimetric measurements. Students slide the wedge between the light source and detector. They measure output as a function of sulfate solution thickness. Some adjustment of solution strength may be needed.

Alternative approaches Ghost tape provides an appropriate reduction intensity for some light sources and sensors (TLS 250 and a 36 W lamp). Here discrete values of tape enable exponential thinking to get to grips with the problem. If one piece lets through X%, then two pieces will let through X% squared, etc

Social and human context Fibre optics has only become possible because of the huge advances in the manufacture of low attenuation glasses.

Safety Copper sulfate solution can be harmful. At this concentration, no washing label is required. If you get the solution on your hands or elsewhere wash it off immediately

External reference This activity is taken from Advancing Physics chapter 4, 240E

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INDEX

TAP 511- 4: Absorbing radiations

Shielding from x-rays and gamma rays intensity I penetrating thickness x

fraction absorbed ∝ extra thickness dI/I = –µ dx intensity decreases exponentially I = exp (–µx) I0 ln 2 half-thickness x1/2 = µ

exponential decrease of intensity with thickness

Ways photons are absorbed or lose energy 1. ionise atom or put atom in higher energy level

thickness x

2. scattering from electrons half-thickness of absorber

3. electron–positron pair production (important at high energies) Half-thicknesses

Absorption by successive thicknesses

photon energy 200 keV x-ray

intensity reduced to one half by each block

1 MeV γ ray

concrete 25 mm

50 mm

Here shielding is related to the absorption of a constant fraction of the radiation in each thickness of absorber traversed.

97

lead 0.5 mm

10 mm

INDEX

External reference This activity is taken from Advancing Physics chapter 18, 10 O (O stands for ‘O.H.T’.)

98

INDEX

TAP 511- 5: Absorption in biological materials Closer to home Relating radiation to risk means thinking about how ionising radiations interact with you and other biological materials. Simple measurements could show you how much the radiation penetrated, and how much energy it deposited in each cubic metre of you. Although these two are related, you can do a fairly simple sum, based on the energy per particle for a known source, and the activity, to suggest that the sources available to you in the laboratory are very un-likely to produce temperature rises that you can easily measure. So the energy delivered is probably beyond what can easily be achieved in the school laboratory. Here you measure how the quantity of radiation varies as you alter the thickness of biological materials though which it has to pass. For ethical reasons, use fruit and vegetables.

You will need 9

radiation sensor, beta sensitive

9

samples of fresh fruits and vegetables (marrow, apple etc)

9

an apple corer

9

sharp knife (a plastic one might be advisable)

9

pure beta source and holder

9

forceps or tweezers for manipulating the absorbers

Radioactive sources Follow the local rules for using radioactive sources, in particular do not handle radioactive sources without a tool or place them in close proximity to your body.

Alignment and arrangement Take a sample of the fruit, using the apple corer to obtain a cylindrical specimen, then a knife to trim it to length.

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INDEX

marrow

β source

radiation detector

Take care to maintain the alignment of the source and detector throughout the experiment. You will probably want to take measurements at both extremes (no absorbing material, maximum thickness) to make sure that you are likely to get a reasonable variation in count rate – this should help you decide appropriate lengths of time over which to take each count. Now you will probably need to take perhaps 10 readings to establish a reasonable pattern: for 8 cm of absorber, every 0.5 cm might be suitable for your source.

Making sense See if you can spot a pattern in your results, and then use a sensible analysis to show that the pattern holds. You should draw on other work done on absorption, from chapter 4 onwards, to help suggest likely patterns. Alternatively you may be able to think of sound physical reasons for one pattern. Do not forget about the background count!

Reflections Try again with another fruit, or compare your results with those of another group working with a different fruit. Look for similarities and differences. Can you draw any conclusions from these? Can you test any hunches that you might have? You have worked with beta radiation, as it has suitable properties for use in the school laboratory, allowing measurements with the equipment that you have access to. In what ways would you expect the results to change if you used alpha or gamma radiation? (Your expectations should be based on sound physical reasons!)

You have shown How the count rate varies as beta radiation passes through differing biological tissues.

100

INDEX

101

INDEX

Practical advice To keep the alignment constant you might like to make a jig for the source and detector, with a channel to hold the cylinder of fruit between the two. This channel might usefully be plastic, and removable, to allow for cleaning. The design should also not encourage fruit juices to dribble into either the source or detector. The distance between the source and detector should remain constant, in the range 5–10 cm. The denser the specimen, and the more feeble the beta source, the shorter the cylinder you should use. The fruit or vegetable you choose to use will depend on the season.

Social and human context Many bags of saline solution, as perhaps a suitable description of many biological systems, might be employed as models for humans – our usual concern when discussing risk. You might use this as an introduction to the appropriateness of some choices, reflecting on the modelling process and its assumptions.

External reference This activity is taken from Advancing Physics chapter 18, 100E

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Episode 512: Nuclear equations Now that your students are familiar with different types of radiation, you can look at the processes by which they are emitted.

Summary Discussion: Nuclide notation and N-Z plot. (10 minutes) Student Questions: Practice with notation. (10 minutes) Worked Examples: Equations for alpha, beta and gamma decay. (20 minutes) Student Questions: Practice with nuclear equations. (30 minutes)

Discussion: Nuclide notation Revise nuclide notation:

A Z

X

Discuss how A = mass or nucleon number, Z = charge or atomic number and N = neutron number are related (A = Z + N). Discuss isotopes (common examples: H, D and T, U-235 and U-238, C-14 and C-12). Set the task of finding out the name for nuclides having the same A but different Z (isobars), and the same N but different Z (isotones). Show an N-Z plot (Segrè plot).

Neutron number

n

p + β- + ν

A

ββ+ B p

n + β+ + ν

Proton number 103

INDEX

Student questions: Practice with notation Set some simple questions involving nuclide notation. TAP 512-1: Nuclide notation

A, Z-1

A, Z

A, Z+1

Nucleon number (A)

β-

β+

α A-4, Z-2

Proton number (Z)

Neutron number (N)

β+ N+1, Z-1 N, Z

β-

N-2, Z-2 N-1, Z+1

α

104

Proton number (Z)

INDEX

Grid showing change in A and Z with different emissions Worked examples: Equations for alpha, beta and gamma decay Nuclear decay processes can be represented by nuclear equations. The word equation implies that the two sides of the equation must ‘balance’ in some way. TAP 512-2: Decay processes You could give examples of equations for the sources used in school and college labs. α sources are americium-241, 241 95

Am

4 Am → 237 93 Np + 2 He.

β- sources are strontium-90, 90 38

241 95

Sr → 90 39 Np +

0 −1

90 38

Sr

e.

The underlying process is: n –> p + e- + ν Here, ν is an antineutrino. Your specification may require you to explain why this is needed to balance the equation. You can translate n –> p + e- into the AZ notation: 1 0

n →11 H +

0 −1

e.

γ sources are cobalt-60 60

decay of the 27 Co . The

60 27

Co . The γ radiation comes from the radioactive daughter 60 28 Ni of the β

60 28

Ni is formed in an ‘excited state’ and so almost immediately loses the

energy by emitting a γ ray. They are only emitted after an α or β decay, and all such γ rays have a well-defined energy. (So a cobalt-60 source which is a pure gamma emitter must be designed so that betas are not emitted. How? – (by encasing in metal which is thick enough to absorb the betas but which still allows gammas to escape.)

Student questions: Practice with nuclear equations TAP 512-3: Practice with nuclear equations The more unusual decay processes (positron emission, neutron emission, electron capture) could be included, and students challenged to write them as nuclear equations.

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INDEX

TAP 512-1: Nuclide notation A periodic table may be needed A

1 Write down the nuclear notation ( Z X ) for: (a)

an alpha particle

(b)

a proton

(c)

a hydrogen nucleus

(d)

a neutron

(e)

a beta particle

(f)

a positron.

A

2 Write down the nuclear notation ( Z X ) for (a)

carbon 13

(b)

nitrogen 14

(c)

neon 22

(d)

tin 118

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INDEX

(e)

iron 54

107

INDEX

Answers 1 4 2

He

(a)

an alpha particle

(b)

a proton 1 H or 1 p

(c)

a hydrogen nucleus 1 H

(d)

a neutron

(e)

a beta particle

(f)

a positron.

1

1

1

1 0

n

0 +1

0 −1

e

e

2 13 6

(a)

carbon-13

(b)

nitrogen-14

(c)

neon-22

(d)

tin-118

118 50

(e)

iron-54

54 26

22 10

C

14 7

N

Ne

Sn

Fe

108

INDEX

TAP 512-2: Decay processes Radioactive decay processes α decay Z N

α

Z–2 N–2

2 fewer protons 2 fewer neutrons

proton number Z β– decay

β–

Z N Z+1 N–1

ν 1 more proton 1 less neutron

proton number Z β+ decay

β+ Z–1 N+1 ν

Z N

1 less proton 1 more neutron

proton number Z γ decay

γ

Z N

same protons and neutrons

proton number Z

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INDEX

External reference This activity is taken from Advancing Physics chapter 18, 90O

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INDEX

TAP 512-3: Practice with nuclear equations 1

The isotope 235U decays into another element, emitting an alpha particle. What is the element? This element decays, and the next, and so on until a stable element is reached. The complete list of particles emitted in this chain is: 235 92 U → [α, β, α, β, α, α, α, α, β, α, β] → X. What is the stable element X? (You could write down each element in the series, but there is a quicker way.)

2

The following fission reaction can take place in a nuclear reactor: 235 1 137 95 1 92 U + 0 n→ 55 Cs + [ ] Rb +[ ]0 n.

Complete the equation, showing how many neutrons are produced in the reaction. What is the significance of the number of neutrons produced?

Why are the products of the reaction, caesium-137 and rubidium-95, likely to be radioactive? What type of decay are these isotopes likely to show?

3

Boron absorbs neutrons with results as follows: 10 5

B + 01 n → 73 Li+ 24 α .

Why is boron suitable for use in a control rod?

4

When the isotope

27 13 Al

is irradiated with alpha particles, the products from each aluminium 30 nucleus are a neutron, and a nuclide that emits positrons to give the stable isotope 14 Si . Write nuclear equations for these two processes.

111

INDEX

5

Complete the following nuclear equations. In each case describe the decay process: 131 53 I

→

Xe +

67 0 31Ga + −1e

11 6C

Zn

B + 01e

→

99m 43Tc

6

→

0 −1e

→

Tc + γ .

The Manhattan Project, the development of the atomic bomb, led to the discovery of the transuranic elements (elements beyond uranium in the periodic table). Plutonium, element 94, is formed by the bombardment of uranium-238 with neutrons. The nuclear equations are: 238 1 239 92 U+ 0 n→ 92 U 239 239 0 92 U→ 93 Np+ −1 e 239 239 0 93 Np→ 94 Pu+ −1 e

Complete the following nuclear equation for the formation of americium: 239 94 Pu

+ 2 01 n →

Am +

0 −1 e.

Curium is produced if plutonium-239 is bombarded with alpha particles. If the curium 242 isotope is 96 Cm , complete the equation 239 94 Pu

4 2α

+

→

If curium is made the target for alpha particle bombardment californium is produced. Complete the nuclear equation to find the atomic number of californium: 242 96 Cm

+

4 245 Cf 2 α→

+ 01 n.

By firing heavier particles such as carbon or boron ions at the target materials heavier elements can be synthesised. Complete the nuclear equation (Lw is lawrencium) 252

Cf +

9 5B

→ Lw + 4 01 n.

One of the transuranic elements is commonly found in the home. Which is this and where is it used?

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Practical advice These questions are intended for homework or practice in class.

Answers and worked solutions 1.

4

The complete decay chain involves the loss of seven alpha particles ( 2 α ) and four beta 0

e

particles ( −1 ). This represents a loss of 7 × 4, i.e. 28, in mass number and (7 × 2 – 4), i.e. 10, in atomic number. X is therefore an isotope with mass number (235 – 28), i.e. 207, and 208 atomic number (92 – 10), i.e. 82. This is lead, 82 Pb .

2.

235 92 U

1 137 0 n→ 55 Cs

+

+

95 37 Rb

+ 3 01 n.

The reaction produces more neutrons than it absorbs, this will cause a ‘chain reaction’. To see why the products of the reaction are likely to be radioactive you need to consult the plot of neutron number against atomic number for the known stable nuclei. For elements with atomic numbers up to about 30 the number of neutrons in the nucleus is the same as the number of protons if the nucleus is stable. For higher atomic numbers the ratio of 137 neutrons to protons gradually increases to 1.5. Look at the position of both 55 Cs and 95 37 Rb

on the plot and you will see that they both have a considerable excess of neutrons. They are therefore likely to be radioactive. To become more stable the nuclides need to decrease the neutron to proton ratio. The emission of a beta particle does this, increasing the number of protons by one and decreasing the number of neutrons by one. These isotopes are therefore likely to decay by emitting beta particles. 3.

When boron captures a neutron it is transformed into a stable isotope. If the control rods are pushed into the reactor more neutrons are absorbed, causing the chain reaction to slow down. If they are pulled out the chain reaction will proceed more vigorously. 4. 27 13 Al 30 15 P

+

=

4 2α

30 15 P

=

30 14 Si

+

+

1 0n

0 1e

5. 131 131 53 I→ 54 Xe

11 11 6 C→ 5

+

B +

0 −1e

beta decay

0 1e

positron emission

67 0 67 electron capture 31 Ga + −1 e → 30 Zn 99 m 99 nuclear rearrangem ent 43 Tc → 43 Tc

6. 239 1 241 94 Pu + 2 0 n→ 95 Am 239 4 242 94 Pu + 2 α → 96 Cm 242 4 245 96 Cm + 2 α → 98 Cf

252 98 Cf

+

9 257 5 B→ 103 Lw

+ + +

0 −1e 1 0n 1 0n

+ 4 01 n.

Americium is commonly found in homes where it is used in smoke detectors.

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External reference This activity is taken from Advancing Physics chapter 18, 210S

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Episode 513: Preparation for exponential decay topic In this section, we are not concerned with the products of radioactive decay. Rather, we are looking at the random, spontaneous nature of radioactive decay and its consequences. The key problems students have are (i) thinking erroneously that after two half lives, everything is over, and (ii) understanding the connection between what individual nuclei do, and how a large sample of radioactive nuclei behave.

Depending upon the specification you are following, some of the material is part of the first year or the second year of post-16 level course. You will need to adjust the level of mathematics accordingly.

Episode 514: Patterns of decay Episode 515: The radioactive decay formula Episode 516: Exponential and logarithmic equations

Main aims Students will: 1. Define the term half-life. 2. Make calculations involving numbers of half-lives. 3. Relate half-life to decay probability λ. 4. Measure the half-life of a fast-decaying nuclide. 5. Use exponential and logarithmic equations for radioactive decay.

Prior knowledge Basic, descriptive radioactivity should already have been covered. Students will have previously been introduced to the term half-life, but are unlikely to be confident in using the quantity in calculations.

Where this leads The mathematics of exponential decay parallels that of capacitor discharge, damped SHM etc, so these would be suitable topics to tackle next. If students have already met these topics, you could usefully spend time drawing the parallels between them.

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Episode 514: Patterns of decay This episode assumes that students already have an idea of half-life, and links it to the empirical decay curve.

Summary Discussion and demonstration: Measuring half life. (40 minutes) Worked examples: Involving whole numbers of half lives. (30 minutes) Student example and discussion: Plotting a graph. (30 minutes) Student questions: Calculations. (30 minutes)

Discussion + demonstration: Measuring half life Each radioactive nuclide has its own unique half-life. Values range from millions of years (e.g. uranium-238 at 4.47×109 years) to minute fractions of a second (e.g. beryllium-8 at 7×10-17 s). Sealed school/college sources have half-lives chosen to ensure that they will remain radioactive over a period of years (though Co-60 will become significantly less active year-by-year):

Radiation

Source

Nuclide

Notes

T1/2

α

americium-241

241 95

Also emits gammas

458 y

Sr

The energetic betas come from the daughter Ys-90.

28.1 y

Co

Also emits betas, but these may be absorbed internally

5.26 y

β

γ

strontium-90

cobalt-60

Am

90 38

60 27

You need something else if you are to measure half-life in the course of a lesson. A short half-life source commonly available in schools and colleges is protactinium-234, T1/2 = 72s. Another is radon-220, T1/2 = 55s.

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Demonstrate the measurement of half-life for one of these. (There is a data logging opportunity here.) It is vital to correct for background radiation. Explain how to find T1/2 from the graph. (Students could repeat this experiment for themselves later.) Activity (Bq)

200

(Diagram: resourcefulphysics.org)

100

T

5

2T

3T

1

4T

Time (s)

Demonstration of half life of protactinium: Measuring half life TAP 514-1: Half life of protactinium

Worked examples: Involving whole numbers of half-lives Here is a set of questions involving integral numbers of half lives. You could set them as examples for your students, or work through them to explain some basic ideas. Question: The nuclear industry considers that after 20 half lives, any radioactive substance will no longer present a significant radiological hazard. The half life of the fission product from a nuclear reactor, caesium-137, is 30 years. What fraction will still be active after 20 half lives? Answer: Use the long method. Calculate 1/2 of 1/2 and so on for twenty steps: 1/2, 1/4, 1/8, 1/16, 1/32, …, 1/1024 after 10 half lives 1/2048, …, 1/1048576 after 20 half lives i.e. less than one millionth of the original quantity remains radioactive. A quicker method is to calculate (1/2)20. Show how this is done with a calculator, using the yx key. Question: How many years into the future will Cs-137 be “safe”? Answer: 20 × 30 = 600 years

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Question: If after ten half lives the activity of a substance is reduced to one thousandth of its original value, how many more half lives must elapse so that the original activity is reduced to one millionth of its original value? Answer: Ten half lives reduces activity by a factor of 1/1000. One millionth = 1/1000 × 1/1000, so ten more half lives are needed.

Student example + discussion: Plotting a graph Provide your students with the following data. Time /s

0

5

10

15

20

25

30

35

40

Activity /Bq

10057

6253

3648

2296

1403

798

508

307

201

Ask them to inspect the data and estimate the half-life. (It lies between 5 and 10 seconds.) Now ask them to plot a graph and thereby determine the half life of the substance. They should deduce a total time of 35 seconds for five halvings of the original activity, gives a half life of 7 seconds. Emphasise the definition of half life T1/2. The half life of a radioactive substance is the time taken on average for half of any quantity of the substance to have decayed. (Check the precise wording of your specification.) Introduce the term exponential to describe this behaviour, in which a quantity decreases by a constant factor in equal intervals of time. TAP 514-2: Half life Exponential behaviour in nature is very common. It appears in the post-16 specification several times, so this may not be the first time your students have met it. (The equalisation of electric charge on a capacitor C whose plates are connected via a resistor R; the reduction in amplitude of a damped simple harmonic oscillator; the absorption of electromagnetic radiation passing through matter (e.g. γ rays by lead, visible light by glass); Newton’s Law of Cooling. So either refer back to previous situations and make the analogy explicit, or flag forward to other examples to come.

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TAP 514- 1: Measuring half life of protactinium Following your school or college’s safety procedures carefully, measure the half-life of a short-lived radioactive isotope. Plot a suitable graph showing how the activity of the sample changes with time, and hence show that the decay is indeed exponential. Have a spills tray available.

Radioactive sources Follow the special local rules for using the protactinium source. (Do not open the bottle! To avoid contamination if the plastic bottle should leak, keep the bottle in a tray and wear suitable gloves, a lab coat and eye protection.)

Introduction The isotope of protactinium 234Pa has a half-life of several tens of seconds. You can monitor its decay using a GM tube connected to a ratemeter or scaler or to a data logging interface and computer, and hence determine its half-life experimentally. The notes here are aimed at explaining how to make the right measurements, but do not proceed until you are sure how to handle the source safely. The precise details of your experiment will vary depending on your school or college’s local rules for handling radioactive sources. Make sure that you are aware of these rules and follow them carefully. The sample of 234Pa is produced by the decay of 238U. Pa234 is the grand-daughter of U238: alpha decay to Th234 followed by beta decay to Pa234, as described later. The 238U is in the form of uranyl nitrate dissolved in water and is contained in a sealed plastic bottle. The bottle also contains an oily solvent that floats above the water. The bottle will already have been prepared for you. Do not open the bottle! To avoid contamination if the plastic bottle should leak, keep the bottle in a tray and wear rubber gloves, a lab coat and eye protection. When the bottle is shaken some of the 234Pa in the watery layer dissolves into the oily layer. Once the two layers have separated out no more 234Pa moves into the oily layer, so we have a fixed sample of 234Pa in the oily layer. 234Pa emits energetic beta radiation, which can penetrate the plastic bottle and travel some distance in air. This is the radiation that you will monitor.

Procedure Without shaking the bottle, set up the apparatus as shown below.

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Measure the background radiation (it comes from the bottle and the lab), counting for a suitably long period. Shake the bottle for about 30 s and place it back in position. Wait for the layers to settle and start the clock. The GM tube window should be close to the oily layer and is best mounted on a retort stand. Take readings from the GM tube as frequently as you can accurately manage until the count rate is the same as the background count.

Analysis of results Subtract the effects of background radiation from your readings, and hence calculate the counts per second from the 234Pa in the oily layer. This is the corrected count rate (CCR); it corresponds to the activity of the sample. Plot a graph of corrected count rate against time and a graph of log(CCR) against time. Do your graphs correspond to what you already know about exponential decay? Use each graph to calculate the half-life of the 234Pa. (Hint: one of these involves using the gradient.) The accepted value for the half-life of 234Pa is 72 s. Discuss any disparity between your measurement and this accepted value. How great was the uncertainty in your measurement? Which of your two ways of determining T1/2 was the most reliable? What were the main causes of uncertainty in your result? How could you improve on your procedure?

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Practical advice This is a standard demonstration, the precise details of which will depend on the apparatus you have available. If data-logging is available, then we strongly recommend that you use it. It is vital that students correct the readings for background count. Make sure that students are clear that the half-life of an exponential decay is not exactly the same as the time constant, the latter is the time for a quantity to decay to 1/e of its initial value, not 1/2, and so the half-life is always a bit shorter than the time constant. A discussion of the units for λ and T1/2 can bring out several useful teaching points: •

There is an enormous range of half-lives.

•

It is conventional to quote half-lives in whatever units are most suitable, and it would be ludicrous always to insist dogmatically on SI units (seconds) to express very long half-lives.

•

The decay constant is usually expressed in units that tally with those used for the half-life (e.g. s-1, yr-1 as appropriate).

•

It is vital always to quote the units of half-life or decay constant – a number on its own would be highly ambiguous.

•

While the half-life and decay constant may have any one of several units, they always have dimensions of time and time -1, respectively.

Some details of the decay may be of use and could be used to give students more experience of writing a set of decay equations. Decays by Decay chain Half Life

α 238

U

4.5 x 109 /years

–>

β 234

Th

–>

β 234

Pa

–>

234

U

24.1

72

2.5 x 105

/days

/s

/years

234

Pa is extracted from a solution containing the parent 234Th and ’grandparent’ 238U with which it is in equilibrium. Once extracted the decay of 234Pa can be recorded. The recovery of 234Pa can be followed in the aqueous layer if desired. About 95% of the protactinium is removed together with some uranium when the bottle is shaken. The GM tube does not detect the alpha from the 238U or the weak beta particles (about 0.2 MeV) from the 234Th but detects only the energetic (2.32 MeV) beta radiation from the 234Pa.

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Technician’s notes

Radioactive sources Follow the special local rules for using the protactinium source. (Do not open the bottle! To avoid contamination should the bottle leak, keep the bottle in a lined tray and wear suitable gloves, a lab coat and eye protection.)

Apparatus 9

Thin walled polythene bottle (50 cm3) containing: uranyl nitrate, 2.5 g, dissolved in 7.5ml pure water, concentrated hydrochloric acid, 17.5 ml, 25 ml of amyl acetate (pentyl ethanoate)

9

GM tube + holder

9

Scaler/ratemeter and stopwatch/clock or datalogger

9

Tray lined with paper towel in case of spillage

9

Retort stand, boss and clamp to align the GM tube with the appropriate part of the bottle

Dissolve the uranyl nitrate in 7.5 ml water and add the hydrochloric acid. Then add the amyl acetate. Pour into the polythene bottle and screw the cap securely. Putting thin plastic film (e.g. cling-film) between cap and bottle can help make a better seal. Wear ‘chemical splash’ rubber gloves, a lab coat and eye protection, follow local radiation rules and dispose of any contaminated material safely. A source in a polythene bottle should not be kept for more than 12 months. Polypropylene bottles last longer. Teflon bottles (provided the wall is thin enough to let the betas through) will last 10 years. A recipe based on a 30 ml bottle is on CLEAPSS Recipe Card 57. The bottle should be labelled with the radiation symbol and also TOXIC and IRRITANT It is advisable to check the activity of the solution each year. Some teachers prefer to use fresh solution each time. Uranyl nitrate can be obtained from: TAAB Laboratory Support Limited, 3 Minerva House, Aldermaston RG7 8NA. Tel: 0118 981 7775 E-mail: [email protected] Prices as of September 2005: 1g for £7.99 and 25g for £69.50.

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External references This activity is taken from Salters Horners Advanced Physics, section RftS, Activity 9, which was an adaptation of Revised Nuffield Advanced Physics experiment F6

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TAP 514- 2: Half Life The half-life of strontium-90 is 27 years. The half-life of sodium-24 is 15 hours. 1.

Sketch two curves on one set of axes to show how the number of atoms of each change with time.

Two samples are prepared, one containing 1020 atoms of strontium and the other containing 1020 atoms of sodium. 2.

Which of the two samples has the highest activity?

3.

A sample of iodine-131, with half-life 8.04 days, has an activity of 7.4 x 107 becquerel. Calculate the activity of the sample after 4 weeks?

4.

234

(a)

What fraction of a sample remains after 96.4 days?

(b)

What fraction of a sample remains after 241 days?

Th has a half life of 24.1 days.

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Answers and worked solutions 1. strontium

1020

sodium

0

15

30

45

60

t/h

2.

Sodium, since a longer half-life means the source is less active.

3.

First find the number of half-lives in 4 weeks: 4 × 7 days = 3.48. 8.04 days

The activity will be: 7.4 × 10 7 Bq 2

3.48

= 6.6 × 10 6 Bq.

4 (a)

96.4/24.1=4 so 4 half lives so (1/2)4 = =1/16 or .0625

(b)

241/24.1=10 so 10 half lives (1/2)10 = 1/1024 or 9.8 x 10-4

External references This activity is taken from Advancing Physics chapter 10, 20S which was an adaptation of Revised Nuffield Advanced Physics question 15 section F.

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Episode 515: The radioactive decay formula Here, the key idea is the random nature of the decay. Avoid simply pulling pull equations out of the air – at least make them plausible.

Summary Discussion: The meaning of the decay constant λ. (15 minutes) Discussion: The link with half-life. (15 minutes) Student experiments: Analogue experiments linking probability with decay rates. (20 minutes)

Discussion: The meaning of the decay constant λ. Start from the definition of the decay constant λ: the probability or chance that an individual nucleus will decay per second. (You may like to comment on the problem with notation in physics. λ is used for wavelength as well as the decay constant. The context should make it unambiguous.) Units: λ is measured in s-1 (or h-1, year-1, etc). If you have a sample of N undecayed nuclei, what will its activity A be? In other words, how many of the N will decay in a second? A = λN (because the probability for each of the N is λ).

As time passes, N will get smaller, so A represents a decrease in N. To make the formula reflect this, it needs a minus sign. A = -λN

Plausibility: The more undecayed nuclei you have, and the greater the probability that an individual one will decay, the greater the activity of the sample. In calculus notation, this is dN/dT = -λN Ask your students to put this equation into words. (The rate of decay of undecayed nuclei N is proportional to the number N of undecayed nuclei present.) This is the underlying relationship in any process that follows exponential decay. More generally: if the rate of change is proportional to what is left to change, then exponential decay follows. Conversely if data gives an exponential decay graph, then you know something about the underlying process. TAP 515-1: Smoothed out radioactive decay. TAP 515-2: Half-life and time constant.

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Discussion: The link with half-life. How are λ and T1/2 linked? If a nuclide has a large value of λ, will it have a long or short T1/2? (High probability of decay => it will decay quite quickly => a short half life, and vice versa.) What type of proportionality does this suggest? (Inverse proportion.) Thus λ ~ 1/T1/2. In fact, T1/2 = ln 2/λ = 0.693/λ This is a very important and useful formula. Depending which of λ or T1/2 is easiest to measure experimentally, the other can be determined. In particular it allows very long half lives (sometimes millions of years) to be determined. How could we measure the 4.5 billion year half-life of uranium-238? (Take a sample of a known number of U-238 atoms; count how many decay in one second. This gives λ, and we can calculate half-life.)

Student experiments (or demonstrations): Analogue experiments linking probability with decay rates. Explore some analogue systems to reinforce the way in which decay probability is related to half-life. Each gives a good exponential graph. For example: Throw a large number of dice. A 6 represents ‘decayed’, and this dice is removed. Each throw represents the same time interval. Each face has a probability of 1/6 of being upwards on each throw. (A quick way to find out how many remain to decay is to weigh them.) TAP 515-3: Modelling radioactive decay

Cubes with differently coloured faces instead of dice can incorporate 3 different ‘decay constants’ e.g. colouring three faces red gives a chance of a red face being uppermost representing decay has a probability of decay of ½; colour two faces blue (chance of decay = 2 x 1/6 = 1/3), and one face yellow (chance of decay 1/6). Sugar cubes can be used, painted with food colouring. Throw drawing pins: decay = point upwards. Safety: beware sharp points! Is the chance of decay = ½? The drop in height of the head on a glass of beer usually shows exponential behaviour – opportunity for a field trip? Water flowing out through a restriction: at any instant the water remaining represents the undecayed nuclei; the water that has flowed out represents the decayed nuclei. Flow rate depends upon the pressure head ~ the height of water remaining ~ quantity of water remaining.

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TAP 515-1: Smoothed out radioactive decay Here is the first step in making a model – idealisation and simplification of a messy process

Smoothed out radioactive decay Actual, random decay

ΔN ΔN

Δt

Δt

time t probability p of decay in short time Δt is proportional to Δt: p = λ Δt average number of decays in time Δt is pN Δt short so that ΔN much less than N change in N = ΔN = –number of decays

ΔN = –λN Δt

ΔN = –pN ΔN = –λN Δt

Simplified, smooth decay

rate of change = slope dN = dt

time t Consider only the smooth form of the average behaviour. In an interval dt as small as you please: probability of decay p = λ dt number of decays in time dt is pN change in N = dN = –number of decays dN = –pN dN = –λN dt

dN = –λN dt

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Practical advice The diagram could be used for an OHT

External reference This activity is taken from Advancing Physics chapter 10, display material 10O

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TAP 515-2: Half-life and time constant Here is the relationship between these two, explained

Radioactive decay times dN/dt = –λ N

N/N0 = e–λt

N0

N0 /2 N0 /e

0 t=0

t = t1/2 t = time constant 1/λ

Time constant 1/λ at time t = 1/λ N/N0 = 1/e = 0.37 approx. t = 1/λ is the time constant of the decay Half-life t1/2 at time t1/2 number N becomes N0/2 N/N0 =

1 2

= exp(–λ t1/2)

In 12 = –λt1/2 t1/2 = ln 2 = 0.693 λ λ In 2 = loge 2

Half-life is about 70% of time constant 1/λ. Both indicate the decay time

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Practical advice The diagram could be used for an OHT

External reference This activity is taken from Advancing Physics chapter 10, display material 40O

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TAP 515-3: Modelling radioactive decay Use dice, coins or a computer model to simulate the random decay of radioactive nuclei. Plot a graph showing how the number of remaining ‘nuclei’ changes with time. The chance throw of dice illustrates radioactive decay very clearly. You need a large number (over 100) dice for this activity to work. If you don’t have enough dice, then you will be able to imagine the sort of results you might get, so keep reading. Let’s suppose that an atom of an isotope has a 1 in 6 chance of decaying in any one minute. We can simulate this by throwing a die representing the atom; if a six comes up, we can say that the atom has decayed. Now with 100 or more dice we have a model of a (very small) sample of radioactive isotope. Each time we shake the dice assume that another minute of the sample’s life has passed. In this time more ‘nuclei’ have decayed – the ones that came up six. Shake the dice. Count and remove the sixes. Repeat the procedure, completing a table like table below, until all the dice have ‘decayed’.

If you have time, repeat the whole process. Plot a graph showing how the number of surviving dice changes with time. Plot a graph of log (N ) against t and hence decide whether the decay really is exponential. You could also plot graphs to show how ‘activity’, i.e. the number of decays per minute, varies with time. Use suitable tests to decide whether activity changes exponentially.

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Practical advice In previous episodes, students may have met examples of exponential change that are all determined by equations that describe the particular situation exactly. Radioactive decay, on the other hand, involves a degree of randomness – this is clear from simple observations such as the production of tracks in a cloud chamber, or listening to ‘clicks’ from a GM counter. The aim of this activity is to demonstrate that a decay process that is governed by independent probabilities can (indeed must) be exponential – an important point, which should not be glossed over. There are a number of simulations of radioactive decay in physics software packages and any of them could be used at this stage. However, students readily understand the laws of probability that govern the throw of dice, so the one outlined on in TAP 515-3 remains an excellent demonstration. The ‘experiment’ should take only a few minutes, plus graph-plotting time. Another good analogy for radioactive decay is liquid draining from a cylindrical container through a capillary tube. The flow-rate is proportional to the pressure due to the liquid, which is proportional to its depth. The rate at which the depth of liquid decreases is thus proportional to the depth itself. A graph of depth against time will be an exponential decay curve, and a log-linear graph will be a straight line. Obtain data by marking the position of the water surface at regular intervals as the water slowly drains from the cylinder.

Technicians’ notes Apparatus dice model 9

Dice (identical if possible) or small wooden cubes (approx 1 cm3) with one face coloured

9

At least 100 dice or cubes are needed.

Water- flow model 9

wide glass tube (4–5 cm diameter), pierced bung to fit tube capillary tube to fit bung, rubber tube to fit capillary

9

Hoffmann clip

9

clamp, boss and stand

9

marker pen

9

stopwatch

9

container to catch water draining from tube

9

Assemble the apparatus as shown in below, and arrange for the outlet to be over a large container, or sink if conveniently located. Ensure the Hoffmann clip is closed, and fill the wide tube with water.

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Above: Modelling radioactive decay using water flow Alternatively, a computer running a simulation of radioactive decay may be requested.

External reference This activity is taken from Salters Horners Advanced physics, A2, STA activity 10.

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Episode 516: Exponential and logarithmic equations Students may find this mathematical section difficult. It is worth pointing out that they have already covered the basic ideas of radioactive decay in the earlier episodes.

Summary Discussion: The exponential decay equation. (15 minutes) Student question: An example using the equation (20 minutes) Discussion: The logarithmic form of the equation. (15 minutes) Worked example: Using the log equation. (20 minutes) Student questions: Practice calculations. (60 minutes) Discussion (optional): Using Lilley’s formula. (15 minutes)

Discussion: The exponential decay equation Explain that the equation N = N0 e-λt can be used to generate an exponential decay graph. Work through a numerical example, perhaps related to the dice-throwing analogue (N0 = 100; λ = 1/6). Make sure that your students know how to use the ex key on their calculators. Emphasise that similar equations apply to activity A (A = A0 e-λt) and count rate C (C = C0 e-λt). (Not all the radiation emitted in all directions by a source will collected by a detector lined up in one direction from a source)

Student question: An example using the equation Set students the task of drawing a graph for a lab source, e.g. Co-60 (λ = 0.132 y-1, C0 = 200 counts s-1). They should first calculate and tabulate values of C at intervals of 1 year, and then draw a graph. From the graph, deduce half-life. Does this agree with the value from T1/2 = ln 2/ λ = 0.693/ λ?

Discussion: The logarithmic form of the equation Point out that a straight line graph is usually more useful than a curve, particularly when dealing with experimental data. Introduce the equation lnN = lnN0 - λt. Emphasise that this embodies the same relationship as the exponential equation. Use a sketch graph to how its relationship to the straight line equation y = mx + c (intercept = lnN0, gradient = −λ).

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Worked example: Using the log equation Start with some experimental data (e.g. from the decay of protactinium), draw up a table of ln (count rate) against time. Draw the log graph and deduce λ (and hence half-life). The experimental scatter should be obvious on the graph, and hence the value of a straight line graph can be pointed out. You will need to ensure that your students can find natural logs using their calculators.

Student questions: Practice calculations Your students should now be able to handle a range of questions involving both ex and ln functions. It is valuable to link them to some of the applications of radioactive materials (e.g. dating of rocks or ancient artefacts, diagnosis and treatment in medicine, etc). TAP 516-1: Decay in theory and practice TAP 516-2: Radioactive decay with exponentials TAP 516-3: Radioactive decay used as a clock

Radio Carbon Dating

(Diagram: resourcefulphysics.org)

TAP 516-4: Two important dating techniques

Discussion (optional): Using Lilley’s formula Some students may benefit from a simpler approach to the mathematics of radioactive decay, using ‘Lilley’s formula’ f = (1/2) n.

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When first introduced at pre-16 level, radioactivity calculations are limited to integral number of half lives. After 1, 2, 3, …, half-lives, 1/2, 1/4, 1/8, … remains. The pattern here is that after n half lives, a fraction f = (1/2) n remains to decay. This formula works for non integral values of n; i.e. it also gives the fraction remaining yet to decay after any non-whole number of half-lives (e.g. 2.4, or 3.794). To use this formula, a little skill with a calculator is all that is required. For example: The T1/2 of 14C is 5730 yrs. What fraction f of a sample of 14C remains after 10 000 years? Answer: f = (1/2) n

The number of half lives n = 10 000/5730 = 1.745 Thus the fraction remaining f = (1/2)1.745 And using the yx button on a calculator gives f = 0.298. If students know how to take logs (or can lean the log version of the formula), they can solve other problems: How many years will it take for 99% of 60Co to decay if its half life is 5.23 yr? The fraction remaining = 1%, so f = 0.01. Hence 0.01 = (1/2) n Taking logs of both sides gives ln (0.01) = n ln (1/2) giving n = 6.64 half lives, and so the number of years = 6.64 × 5.23 = 34.7 years.

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TAP 516-1: Decay in theory and practice Fractions and activity 1.

In exponential decay the number of atoms remaining after a given interval of time is always the same fraction of the number present at the beginning of the time interval. Therefore, someone might say, however long you wait there will always be some left. A sample of radioactive atoms will last for an infinite time. Is this reasonable? Explain your answer.

2.

The decay constant for caesium-137 is 7.3 × 10–10 s–1. Calculate the number of atoms in a sample of caesium-137 that has an activity of 2.0 × 105 Bq.

Experimental exponentials No ordinary place on the Earth’s surface is free from ‘background radiation’. In making a measurement of this background a student set up a GM tube and counter and recorded the counter reading every 30 s.

Time / s

Count

0

0

30

13

60

31

90

48

120

60

150

75

180

88

210

102

240

119

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INDEX

3.

Using these data, determine the background count per minute. What would you expect the total count to be after a further 30 seconds?

A radioactive source is set up in front of the same GM tube in the same laboratory. Counter readings of the activity of the source are taken for 1 minute at time intervals of 1 hour.

Time / h

A / counts min–1

0

828

1

510

2

320

3

202

4

135

5

95

6

70

7

51

8

41

9

38

10

37

11

31

12

33

13

34

14

29

15

35

4.

What action should be taken to deal with the background count?

5.

Plot a graph to show how the activity of the radioactive substance changes with time and derive from it three different values of the half-life. Calculate the mean of these.

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6.

Worried by the obvious fluctuations in the final hours of the count, one student suggests that it might be wise to continue counting for at least as long again. Is this a good idea? Explain your answer.

7.

There is a slight increase in the count rate towards the end of the experiment. Is this significant? Explain your answer.

8.

Suggest steps you might take to improve the experiment.

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Practical advice This question compares the theoretical exponential decay with the messy practical situation. This comparison can be used to bring out some of the characteristics of the models.

Answers and worked solutions 1.

No. The smooth exponential model provides a good description only when there are a very large number of atoms. However, when the sample is reduced to a few atoms, the model is a poor fit. How long an atom will last on average is known, but it is not possible to say how long one particular atom will last.

2. dN = −λN dt N=

– dN / dt activity 2.0 × 10 5 Bq = = = 2.7 × 10 14 atoms 10 1 − − λ λ 7.3 × 10 s

3.

119 counts = 30 counts min −1 4 minutes After further half minute count = 119 + 30 / 2 = 134 counts. background count =

4.

Subtract the background count from each of the tabulated counts min–1.

5.

The half-life is about 80 minutes.

6.

The count rate would not change significantly if the counting were continued for a longer period. Very little of the substance remains after 8 hours.

7.

No; random fluctuations in the background count rate are probably responsible for this increase.

8.

The experiment could be repeated and readings taken at intervals closer than 1 hour during the first few hours.

External reference This activity is taken from Advancing Physics chapter 10, 30S

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TAP 516-2: Radioactive decay with exponentials 1.

The half-life of one radioactive isotope of sodium is 2.6 years. Show that its decay constant is 8.4 × 10–9 s–1.

2.

Calculate the activity of a sample containing one mole of the sodium. (One mole contains 6.02 × 1023 atoms.)

A scientist wishes to find the age of a sample of rock. Realising that it contains radioactive potassium, which decays to give a stable form of argon, the scientist started by making the following measurements: decay rate of the potassium in the sample = 0.16 Bq mass of potassium in the sample = 0.6 × 10–6 g mass of argon in the sample = 4.2 × 10–6 g 3.

The molar mass of the potassium is 40 g. Show that the decay constant λ for potassium is 1.8 × 10–17 s–1 and its half life is 1.2 × 109 years.

4.

Calculate the age of the rock, assuming that originally there was no argon in the sample and the total mass has not changed. Show the steps in your calculation.

5.

Identify and explain a difficulty involved in measuring the decay rate of 0.16 Bq given above.

6.

Iodine 124, which is used in medical diagnosis, has a half-life of 4.2 days. Estimate the fraction remaining after 10 days.

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7.

Explain how you would find the half-life of a substance when it is known to be more than 10 000 years. Assume that a sample of the substance can be isolated.

In an experiment to find the half-life of zinc-63, a sample containing a sample of the radioactive zinc was placed close to a GM tube and the following readings were recorded. The background count rate was 30 min–1.

Time / hours

Counts / min–1

0

259

0.5

158

1.0

101

1.5

76

2.0

56

2.5

49

3.0

37

8.

Plot a graph of count rate against time and use this to find the average time for the count rate to fall to one-half of its previous value.

9.

Plot a second graph, ln (count rate) against time, and use it to find the half-life.

10.

Discuss which method, 8 or 9, provides a more reliable value.

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Practical advice The questions provide a variety of kinds of practice, at post-16 standard, with radioactive decay, half life and decay constants.

Answers and worked solutions 1.

λ=

ln 2 0.693 = = 8.4 × 10 −9 s ? t 12 2.6 y × 3.16 × 10 7 s y ?

A=

dN = – λN dt

2.

= (8.4 × 10 −9 s −1 ) × (6.02 × 10 23 ) = 5.1 × 1015 Bq

3.

λ= =

dN / dt N 0.16 Bq [(0.6 × 10

−6

g)/40 g] × (6.02 × 10 23 )

= 1.8 × 10 −17 s −1 t1 2 = =

ln 2 λ 0.693 1.8 × 10 −17 s −1

= 3.85 × 1016 s = 1.2 × 109 years 4.

0.6 × 10 −6 g 1 1 so there have been three half-lives. = = 4.8 × 10 −6 g 8 2 3 3 × 1.2 × 10 9 years = 3.6 × 10 9 years

5.

The background count rate is likely to be higher than 0.16 Bq.

6.

There have been

10 days 1 = 2.38 half-lives. The ratio remaining is 2.38 = 0.19. 4.2 days 2

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7.

Place a GM tube near the sample and measure the count rate over a long period of time. Determine a value for dN/dt. Determine N by chemical means, and from that find λ and T1/2.

8.

Subtract the background count rate from all readings. The half-life is about 40 minutes. first sample half life 200 to 100 in 0.6h + second sample half life 160 to 80 in 0.65h

200

third sample half life 120 to 60 in 0.7h +

average time = 0.65 h ≈ 40 min

0.6 h

100

0.65 h

+ 0.7 h

+ + + 0 0

0.5

1.0

1.5 2.0 t/h

2.5

+ 3.0

9.

Corrected count rate / min–1

ln corrected count rate / min–1

229

5.43

128

4.85

71

4.26

46

3.83

26

3.26

19

2.94

7

1.95

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5

gradient = –λ =

+

t1 =

+

4

+

2

+

3

ln 2 ln 2 × 2.0 h = = 0.60 h 2.3 λ = 36 min +

2.7 – 0.7 = 2.0 h

2

–2.3 2.0 h

+

1 0 0

10.

0.5

1.0

1.5 2.0 t/h

2.5

3.0

The second method is more reliable. By drawing a straight line of best fit in question 9, all the data are being averaged rather than just the three pairs taken from the curve in question 8.

External reference This activity is taken from Advancing Physics chapter 10, 90S

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TAP 516-3: Radioactive decay used as a clock Because of the predictability of the random behaviour of large numbers of atoms, activity can be used as a clock.

Clocking radioactive decay Activity

Half-life N0 number N of nuclei halves every time t increases by half-life t1/2

N0 /2

slope = activity =

dN dt

halves every half-life

N0 /4 N0 /8 t1/2

t1/2

t1/2

t1/2

t1/2

time t

time t

Radioactive clock In any time t the number N is reduced by a constant factor

Measure activity. Activity proportional to number N left

In one half-life t1/2 the number N is reduced by a factor 2

Find factor F by which activity has been reduced

In L half-lives the number N is reduced by a factor 2

L

(e.g. in 3 half-lives N is reduced by the factor 23 = 8)

149

Calculate L so that 2L = F L = log2F age = t1/2 L

INDEX

Practical advice This diagram is reproduced here so that you can talk through it, or adapt it to your own purposes.

External reference This activity is taken from Advancing Physics chapter 10, Display material 20O

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TAP 516-4: Two important dating techniques Radiocarbon dating This technique was devised by American physicist W. F. Libby in 1949. Carbon-14 (14C) is a naturally occurring radioactive isotope of carbon which is produced when neutrons associated with cosmic rays collide with nitrogen (14N) in the atmosphere. The 14C is taken up by living organisms (plants, animals); the ratio of 14C to 12C in living tissue remains constant during the life of the organism and depends only on the relative proportions of 14C and 12C present in the atmosphere. (12C is the `ordinary', stable, abundant isotope of carbon.) When the organism dies, the ratio of 14C to 12C decreases because the 14C decays to produce nitrogen:

After about 5700 years (the half-life of 14C), the ratio of 14C to 12C falls to half its initial value; after a further 5700 years the ratio halves again ± and so on. The 14C /12C ratio therefore decreases in a known and predictable way as the sample ages. If the original ratio of 14C /12C is known for an organic sample, say bone, and the present ratio can be measured, then an accurate value for the age of the sample can be calculated. To measure the ratio of 14C to 12C, the sample needs to be ground up, thus destroyed. The 14C in a sample is detected and measured via its radioactive emission, and the concentration of carbon is measured chemically. The method works best on samples containing large quantities of carbon. Thus wood, charcoal, bone and shells of land and sea animals are good archaeological samples. Around 100 g of wood or 30 g of charcoal is required to obtain a date. Bone contains a smaller proportion of carbon so more bone is required than wood (around 1 kg). Around 100 g of shell is required for dating. The precision of the age measurement decreases with age of the sample because the amount of 14C decreases with time. The half-life of 14C is 5730 ± 40 years so the method is most reliable for dating samples no more than a few thousand years old. For a sample about 50 000 years old the uncertainty is about ± 2000 years. The range of radiocarbon dating is at most 100 000 years. The method rests on the assumption that the sample gains no 14C after death, and that the atmospheric ratio of 14C /12C has remained constant with time. It is therefore necessary to compare radiocarbon dates against other independent methods of dating such as dendrochronology. For an article on the radiocarbon method, see:

Radioactive decay. Dobson, K., Physics Review, Vol. 4, No. 2, pp. 18-21 A good World Wide Web site for details on the method and much besides is: http://www.c14dating.com/

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Dendrochronology The name of this method comes from the Greek `dendron' (tree) and `chronos' (time). The cross-section of a tree trunk shows a series of concentric circles known as growth rings. Each growth ring represents one year of the life of the tree. The thickness of each ring depends on the climatic conditions in the year that the ring was laid down. Thus the rings show a distinctive pattern of varying thickness. The same pattern can be seen in timber of different ages and so a pattern covering a considerable period of time can be constructed. Once the pattern has been established and tied to known dates, a pattern in a timber of unknown date can be matched against the established pattern and its date determined. The first person to suggest dendrochronology as a dating method was Thomas Jefferson, one of the first American Presidents, who suggested its use in dating Native American burial mounds. The method can be used with wood grown since the last Ice Age, about 10 000 years ago. One of the longest patterns established is that using the Bristlecone Pine which grows in California. The arid conditions of the area allow samples of the pine to survive in good condition for thousands of years. Apart from its use as an absolute method of dating wood, dendrochronology is also used to calibrate radiocarbon dates. When radiocarbon dates are matched against the tree ring dates, it is found that radiocarbon `years' do not equate directly with calendar years because the amount of 14 C in the atmosphere varies slightly. In general, radiocarbon dating gives a young age for older samples (e.g. a radiocarbon date of 4100 BC may be closer to a real date of 5000 BC). The method does have its limitations. Climatic conditions vary from place to place and so a particular pattern can only be used locally. The wood has to be in a good state of preservation; such wood is difficult to find on archaeological sites. The wood on a site may not indicate the true age of the site. For example, the wood could have been cut down many years before its use on a particular site. And not all types of wood can be used since some types have growth rings of uniform thickness regardless of climatic condition

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Practical advice This activity summarises two important dating techniques used in modern archaeology.

External reference This activity is taken from Salters Horners Advanced Physics, AS, DIG additional activity 10

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Episode 517: Preparation for accelerators and detectors topic Accelerators and detectors There are over 10,000 accelerators worldwide: 5000 each in medicine (therapy, isotope production) and industry (e.g. used for ion implantation in semi conductors and into metal for hardening etc), plus over 100 used for basic research by a whole range of scientific disciplines. The topic of particle acceleration and detection is useful as it illustrates many aspects of physics which students are likely to have studied already, and it introduces them to an area in which fundamental physics is making progress. There are opportunities for demonstrations that illustrate the basic ideas of particle acceleration and detection.

Episode 518: Particle accelerators Episode 519: Particle detectors

Main aims Students will: Apply their knowledge of the motion of charged particles in electric and magnetic fields to particle accelerators and detectors.

Prior knowledge Students should know about forces on charges in electric and magnetic fields, and their effects on motion. In particular, they should know about how circular motion arises when a charged particle moves in a uniform magnetic field. They should be familiar with the basic conservation laws (charge, energy, momentum). An alternative approach would be to combine a study of accelerators and detectors with a study of electric and magnetic fields.

Where this leads If your specification requires it, you could go on from here to study the Standard Model (quarks, leptons and fundamental forces).

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Episode 518: Particle accelerators This episode requires students to apply their knowledge of charged particles and fields.

Summary Discussion and worked example: Acceleration in an electric field. (15 minutes) Student activity: Researching accelerators. (30 minutes) Demonstration: Electrical breakdown. (15 minutes Discussion: How a linear accelerator works. (15 minutes) Discussion: Particles in a magnetic field. (10 minutes) Demonstration: Fine beam tube. (20 minutes) Student questions: Calculations. (30 minutes) Discussion (optional): Relativistic effects and Bertozzi’s experiment. (15 minutes) Visit (optional): Take a trip to CERN. (A long weekend)

Discussion + worked example: Acceleration in an electric field Why accelerate particles? Following Rutherford’s alpha-scattering experiment, physicists wanted to probe matter with beams of particles that were more energetic, more intense and ‘purer’. How can particles be speeded up? (use an electric field.) Won’t a magnetic field do? (particles are accelerated, but the force is centripetal, so their energy does not increase.) Calculate the speed of an electron (or proton) accelerated through 10 kV. What equation to use? (½) mv2 = eV

v=

e = 1.6 x 10-19 C and m = 9.1 x 10-31 kg

2 qV ≈ 6 x10 7 m s −1 m

Take care! This is approaching speeds where relativistic effects need to be taken into account. Will a proton travel faster or slower than this? (slower, because charge is the same but mass is greater.) In the largest research accelerators, energies are so great that they recreate the conditions minuscule fractions of a second after the Big Bang (typically 10-10 s for LEP and a planned 10-12 s for the Large Hadron Collider (LHC) opening in 2007).

Student activity: Researching accelerators Find out about the development of linear and circular accelerators. Identify important spin-offs (e.g. the development of www, computer graphics, body scanner magnets, isotope production for medicine and industry, material processing etc.)

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TAP 518-1: Some information about LEP at CERN

Demonstration: Electrical breakdown In linear accelerators, the approach is to get as large a voltage as possible, and to apply it to the particles several times. A practical limit to voltage difference is set by the ability of materials to withstand the electric fields involved. You can demonstrate electrical breakdown. TAP 518-2: Electrical breakdown. high frequency (3000 MHz) a.c voltage

drift tube (Diagram: resourcefulphysics.org)

particle beam

Discussion: How a linear accelerator works Explain the construction of the linear accelerator. The drift tubes get longer as the particles move faster. But at the highest speeds approaching that of light, increase in energy makes very little difference to the speed, so the drift tubes are the same length. TAP 518-3: The linear accelerator

Discussion: Particles in a magnetic field There is an advantage in making the particles travel around in a circular path – they can be accelerated time and again. Discuss how the particles trajectories are bent into a circular path with a magnetic field to bring them back to the accelerating electrical field many times. Compare with an electric field. Recap the equation for this (mv2/r = Bqv).

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TAP 518-4: How a magnetic field deflects an electron beam TAP 518-5: How an electric field deflects an electron beam

Demonstration: Fine beam tube Do this if you haven’t previously done so in episode 413 TAP 413-2: Measuring the charge to mass ratio for an electron Show the fine beam tube with Helmholtz coils to provide a magnetic field. TAP 518-6: The fine-beam tube

Student questions: Calculations Your students now know the equations needed to solve many problems relating to accelerators. You may have covered these questions in Episode 413, if not students should try them now. TAP 413-3: Deflection with electric and magnetic fields TAP 413-4: The cyclotron TAP 413-6: Charged particles moving in a magnetic field Also try: TAP 518-7: Fields in nature and in particle accelerators

Discussion (optional): Relativistic effects and Bertozzi’s experiment Your students should be aware that, at relativistic speeds, things become more complicated. One way to present this is to discuss Bertozzi’s experiment.

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Accelerators such as the synchrotron are designed to compensate for the effective increase in m by controlling the frequency of the accelerating voltage as the particles speed up. TAP 518-8: The ultimate speed – Bertozzi's demonstration TAP 518-9: Principle of the synchrotron accelerator

Visit (optional): Take a trip to CERN (a long weekend) You can organize a trip to CERN. http://www.pparc.ac.uk/Pbl/Cern.asp If you can’t make the visit this year borrow a video http://teachingphysics.iop.org/resources/video/video_book.doc

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TAP 518-1: Some information about LEP at CERN This information is provided for interest, perhaps to stimulate further research. •

First experiments: 1989

•

Particle collisions: electrons and positrons

•

Maximum beam energy: 100 GeV

•

Luminosity: 2.4 × 103 s–1

•

Time between collisions: 22 μs

•

Filling time: 20 h

•

Acceleration period: 550 s

•

Injection energy: 550 MeV

•

Bunch length: 1 cm

•

Average beam current: 55 mA

•

Circumference: 27.66 km

•

Dipole (bending) magnets: 3280 plus 24 weaker dipoles

•

RF resonant cavities: 128

•

Peak magnetic field: 0.135 T

•

Vacuum: 10–11 Torr

Between 1983 and 1989 the construction of LEP at CERN was the biggest civil engineering project in Europe. The accelerator tube is 26.67 km in circumference and is shaped to an accuracy of better than 1.0 cm. It runs underground in a specially excavated tunnel inclined at 14° to the horizontal between Geneva airport and the Jura mountains. There are four main experimental stations positioned around the ring. As it enters each of these the beam passes through a large solenoid whose magnetic field squeezes the beam to about 10 μm by 250 μm, increasing the luminosity (and hence the probability of interactions with the oncoming beam). From 1989 to 1995 LEP was used as a Z0 ‘factory’. This was done by setting the collision energy to about 91 GeV (rest energy of the Z0). This allowed physicists to make accurate measurements of the Z0 lifetime. From this they showed that there are only three generations of fundamental particles. If there were more then the lifetime of the Z0 would be lower because it would have more alternative particles into which it could decay. This conclusion agreed with that of cosmologists based on the number of different types of neutrinos needed to explain relative abundances of different nuclei in the early Universe. It is a good example of the growing links between particle physics on the smallest scale and cosmology, the study of the Universe on the largest scale. From 2005 LEP will be replaced by the LHC (large hadron collider, a new accelerator running in the existing LEP tunnel). This will accelerate protons and antiprotons to up to 14 TeV (1 TeV = 1012 eV) about 10 times greater than the Tevatron at Fermilab. Why? Whereas electron–positron collisions can be used to test precise aspects of the Standard Model, more massive particles are used in the hope of detecting rare but exotic events. LHC should reveal the supersymmetric partners of

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ordinary matter particles (as predicted by superstring theory) and may well reveal the Higgs particle – a force-carrier in the hypothetical Higgs field that endows all other particles with mass. The LHC is an amazing project, even by the standards of high-energy physics. The momentum of the high-energy protons and antiprotons is so high that extremely powerful superconducting dipole magnets must be used to keep them in the ring. Their peak field will be about 9 T! To maintain the superconducting properties these magnets must be cooled to 1.9 K. This requires eight cryogenic plants spaced equally around the 27 km ring pumping 70 000 litres of liquid helium through 40 000 leak-proof junctions to cool 31 000 tonnes of equipment!

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Practical advice This information should provide some insight into the engineering challenges to be overcome if the fundamental physics is to be explored.

Alternative approaches Similar information can be gleaned from Web sites dedicated to most large accelerators.

Social and human context The cost of such projects forces collaboration on national governments.

External reference This activity is taken from Advancing Physics chapter 16, 40T

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TAP 518- 2: Electrical breakdown Introduction Air is an insulator under normal conditions. If the potential gradient is sufficiently high, then the air may start to conduct, often with spectacular effect, as in lightning. In this demonstration you will be able to see large sparks produced by an electrostatic generator. On a smaller scale, sparking across a narrow gap can be enhanced by alpha radiation, illustrating the principle behind some radiation detectors.

You will need: 9

van de Graaff generator

9

EHT power supply, 0–5 kV dc

9

spark counter

9

pure alpha source

Wire carefully, EHT supply in use The large protective resistor is in use to prevent dangerous currents passing though humans. The EHT power supply, being current limited will not give a fatal shock, but it can be surprising

The local rules for handling radioactive sources must be complied with. Do not handle radioactive sources or place them in close proximity to your body Use tongs or a source holder to handle the alpha source and put it in a secure place when not in use

Looking at sparks You will be able to see the sparks produced in a high potential gradient. Stray electrons in the air are accelerated and produce secondary ionisation by collisions. An electron shower develops and this allows the air to conduct for a short time so that a spark is seen. The scale of sparking varies considerably from lightning flashes, through laboratory electrostatic generators to the small discharges in radiation detectors. For these smaller events, using an alpha source ionises the air and increases the probability of a spark occurring. 1.

You can charge the sphere of the van de Graaff generator to a high potential by turning it on for a short time. If you bring an earthed sphere close to the main sphere, then a spark will be seen to jump from one sphere to the other once the potential gradient is above about 3×106 V m–1. You should be able to estimate the potential of the main sphere from the distance across which the spark jumps but the breakdown potential gradient can vary considerably from day to day as atmospheric conditions change.

2.

Smaller sparks can be seen in the spark counter. This requires a lower potential difference because the gap between the wire and the earthed plate is only a couple of millimetres. If

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you connect the wire to the positive terminal of the EHT power supply and the plate to earth and turn the supply voltage up to maximum (5 kV), you will be able to see and / or hear occasional sparks between the wire and the plate (see the diagram above). It may be necessary to work in a darkened area of the laboratory. These sparks are initiated by stray electrons or ions passing close to the gap between the wire and the plate. Radioactive source and holder held by tongs

0V

+ 5000 V

gauze

sparking here

thin wire

3.

Alpha particles produce a high density of ionisation along their path. If you hold a source a short distance (2 or 3 cm) from the wire, there will be a considerable increase in the rate of sparking because the additional electrons are likely to produce secondary ionisation leading to breakdown. Many radiation detectors use this as a way of detecting particles.

You have seen 1.

How sparks can be produced in a strong electric field and how this can be used in radiation detectors.

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Practical advice If students have not seen the sparks produced by a laboratory electrostatic machine, then they should now. The demonstration of the small discharges in a spark detector follows naturally and links well with the discussion of the fields found in Geiger–Müller tubes. Commercial spark gaps are available, and are less likely to give shocks, but they are expensive. (Some schools may have a spark counter but not recognise it.) TAP 509-4: Rays make ions Episode 519: Particle detectors

Social and human context Lightning has a place in discussions of extreme weather conditions on Earth and on other planets. Discharges from overhead power lines can be a reason for energy loss as well as a potential hazard.

Wire carefully, EHT supply in use The large protective resistor is in use to prevent dangerous currents passing though humans. The EHT power supply, being current limited will not give a fatal shock, but it can be surprising

The local rules for handling radioactive sources must be complied with. Do not handle radioactive sources or place them in close proximity to your body Use tongs or a source holder to handle the alpha source and put it in a secure place when not in use

External reference This activity is taken from Advancing Physics chapter 16, 180D

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TAP 518- 3: The linear accelerator This shows the principles of the linear accelerator, although not all of the engineering complexities.

Principle of linear accelerator The accelerating field field between electrodes

negative electron accelerated

–

–

zero field inside tube

+

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Principle of linear accelerator Switching p.d.s to keep accelerating electrons alternating high frequency p.d.

at one instant

–

+

+

–

–

– +

– +

bunches of electrons between electrodes are accelerated a little later

zero p.d.

bunches of electrons drift through tube a little later still

+

–

–

+

+

+ –

+ –

bunches of electrons between electrodes are further accelerated

electrodes must be longer because electrons are going faster

The alternating p.d. switches back and forth so that the electrons are accelerated as they pass between successive electrodes

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Practical advice This diagram is here so that you can discuss it with your class.

External reference This activity is taken from Advancing Physics chapter 16, 40O

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TAP 518- 4: How a magnetic field deflects an electron beam The plane in which the deflection occurs is important, causing circular motion.

Magnetic deflection

positive ions +

force B-field beam velocity

negative charges – e.g. electrons

force on positive charge B v force at right angles to field and velocity of charge, F = qvB

Magnetic fields deflect moving charged particles in circular paths

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Practical advice This diagram is reproduced here so that you can discuss it with your class.

External reference This activity is taken from Advancing Physics chapter 16, 120O

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TAP 518-5: How an electric field deflects an electron beam The role of electric fields is central here.

Deflections of electron beam by electric field

deflection plates

electron gun + anode

hot cathode

+V force

electric field E =

V d

_ spacing d

vertical force F = eE = e

zero potential

accelerating potential difference

F=e _ horizontal acceleration

_

_

V d

vertical acceleration constant horizontal velocity

constant velocity

171

_ resultant velocity

V d

INDEX

Practical advice This diagram is reproduced here so that you can discuss it with your class.

External reference This activity is taken from Advancing Physics chapter 16, 100O

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TAP 518-6: The fine-beam tube Use a fine-beam tube to explore how the accelerating voltage and the magnetic field both affect the radius of the electrons’ path.

You will need: 9

fine-beam tube with Helmholtz coils

9

power supply (6.3 V) for cathode

9

ht. supply (0–300V dc) with voltmeter

9

low voltage supply (0–12 V) for Helmholtz coils

9

ammeter

9

connecting leads with fixed or sprung shrouds over the 4mm plugs

Safety Wire carefully, no bare conductor above 40 V. HT supplies are always dangerous, especially so in the dark. Shrouded plug leads MUST be used. No connections should be changed with the power switched on. We can observe the motion of electrons in a magnetic field using a fine-beam tube. There is a small amount of gas at low pressure in the tube. When the gas molecules are struck by electrons, the molecules are excited and emit visible light. In this way we can see the path of the electrons. A pair of coils can be used to apply a magnetic field perpendicular to the motion of the electrons. With no current in the coils, the electron beam is undeflected. •

Use a compass to determine the direction of the magnetic field.

Here are some things to try. Make a prediction before you try each one. •

Apply the left-hand rule to the electrons to predict the direction in which they will be deflected when the magnetic field is switched on.

•

What will happen if the speed of the electrons is increased (by increasing the gun voltage)?

•

What will happen if the magnetic field strength is increased (by increasing the current in the coils)?

•

What will happen if an electric field is applied along the direction of the magnetic field, or if the beam enters the magnetic field at an angle other than a right angle

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Practical advice If you measured e/m using a fine beam tube it would be unnecessary to repeat the fine beam tube here unless you require a quick reminder. TAP 413-2: Measuring the charge to mass ratio for an electron To avoid damage to the tube, it is desirable to set this experiment up in advance. At least partial blackout is required. It is also a good idea to cover the back of the tube in black cloth. You will probably get a spread of circles, perhaps due to the field being non-uniform (a Helmholtz pair only produce an approximately uniform field in the central third of the volume they enclose), due to the electrons provided by the electron gun having a range of velocities, and due to random thermal motion superimposed on the velocity acquired from the action of the accelerating pd. This activity is a more quantitative look at the fine-beam tube, and also provides an opportunity to revise ideas about charge, voltage and energy. Students should be able to say that increasing the accelerating voltage will increase the speed of the electrons, and so their orbital radius in a given magnetic field will increase. They should also be able to predict that if the current in the coils in increased, the field becomes stronger, which will steer the electrons into a tighter orbit. You might like to show that tilting the tube sends the electrons into a spiral (corkscrew-shaped) orbit; they continue to move with constant velocity along the field direction, while being deflected into circular motion around the field direction. In this context Helmholtz coils are a pair of coils radius R and distance R apart.

Safety Wire carefully, no bare conductor above 40 V. HT supplies are always dangerous, especially so in the dark. Shrouded plug leads MUST be used. No connections should be changed with the power switched on.

External reference This activity is taken from Salters Horners Advanced Physics, section PRO, activity 27

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TAP 518- 7: Fields in nature and in particle accelerators Instructions and information Write your answers in the spaces provided. The following data will be needed when answering these questions: •

electronic charge = – 1.6 × 10–19 C

•

mass of electron = 9.1 × 10–31 kg

•

ε0 = 8.9 × 10–12 F m–1

Questions Suppose that a thundercloud has a flat horizontal base of area 1 km2. The base of the cloud is 200 m above the Earth’s surface so that a uniform electric field is formed between the cloud and the Earth. A field of 106 V m–1 between the base of the cloud and the Earth is sufficient to cause a lightning flash. 1.

2.

Calculate the potential difference between the cloud and the ground at the moment the lightning flash begins.

The charge on the base of the cloud is given by the formula Q = ε0 A E where A is the area of the base of the cloud and E is the electric field strength between the base and the Earth. Calculate the charge on the base of the cloud.

3.

The cloud and the Earth can be thought of as a parallel plate capacitor that stores energy when charged. Assuming that the potential difference immediately after the flash is very small compared with the potential difference at the beginning of the flash, calculate a value for the energy released during the flash.

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Electrons accelerated through a potential difference of 200 V enter a uniform magnetic field of 0.001 T perpendicular to the direction of motion. 4.

Calculate the speed of the electrons when they enter the magnetic field.

5.

Use your answer to question 4 to calculate the radius of the orbit in the magnetic field.

In an electron tube, electrons were passed through a region containing a vertical electric field E and a horizontal magnetic field B. When the forces on the electron were balanced the electrons passed through the tube undeflected. horizontal magnetic field

undeflected path of electrons

vertical electric field

6.

Show that the electrons of charge e pass undeflected when they have a velocity v = E/B.

The separation of the deflector plates was 24 mm and no deflection was observed when the voltage across the plates was 3.2 kV and the magnetic field was 8.2 × 10–3 T. 7.

Calculate the velocity of the electrons.

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The voltage used to accelerate the electrons to this velocity was 750 V. 8.

Use your answer to question 7 to calculate the ratio e/m for electrons where m is the mass of an electron.

A proton joined to a neutron is known as a deuteron or deuterium ion and is used in nuclear scattering experiments. A deuteron has a mass of 3.3 × 10–27 kg and a charge of + 1.6 × 10–19 C. 9.

Calculate the voltage required to accelerate a deuteron from rest in a vacuum to a velocity of 9 × 106 m s–1 (3% of the speed of light).

In an early form of particle accelerator, deuterons were made to move in a circular path within a toroidal tube of diameter 1 m. A toroidal tube is like a hollow ring. 10.

Calculate the magnetic field required to constrain a deuteron within the tube at the velocity of 9 × 106 m s–1.

Hints 4.

The charge and mass of the electron are given in the ‘Instructions and information’.

6.

The electric and magnetic forces are equal in magnitude.

7.

You will need to change the units before calculating the electric field strength.

8.

You will need to consider the equation for the electron gun used to accelerate the electrons.

9.

You are used to calculating the speed of electrons when accelerated. The mass of a deuteron is not the same as the mass of an electron.

10.

The question gives the diameter of the orbit and not the radius.

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Practical advice In questions 7 and 8 there are several stages in the calculation so excessive rounding in the earlier parts of the question would lead to errors in the final answer. All answers have been rounded to two significant figures, but three significant figures have been used for numerical values that have been carried through to the next stage of the calculation.

Answers and worked solutions 1. V = Ed = 10 6 V m −1 × 200 m = 2 × 10 8 V.

2. Q = ε 0 AE = (8.9 × 10 −12 F m −1 ) × (1000 m) 2 × 10 6 V m −1 = 8.9 C.

3. 1 QV 2

=

1 2

× 8.9 C × ( 2 × 10 8 V )

= 8.9 × 10 8 J. 4. eV =

1 2

mv 2

so v =

2 eV m

2 × (1 . 6 × 10

=

9 . 1 × 10 6

= 8 . 39 × 10 ≈ 8 . 4 × 10

− 19

6

m s

C ) × 200 V

− 31

kg

−1

m s −1 .

5. Bev = mv 2 / r

so r =

mv Be

=

= 0 . 048

( 9 . 1 × 10

− 31

0 . 001

kg ) × ( 8 . 39 × 10 T × ( 1 . 6 × 10

m.

6. eE = Bev

so

178

− 19

6

m s

C)

−1

)

INDEX

v = E / B. 7. E =

3200 V V = d 0 . 024 m

= 1 . 33 × 10 v =

5

−1

V m

.

5

1 . 33 × 10 V m E = − B 8 . 2 × 10 3 T

= 1 . 63 × 10 ≈ 1 . 6 × 10

7

7

m s

m s

−1

−1

−1

.

8.

eV =

1 2

mv 2

so

(1 . 63 × 10 7 m s − 1 ) 2 e v2 = = m 2V 2 × 750 V = 1 . 8 × 10 11 C kg −1 . 9.

qV =

1 2

mv 2

1 2

mv

so V =

2

=

q

= 8 . 4 × 10

5

( 3 . 3 × 10

− 27

kg ) × ( 9 × 10

2 × (1 . 6 × 10

− 19

6

m s −1 ) 2

C)

V.

10. Bqv = mv 2 / r

so B =

mv qr

=

( 3 . 3 × 10

− 27

kg ) × ( 9 × 10

( 1 . 6 × 10

− 19

6

m s

C ) × 0 .5 m

= 0 . 37 T .

External reference This activity is taken from Advancing Physics chapter 16, 160S

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−1

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TAP 518- 8: The ultimate speed – Bertozzi's demonstration This is a presentation of some experimental data, gathered by Bertozzi that illuminates debates about motion when Einstein’s theory of relativity is considered.

The ultimate speed: Bertozzi’s demonstration

8.4 m drift space tube detects electrons passing accelerated electrons

bunch of electrons

aluminium plate detects electrons arriving. Rise in temperature checks energy

v

time

oscilloscope

The results: speed calculated 1 from 2 mv2 = qV

9

6

The difference made by relativity As particles are accelerated speed v reaches a limit, c kinetic energy EK increases without limit momentum p increases without limit At all speeds EK = qV

speed of light

3 actual speed of electrons

0 0

4 2 accelerating p.d./MV

At low speeds p ≈ mv EK ≈ 12mv2

At high speeds v≈c EK ≈ pc

Powerful accelerators can’t increase the speed of particles above c, but they go on increasing their energy and momentum

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Practical advice This diagram is here so that you can discuss it with your class.

External reference This activity is taken from Advancing Physics chapter 16, 80O

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TAP 518- 9: Principle of the synchrotron accelerator Here both electrical and magnetic fields are harnessed to the task of deflection and accelerating beams.

Principle of synchrotron accelerator

inject beam at v ≈ c from smaller accelerator

electrostatic deflector

magnets to bend beam

radio frequency cavity to accelerate beam

magnets to focus beam

electrostatic deflector to extract beam and direct it into targets

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Practical advice This diagram is here so that you can discuss it with your class.

External reference This activity is taken from Advancing Physics chapter 16, 150O

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Episode 519: Particle detectors Summary Discussion: The idea of particle detectors. (10 minutes) Demonstration: Cloud chamber (and spark detector). (15 minutes) Discussion: Explaining tracks. (10 minutes) Student questions: Interpreting tracks. (20 minutes) Particle tracks: Try analyzing some particle tracks. (time permitting) felt ring

Discussion: The idea of particle detectors What particle detectors do you know of? (Spark Counters, Geiger counters.) What do these tell us? (They count particles of ionizing radiation.)

air saturated with meths vapour

alpha source solid carbon dioxide

What else might we want to know about particles? How could we tell which particle we have detected? (Need to know a range of properties: e.g. mass/energy, electric charge, momentum, lifetime, etc.)

foam rubber (Diagram: resourcefulphysics.org)

In general, detectors work by analyzing particle collisions using conservation laws (momentum, energy, charge).

Demonstration: Cloud chamber (and spark detector). Cloud & bubble chambers make visible the invisible: alpha diameter ~ 10-14 m gives a visible track 0.1 mm wide, a factor of 1010 increase in size! If you have access to a spark detector, you could also demonstrate this at this point. You may have demonstrated a spark counter before? TAP 509-4: Rays make ions TAP 518-2: Electrical breakdown TAP 519-1: Range of alpha particles with a cloud chamber

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Discussion: Explaining tracks Students may be familiar with the patterns made by particles in detectors, and so you could discuss the basic ideas behind analysing the tracks. The length of track is related to the energy of the particle, and also to its lifetime if it decays with a very short half life. Magnetic fields deflect charged particles and so bend their tracks. The curvature depends on momentum, charge and the strength of the field. How could you tell whether a particle had positive or negative charge? (Curving to left or right; Fleming’s rule.) If a track is a spiral, what does this tell you about the particle’s motion? (It is slowing down; charged particles radiate as they are accelerated, so they slow down.) TAP 519-2: Measuring the momentum of moving charged particles.

Student questions: Interpreting tracks Students can apply their knowledge to the interpretation of tracks from a bubble chamber. TAP 413-6: Charged particles moving in a magnetic field

Particle tracks: If time permits try analyzing some particle tracks at: http://hepwww.ph.man.ac.uk/~wyatt/events/home.html TAP 519-3: Particle tracks

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TAP 519- 1: Range of alpha particles with a cloud chamber Using a cloud chamber The alpha particle is one type of emission that is possible from the nuclei of some atoms. This activity allows you to investigate how far these alphas can travel in a gas.

You will need 9

cloud chamber; expansion or diffusion type

9

pure alpha source for cloud chamber

9

optional: a beta source and forceps for holding it

Ethanol is highly flammable. Do not bring the stock bottle into the lab until all sources of ignition have been removed. Solid carbon dioxide (dry ice) can cause burns. Wear eye protection and leather gauntlet-style gloves while preparing and using it.

Radioactive sources Follow local rules for handling radioactive sources.

What to do 1.

Set up the cloud chamber. You will be given instructions on how to do this.

2.

If the chamber is a diffusion type (containing dry ice), allow it to settle down – this may take some minutes. Keep watching and you should begin to see tracks radiating from the alpha source in the chamber. If the chamber is an expansion type, operate the pump and look in the chamber immediately after the exhaust stroke. You should be able to see tracks radiating from the alpha source.

3.

The tracks are formed when the supersaturated moist air in the chamber condenses onto the ions that are produced by the alphas as they move through the gas. The tracks show where the alphas have been, not their current position.

4.

Try to make a sketch of these tracks. Pay particular attention to the track length and the shape of the track.

5.

You may occasionally see other tracks that cross the chamber and do not appear to come from the source. These may be cosmic rays or other particles that help to make up the background radiation.

6.

If you are very, very lucky you may see a forked track; this may have been a reaction in the chamber that resulted in the formation of two or more particles. This is the type of event that experimental nuclear physicists use in their work.

6b.

0bserve tracks from a beta source if possible.

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7.

Explain your observations in terms of (i) the initial energy of the alpha particles as they leave their parent nucleus and (ii) the mass of the alphas relative to the gas of atoms through which they move in the chamber.

You have seen that 1.

Alpha tracks are straight. This is because the alpha particles are not greatly deflected as they ionise the gas atoms in the chamber.

2.

Alpha tracks from the same isotope species are of the same length. This is because they all begin with the same initial energy and lose this energy at approximately the same rate.

3.

Beta tracks are not straight but ‘wander’.

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Practical advice There are various types of chamber on the market. This activity does not provide precise instructions for each, you will need to instruct students in the use of the chamber, or set it up for them. You need to be clear that the students are not inadvertently exposed to the radiation.

Alternative approaches The range of alphas can also be assessed using a spark detector.

External reference This activity is taken from Advancing Physics chapter 18, 80E

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TAP 519- 2: Measuring the momentum of moving charged particles Analysing the motion of a charged particle in a magnetic field can lead to a determination of its momentum.

Measuring the momentum of moving charged particles velocity v

×

×

×

×

× force F

×

circular path, radius r

×

×

magnetic field B into screen

×

×

×

×

×

×

×

×

charge q

velocity v

motion in circle

magnetic force

m v2 = force F r

force F = qvB

m v2 = qvB r p = mv = qrB

at relativistic speed p = qrB is still true but p > mv

Momentum of particle proportional to radius of curvature of path

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Practical advice This diagram is here so that you can discuss it with your class.

External reference This activity is taken from Advancing Physics chapter 16, 140O

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TAP 519- 3: Particle tracks Seeding cloud trails For many years, the best way to record the particles involved in a subatomic interaction was to take a holiday snap. To do that, the particles had to be persuaded to pose for a picture. The earliest device for picturing the tracks of particles was the cloud chamber. Schools often have small cloud chambers running on ‘dry ice’ (solid carbon dioxide), which show tracks of alpha particles easily. The idea is that a fast-moving charged particle leaves behind it a trail of charged ions from gas molecules it has collided with and disrupted. Typically the path contains hundreds of thousands of such ions. If the gas is kept wet with water or alcohol, droplets of liquid condense around the ions. So the particle track is revealed by a cloud trail. The wet gas is made ready to condense by cooling it with a sudden expansion.

The successor to the cloud chamber was the bubble chamber. Instead of a gas, there is a liquid though which the particles pass. Instead of droplets along the path there are tiny bubbles in the liquid. The bubbles form round the ions left behind by the charged particles, just as the droplets do in a cloud chamber. One liquid often used is liquid hydrogen, because its protons also make good targets for collisions. The liquid is made ready to boil by suddenly reducing the pressure on it. Then vapour droplets first form around the ions.

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Getting just the snap you want The trouble with holiday snaps is that often you don’t have your camera ready when something interesting happens. The trouble with cloud and bubble chambers is that they have to be ‘primed’ to be ready (by expanding the gas or reducing the pressure on the liquid). The event they happen to see may not be the one you want. A way round this is to detect the particles electronically as well, pick out an interesting event, and get the electronics to ‘fire’ the chamber to take its picture. It doesn’t matter that the event is already over: the ion trails it left behind in the gas or liquid are there for several milliseconds, so the picture can still be made after the event. An example would be an event which sends out two particles in opposite directions. Counters can pick them up and detect the coincidence in time. The idea was much used in early research on cosmic rays, because their arrival is so unpredictable. But of course all quantum events are inherently random, so the idea has very general use. In this way, the experimenters just leave the chambers to work automatically. In fact automation is increasingly taking over the business of particle detection.

Using magnetic fields From the beginning physicists have used magnetic fields to tell particles apart. Positively charged particles curve one way, and negatively charged particles curve the opposite way. This is how beta particles were shown to be negatively charged. It is also how Anderson showed that he had seen the first antiparticle, the positron.

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Measuring momentum Suppose a particle makes a curved track moving at right angles to a magnetic field B. Just measuring the radius r of the track tells you the momentum p of the particle, if you know its charge q. The relation is simply p = qrB. This works for particles moving at any speed, including close to the speed of light. If you also know the energy of the particle, you can find its mass, which helps you to tell what it is.

Measuring energy There are many ways of measuring the energy of particles. One of the most fundamental is simply to let them be absorbed in some material, and see how much hotter the material gets. The large detectors used at CERN and other accelerator laboratories may have several layers of absorber. One layer, for example made of lead glass, stops all the electrons, positrons and photons, recording their total energy. Another layer further out made of iron stops more massive particles and records their energy too. But some particles, such as neutrinos, aren’t absorbed at all. Their energy has to be found by looking at the difference between the energy that went into the collision and the energy that came out carried by other particles. So even these ‘ghostly’ particles get accounted for.

Some things to find out about 1.

What is the connection between bubble chambers and bottles of beer? Who invented the bubble chamber?

2.

What have coincidences to do with the Italian physicist Bruno Rossi? Why was he in England and not in Italy at the time?

3.

What is the advantage of using superconducting coils to produce the large magnetic fields needed in big particle detectors?

4.

What are DELPHI, ALEPH and OPAL? (Look under CERN.)

5.

Photographic film is also used to detect ionising radiation. Find out about one such use.

6.

Another kind of detector is named after the Russian physicist Cerenkov. What is Cerenkov radiation and how does a Cerenkov detector work?

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Annihilation and pair production: bubble chamber pictures Below:

Pair production and a processed image with some tracks removed

Both images can be interpreted by assuming two photons to be entering from the top of picture, leaving no track. One (top) has created a positron / electron pair and a ‘knock on’ electron from within an atom. The other has simply produced a positron / electron pair.

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Practical advice This reading, is provided for teachers or interested students. The reading provides a good opportunity to revise topics such as ionisation, forces on charges in electric and magnetic fields, energy and momentum, together with thought about qualities of measuring instruments, particularly resolution and response time. By adding questions, you could convert this reading into a comprehension exercise. The reading extends beyond exam specifications See also Episode 533: The particle zoo

Alternative approaches You could select types of detector and assign students to find out about them.

Social and human context The sheer scale of some detectors, and the hundreds of people involved in the teams running them, underline effectively the interdisciplinary nature and the cost of high-energy research facilities.

External reference This activity is taken from Advancing Physics chapter 17, 80T

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Episode 520: Preparation for Rutherford scattering topic The nuclear atom Students will be familiar with the nuclear model of the atom, in which the atom is pictured as a miniature solar system. They may have been introduced to Rutherford scattering and how this leads to the nuclear model. Here you have the opportunity to deepen their understanding, making use of ideas about electric fields. You can also introduce them to other evidence that tells us about the size of the nucleus.

Episode 521: Rutherford’s experiment Episode 522: The size of the nucleus

Main aims Students will: 1. Describe Rutherford’s experiment and explain why it leads to the nuclear model of the atom. 2. Use Coulomb’s law to estimate the size of the nucleus. Electron cloud

3. State the approximate sizes of atom and nucleus.

Prior knowledge There is a lot of Physics knowledge that can contribute to this topic: collisions and momentum, Coulomb’s law, and wave-particle duality.

Alpha particle

nucleus

If you have not covered all of these topics already, you will have to modify the suggested approach to take account of this.

Where this leads Once the idea of the nuclear atom is established, you can go on to look at nuclear structure, particle accelerators, the Standard Model and the whole of particle physics. This topic also provides a good opportunity to discuss the use of models in physics, including both mechanical and mathematical models.

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Episode 521: Rutherford’s experiment In this episode, students look in detail at Rutherford’s experiment and relate it to a mechanical analogue.

Summary Discussion: Recollecting the significance of Rutherford’s experiment. (10 minutes) Discussion: Rutherford’s experiment. (20 minutes) Demonstration: Collisions and momentum. (10 minutes) Discussion: Rutherford scattering and Coulomb’s law. (10 minutes) Student experiment or demonstration: The Chinese hat analogue. (30 minutes) Student questions: Rutherford experiment and atomic structure (optional) (20 Minutes) Question: Rutherford’s results (optional) Discussion: Models in physics. (10 minutes)

Discussion: Recollecting the significance of Rutherford’s experiment As a preparatory task, ask your students to revise what they have previously learned about Rutherford’s α-scattering experiment. What idea of the atom did it suggest? (The nuclear model.) What model of the atom did this replace? (Thomson’s ‘plum pudding’ model, in which atoms are seen as essentially small balls composed of a mixture of positive and negative electric charge, with no concentration of charge at any particular position.) Is ‘plum pudding’ a good name for the model? (Yes, if you see the negative electrons dispersed throughout a spherical lump of continuous positive charge, not so good if the volume of the atom has both positive and negative particles continuously distributed through it – students may well be recalling different pictures from different sources.)

Discussion: Rutherford’s experiment Now you can present a more advanced exposition of the experiment. Why did Rutherford ask for the experiment to be done? Experiments on the absorption of β particles had also shown that sometimes the β particles were ‘back scattered’. Rutherford suggested that Geiger and Marsden should try looking for similar behaviour with α particles. Rutherford thought it was highly unlikely; because α particles are relatively massive compared with electrons, it was predicted that the αs would simply suffer a series of small deflections. They were expected to travel more or less straight through the absorber. However, Rutherford’s main concern was to give Geiger & Marsden something to do that would occupy them and get them some useful hands-on experience, rather than expecting them to get any very exciting results. Show a diagram of the apparatus. The absorber was a thin gold metal foil. Why use gold? (Thin gold foils, typically 250 atoms thick, were easy to make and readily available.) Why thin? (αs are easily absorbed.)

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As expected, virtually all the αs went straight through, but about 1:8000 were turned through large angles (reflected or back-scattered). An 8 kBq α source gives one large-angle scattering per second. The chance of a series of small deflections resulting in a reflection is far too small to account for what was observed. Rutherford was astonished at the result: “It was quite the most incredible event that ever happened to me in my life. It was as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you!” (You may find other versions of this quote, because Rutherford described his experience on many different occasions.) TAP 521-1: Rutherford scattering TAP 521-2: Alpha particle scattering experiment

Demonstration: Collisions and momentum Use colliding balls to show what happens to a projectile particle hitting a target particle; as the target ball mass gets bigger, the follow through by the projectile gets less. Use a selection of ball bearings, marbles etc on some curtain track, or trolleys loaded with different weights. When the target mass is small relative to the projectile mass, the missile follows through. For equal masses the projectile stops and the target sets off with the speed of the projectile. If the target mass is large, the projectile rebounds. (If your students have already studied momentum, they should be able to predict the outcome of each of these demonstrations.) The back scattering of αs through large angles implies (i) all the positive charge is concentrated together, and (ii) the mass of the concentrated positive charge must be quite a bit larger than of an α particle.

Discussion: Rutherford scattering and Coulomb’s Law Rutherford assumed that (i) Coulomb’s Law was obeyed down to very small distances, and that (ii) most of the mass of the nucleus was concentrated into a very small volume – the nuclear atom that resembles a miniature solar system. (Because the analysis works, we can take this as ‘proof’’ that Coulomb’s Law is valid down to distances about the size of a nucleus.) Show a diagram to explain how the terms impact parameter p and scattering angle φ are defined. Ask: how would you expect the number of αs scattered through angle φ to depend upon (i) the impact parameter p, (ii) the charge on the target nucleus Z, and (iii) the kinetic energy of the α particles? (As p increases φ decreases (force weaker); as Z increases φ increases (greater repulsive force); as energy increases, φ decreases (less ‘interaction’ time).)

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TAP 521-3: Rutherford’s picture of alpha particle scattering.

Student experiment or demonstration: The Chinese hat analogue The ‘Chinese hat’ analogue provides a practical way for students to get a feel for the physics of alpha-scattering. Roll a marble past the hat to see it deflected. You can change the speed and impact parameter. Students can change these parameters systematically and observe the effects. The hat is designed so that, as the slope of the ‘hat’ gets steeper, the component of gravity parallel to the slope opposing the motion of the ball also gets larger. The actual shape is such that at any position on the slope a distance r from the centre of the ‘hill’, the component of a particle’s weight parallel to the slope ~ 1/r2. [Revision of the relationship between 1/r potential and 1/r2 force can be done here if desired] If possible, it’s worth getting several Chinese hats so that (Image: Philip Harris Education – students can work with them in small groups. (The ‘hat’ can resourcefulphysics.org) also be used when studying gravity. Turn it upside down to become a ‘potential well’ so that you can demonstrate orbits, and discuss the difference between bound and unbound ‘particles’.) TAP 521-4: The 1/r hill: Slope and force TAP 521-5: A model for Rutherford scattering A good simulation of alpha particle scattering could be used if desired see this website: http://www-outreach.phy.cam.ac.uk/camphy/nucleus/nucleus_index.htm

Student questions (optional) Rutherford experiment and atomic structure TAP 521-6: Rutherford experiment and atomic structure

Question: Rutherford’s results (optional) Some actual results are given. Students may plot a graph to test Rutherford’s relation for α-scattering TAP 521-7: Rutherford scattering data

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Discussion: Models in physics If time permits, you might have a discussion on the role of models (physical and mathematical) in physics. In what ways is the Chinese hat model similar to Rutherford scattering? In what ways does it differ? In which ways is the solar system a good model for the nuclear atom? What other models do your students know?

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TAP 521-1: Rutherford scattering

gold foil Alpha source

some

Many particles very few

Most α particles travel straight through the gold foil but about 1:8000 were turned through a large angle. The experiment takes place in a vacuum to avoid problems of α absorption by air.

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Practical advice This diagram is here so that you can discuss it with your class.

External reference This activity is taken from Resourceful Physics

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TAP 521-2: Alpha particle scattering experiment Rutherford alpha particle scattering experiment

lead block to select narrow beam of alpha particles

radium source of alpha particles

alpha particle beam

thin gold foil

microscope to view zinc sulphide screen, and count alpha particles

vary angle of scattering observed

scattered alpha particles

zinc sulphide screen, tiny dots of light where struck by alpha particle

The experiment takes place in a vacuum to avoid problems of α absorption by air

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Practical advice This diagram is here so that you can discuss it with your class.

External reference This activity is taken from Advancing Physics chapter 17, 100O

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TAP 521-3: Rutherford’s picture of alpha particle scattering Rutherford’s picture of alpha scattering

nucleus

206

paths of scattered alpha particles

INDEX

Rutherford’s picture of alpha scattering For calculations

force F =

alpha particle scattered

2Ze2 4πε0d

2

charge +2e θ ‘aiming error’ b

scattering angle

d

Assumptions: alpha particle is the He nucleus, charge +2e gold nucleus has charge + Ze, and is much more massive than alpha particles scattering force is inverse square electrical repulsion

gold nucleus charge + Ze equal force F but nucleus is massive, so little recoil

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Rutherford’s picture of alpha scattering TEST: Are slowed-down ‘alpha’ particles scattered more?

Z

reduce alpha energy with absorber

less energetic alpha particle turned around further from the nucleus

lower speed Z

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Rutherford’s picture of alpha scattering

TEST:

Does using nuclei of smaller charge scatter alpha particles less?

Z replace foil by metal of smaller atomic number

alpha particle gets closer to nucleus of less charge and is deflected less

smaller nucleus with less charge, e.g. aluminium

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Practical advice These diagrams are here so that you can discuss them with your class.

External reference This activity is taken from Advancing Physics chapter 17, 120O

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TAP 521- 4: The 1/r hill: Slope and force Mapping potential The 1/r hill is a gravitational model showing how the electrical potential varies round a charged particle. One way up for the variation round a positively charged sphere, invert it for a model of the variation round a negatively charged sphere. As the elevation of the hill above the bench top represents the potential, so the steepness of the hill represents the field.

Remember E=−

dV . dx

You can investigate these field values by looking at the accelerations of a suitably chosen probe - a ball bearing.

You will need: 9

1/r hill

9

ball bearing

Potential varies radially

r

Move across the surface, staying at a fixed gravitational potential energy. What does this correspond to in the electrical case? What shape do you make, as you move across the surface following this rule? Now try moving across the surface so that the potential energy changes as much as possible in as small a distance as possible. What shape do you make now, as you move across the surface?

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Field is potential gradient

r

E=

–dV dr

dV dr

Hold a ball bearing on the surface. Release it and compare the potential gradient with the acceleration. Repeat for several different positions. How does the steepness of the slope fix the acceleration? Why is the acceleration a good measure of the field? Remember that this is just a gravitational model. Can you sketch the corresponding situation for the electrical case?

You have 1. Looked at 1/r variation in potential. 2. Compared a model with the thing being modelled. 3. Thought about the connection between field and potential.

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Practical advice Handling a 1/r hill in the way suggested allows students to show their understanding of some of the more subtle ideas in this chapter physically. Students might usefully have one hill between four and use the questions as a basis to present a mini-lesson on the 1/r hill to their peers, each section being repeated twice.

External reference This activity is taken from Advancing Physics chapter 16, 210P

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TAP 521-5: A model for Rutherford scattering Use a model such as the one shown below to investigate Rutherford scattering. A marble runs down a ramp, and then past a plastic or metal ‘hill’. The closer it gets to the hill, the more it is deflected.

You will need: 9

alpha scattering analogue apparatus

9

talcum powder

9

large sheets of white paper (A3 or larger)

What to do: •

Sprinkle some powder so that the track of the marble shows up.

•

Start by investigating how much the marble is deflected when it is aimed at different distances from the hill. You need to measure two quantities, as shown below, d, the distance of the line of the original track from the centre of the hill and φ the angle through which it is scattered.

•

Make sure that you always release the marble from the same point on the ramp.

•

How could you make the marble travel more slowly? How will this affect its track? Test your ideas.

•

Why does the marble never go over the top of the hill? Use the idea of energy conservation to explain your answer

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Practical advice The ‘hill’ is shaped to reflect an inverse square law of repulsion (its height varies inversely with radius, as required for potential energy in an inverse square law field). You might refer back to this and remind students about studying Coulomb’s law. Talcum powder is useful for showing up the tracks of the marbles. Students should appreciate that, provided the marble never starts higher than the top of the ramp, it will never reach the top of the hill – it will have insufficient energy. If possible, it’s worth getting several Chinese hat demos so students can work with them in small groups of three or four. [The ‘hats’ can also be used when studying gravity. Turned upside down to become a ‘potential well’ you can demonstrate ‘orbits’ and bound or unbound ‘particles’.]

Technicians note The apparatus consists of a plastic or aluminium ‘hill’, a ramp and a steel ball bearing or glass marble (see, for example, Philip Harris item Q88290/4).

External reference This activity is taken from Salters Horners Advanced Physics, section PRO, activity 11

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TAP 521- 6: Rutherford experiment and atomic structure 1.

Describe briefly the two conflicting theories of the structure of the atom.

2.

Why was the nuclear model of Rutherford accepted as correct?

3.

What would have happened if neutrons had been used in Rutherford’s experiment? Explain your answer.

4.

What would have happened if aluminium had been used instead of gold in the alpha scattering experiment? Explain your answer.

5.

What three properties of the nucleus can be deduced from the Rutherford scattering experiment? Explain your answer.

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Practical advice These questions are to help your students to think about the Rutherford ideas.

Answers and worked solutions 1

The English scientist Thomson suggested that the atom, which is a neutral particle, was made of positive charge with ‘lumps’ of negative charge inset in it - rather like the plums in a pudding. For this reason it was known as the Plum Pudding theory of the atom. Rutherford explained it this way. He knew that the alpha particles carried a positive charge so he said that the positive charge of the atom was concentrated in one place that he called the nucleus, and that the negatively charged particles, the electrons, were in orbit around the nucleus. Most of the mass was in the nucleus

2

Rutherford’s prediction using the idea of Coulomb law repulsion was verified by experiment. It also enables experimental values of nuclear charge to be obtained, ie atomic number.

3

They would not have been repelled so it is unlikely that any would ‘bounce back’. Some could be absorbed by the nucleus.

4

The charge on the nucleus is much smaller so deflection would be smaller.

See the equation TAP 521-7: Rutherford scattering data 5

Small, massive and positive.

External reference This activity is taken from Resourceful Physics

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TAP 521-7: Rutherford scattering data The idea of scattering using Coulomb’s law and a small central positive charge for the atom was communicated to the Manchester Literary and Philosophical Society in February 1911. His ideas require that “the scattering due to a single atomic encounter is small” and that “it be supposed that the diameter of the sphere of positive electricity is minute compared with the diameter and sphere of influence of the atom”. The table below shows some of Geiger and Marsden’s results Counting was carried out for the same time at each angle deflected angle φ /

number scattered

degrees 15.0

132,000

22.5

27,300

30.0

7,800

37.5

3,300

45.0

1,457

60.0

477

75.0

211

105

70

120

52

135

43

150

33

The actual formula Number of α particles y falling on unit area deflected by angle φ is given by: -

y=

ntb 2 Q cos ec 4 (φ / 2) 16r 2

,

where Q is the total number of particles falling on the scattering material, t is the thickness of the material, n the number of atoms within unit volume of the material, and b given by the formula below. N is the number of positive charges, e the size of the positive charge, m the mass of an α particle, u their velocity and E the charge of the α particle.

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b=

2 NeE mu 2

One of Rutherford’s conclusions was that the number of scintillations per unit area of zinc sulphide screen is proportional to cos ec 4 (φ / 2)

Maths note cosec (φ/2) = 1/sin (φ/2)

What to do Add extra columns to the table as needed to enable you to draw a graph to test Rutherford’s conclusion that the number of scintillations per unit area of zinc sulphide screen is proportional to cos ec 4 (φ / 2)

As an extension you might like to plot number scattered against 1/φ4 Write down you conclusions from the graph(s)

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Practical advice Some students might like to see Rutherford’s equation and try a test to see how the results come out. This activity is considered optional. Some websites with papers of the time are given below for interest

Alternative approaches A spreadsheet could be used for this activity.

You should find: •

To a reasonable degree y proportional to cos ec 4 (φ / 2)

•

number scattered against 1/φ4 is only proportional at small angles.

External references This activity is based on “The Scattering of α and β Particles By Matter and the Structure of the Atom. By Professor E RUTHERFORD F.R.S., University of Manchester.“ from which the equation is quoted and the section in quotation marks at the top of the page. An abstract of the paper is at: http://dbhs.wvusd.k12.ca.us/webdocs/Chem-History/Rutherford-atom-abstract.html see also: Philosophical Magazine, Series 6, Volume 27 March 1914, p. 488 - 498 http://dbhs.wvusd.k12.ca.us/webdocs/Chem-History/Rutherford-1914.html The paper can also be found in Foundations of Nuclear Physics, Beyer, Robert T (Ed), New York 1949 Dover Publications Inc. pp 111-130. The book also contains papers by Chadwick, Lawrence, Cockcroft, Gamow and Yukawa amongst others. Of interest might also be: On a Diffuse Reflection of the α-Particles, Proc. Roy. Soc. 1909 A vol. 82, p. 495-500 By H. GEIGER, Ph.D., John Harling Fellow, and E. MARSDEN, Hatfield Scholar, University of Manchester http://dbhs.wvusd.k12.ca.us/webdocs/Chem-History/GM-1909.html and The Scattering of the α-Particles by Matter by H. GEIGER, Ph.D. Proceedings of the Royal Society vol. A83, p. 492-504 http://dbhs.wvusd.k12.ca.us/webdocs/Chem-History/Geiger-1910.html

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Episode 522: The size of the nucleus Having established the existence of the nucleus, you can now consider experimental evidence for its size, starting from the Rutherford experiment.

Summary Discussion + worked example: Size of nucleus (15 minutes) Discussion: Atomic and nucleus size (10 minutes) Student questions: Forces and closest approach (30 minutes) Discussion: Atomic number and the charge on a nucleus (5 minutes) Discussion: Upper limit of nuclear size (30 Minutes) Discussion: A puzzle for a future lesson (5 minutes)

Discussion + worked example: Size of the nucleus You can get an idea of the possible size of the nucleus by thinking about Rutherford’s experiment. Ask: What impact parameter will result in the α particle getting closest to the nucleus? (A ‘head-on’ collision with p = 0.) The principle of the conservation of energy is used to calculate the distance of closest approach as a measure for the size of a nucleus. Understanding the calculation that follows depends upon whether the students have covered electric potential and fields. Alternatively it serves as good revision. TAP 522-1: Alpha particle scattering – distance of closest approach TAP 522-2: Distance of closest approach.

Alpha particle (+2e) d

221

Gold nucleus (+79e)

INDEX

When the α is brought momentarily to rest (“having climbed as far as it can up the electrostatic hill”) the work done in bringing it to rest will just equal its initial kinetic energy. When the speed and hence the kinetic energy is zero, all the energy is now electrostatic potential energy. If the α momentarily stops when at a distance d from the (centre of) the nucleus of charge Ze, its electrical potential energy is Eα =

1 Ze 2e 4πε 0 d

This equals the initial kinetic energy of the α particle. Rutherford used an α source given to him by Madame Curie. The α energy was ~ 7.7MeV. For gold, Z = 79. Solving gives d ~ 3 ×10-14 m. Compare this with the diameter of gold atoms ~ 3 ×10-10 m. So a nucleus is at least 10 000 times smaller than an atom. It is important to emphasise that this calculation gives an upper limit on the size of the gold nucleus; we cannot say that the alpha particle touches the nucleus; a more energetic α might get closer still. An atom is mostly ‘empty’ (which is why most αs went straight through – any electrons would hardly impede the relatively massive’ high speed α).

Discussion: Atomic and nucleus size Ask your students to suggest a scale model of the nuclear atom. For example: if a nucleus was 1 mm diameter, an atom would be 10,000 times larger or 10 m in diameter. Choose a suitable position for a 1 mm nucleus (a small ball bearing or ball of Blu-tac). Pace out 5 m (five large steps) to the edge of the atom where the electrons are. NB: textbook diagrams of an atom with a nucleus are not drawn to scale. Reinforce an accurate picture by getting a student to stand up as a ‘nucleus’, estimate their ‘girth’ (40 cm?) and ask where another student would have stand to be at the edge of the ‘atom’. (104 times 40 cm = 4000 m, so the radius of this “atom” is 2 km! Check with a local map to find a named location that students will recognize that is 2 km away. Further reinforcement: in a solid where atoms are close packed, the distance between adjacent nuclei ~ the size of an atom, i.e. equivalent to two students standing 4 km apart! So it’s quite amazing that any αs would ‘hit’ a nucleus at all! Both are a similar size. Cross sectional area presented by a nucleus ~ radius2 = ~ 10-28 m2. Ask: How would you expect the number of reflected αs to depend upon the thickness of the metal foil containing the target nuclei? (Imagine the gold atoms in layers, chance of a deflection increases with thickness, but absorption on the way in or back out of the increasingly thick foil will eventually prevent any further increase in the number reflected and detected. It is of great help if your students can recall the following orders of magnitude:

222

(Diagram: resourcefulphysics.org)

INDEX

Radius of atomic nucleus ~ 10-14 m Radius of atom ~ 10-10 m

Student questions: Forces and closest approach TAP 522-3: Rutherford scattering: Directions of forces TAP 522-4: Rutherford scattering: Energy and closest approach

Discussion Atomic number and the charge on a nucleus Rutherford used his data to find the charge of the gold target nucleus. Further experiments to find the charge of Cu, Ag and Pt foils gave: atomic number

α scattering experiment

Cu

29

29.3 × e

Ag

47

46.3 × e

Pt

78

77.4 × e

So the electric charge on a nucleus is given by the atomic number × e, i.e. Ze. With one exception (hydrogen, H-1), Z is always less than the atomic mass number. So what accounts for the difference? The atom must be electrically neutral. Rutherford proposed the neutron.

Discussion Upper limit of nuclear size Recall that Rutherford’s analysis gives an upper limit on the size on the nucleus (d ~ 1/ (α particle energy). The size you measure depends upon the energy of the α particle you use. So we need another approach to find the size of a gold nucleus. Can you think of a better particle to probe the size of a nucleus? (The neutron – being uncharged it will get closer.) Another technique is the deep inelastic scattering of electrons. Refer back if you have already covered the wave nature of particles (‘de Broglie waves’ λ = h/p), or this topic can be inserted here if desired. The electron diffraction apparatus has a basic similarity with α particle scattering. The electrons are fired at a thin film – in this case of graphite.

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Rutherford was fortunate that the de Broglie wavelength of the α particles (unknown to him) was quite small, and the coulomb repulsion stops αs getting too close – otherwise diffraction effects would have ‘confused’ the data! (Try the calculation if you have already covered λ = h/p.) TAP 506-2: Interpreting electron diffraction patterns TAP 522-5: Deep inelastic scattering TAP 522-6: Electrons measure the size of nuclei

Discussion: A puzzle for a future lesson There is a fundamental problem with Rutherford’s model. Ask your class: How can an atom with a central nucleus can be stable – why doesn’t it collapse? According to classical electrodynamics, the electrons should radiate energy as they orbit, and spiral inwards. (It’s good to leave a class with a puzzle for a future lesson.)

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TAP 522-1: Alpha particle scattering – distance of closest approach

Alpha particle (+2e) Gold nucleus (+79e) d

If an alpha particle with a kinetic energy Eα is fired directly towards a gold nucleus it will feel a repulsion that increases as it gets closer - climbing the potential hill surrounding the nucleus. When all the kinetic energy has been converted to potential energy the alpha particle (charge +2e) has reached its distance of closest approach (d) and comes to rest at that point

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Practical advice The diagram could be used as an OHT and discussed in class

External Reference This activity is taken from Resourceful Physics

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TAP 522-2: Distance of closest approach Distance of closest approach

initial kinetic energy = electrical potential energy

alpha particle with 5 MeV initial kinetic energy

1 variation of potential r

5 MeV distance r

charge + Ze (Z = 79 for gold) d

alpha particle scattered through 180°

alpha particle stops where potential hill is 5 MeV high

Where does the alpha particle stop? Electrical potential energy

Initial kinetic energy = 5 MeV = 5 × 106 eV × 1.6 × 10–19 J eV–1 = 8.0 × 10–13 J

V=

+ 2 Ze2 4πε0d

Z = 79, e = 1.6 × 10–19 C, ε = 8.9 × 10–12 C2 N–1 m–2 0

Alpha particle stops where initial kinetic energy = electrical potential energy + 2 Ze2 8.0 × 10–13 J = 4πε 0 d substitute values of Z, e, ε0: d = 4.5 × 10–14 m

Radius of gold nucleus must be less than of the order of 10–14 m Atoms are 10 000 times larger than their nuclei

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Practical advice The diagram could be used as an OHT and discussed in class

External reference This activity is taken from Advancing Physics chapter 17, 130O

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TAP 522- 3: Rutherford scattering: directions of forces Scattering of alpha particles Rutherford did not have a particle accelerator. Instead he used alpha particles, typically of energy 5 MeV, from radioactive decay. These questions are about the force of the nucleus on the alpha particle. An alpha particle has charge + 2 e, where e is the elementary unit of charge. A nucleus has charge + Ze, where Z is the number of protons in the nucleus (and the number of electrons in the atom).

Directions of forces Path of alpha particle scattered by nucleus C

B

nucleus

A

The diagram shows an alpha particle approaching a massive nucleus from A. Assume that the nucleus recoils negligibly as the alpha particle is scattered. 1.

Add to the diagram an arrow showing the direction of the force on the alpha particle when it is at point A, approaching the nucleus. Label the arrow FA.

2.

Add to the diagram an arrow showing the direction of the force on the alpha particle when it is at point B, at its closest to the nucleus. Label the arrow FB.

3.

What is the ratio of the magnitudes of the two forces, FA / FB given the distances shown in the diagram?

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INDEX

4.

Add to the diagram an arrow showing the direction of the force on the nucleus when the alpha particle is at point B. Label the arrow FN. How does this force compare with the force FA on the alpha particle at the same instant?

5.

At which point, A, B or C, is the alpha particle travelling slowest?

6.

At which point, A, B or C, is the alpha particle travelling fastest?

7.

The nucleus does in fact recoil a little. Add an arrow labelled ‘recoil’ to show the direction of recoil you expect as a result of the passage of an alpha particle along the whole path shown.

Uphill–downhill The electrical potential gradient around the nucleus can be thought of as like the slope of a hill. Imagine that you are riding on the alpha particle as it goes by the nucleus. Are you riding uphill, downhill or momentarily along a contour of the hill: 8.

At A?

9.

At B?

10.

At C?

11.

Is the electric potential at B larger or smaller than the electric potential at A? By what factor?

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Practical advice These are intended as simple ‘start-up’ questions for the discussion of alpha scattering. They are mainly qualitative, asking students to think about the direction of forces, and the nature of changes of kinetic and potential energy. The ‘hill’ metaphor for potential is exploited.

Alternative approaches Alpha particle orbits generated by a computer simulation could be studied is: A Useful web site that includes a simulation of alpha particle scattering http://www-outreach.phy.cam.ac.uk/camphy/nucleus/nucleus_index.htm The following sections are relevant to this topic Plum pudding atoms, α scattering, Geiger & Marsden, Shells off tissue paper, Nucleus

Social and human context The mathematical tools developed by the French mathematicians in the 1700s to deal with planetary orbits in the solar system were just as useful for alpha particle orbits under a repulsive force.

Answers and worked solutions 1.

The repulsive force is along the line joining the alpha particle and the nucleus.

Path of alpha particle scattered by nucleus C

B

FA nucleus

A

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2.

The repulsive force is along the line joining the alpha particle and the nucleus.

Path of alpha particle scattered by nucleus C

FB

B

FA nucleus

A

3.

The force at B is four times as large as the force at A, because the distance is halved and the force varies as 1/r2.

4.

The force on the nucleus is equal and opposite to the force on the alpha particle. But because the nucleus is much more massive, it recoils only slightly.

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Path of alpha particle scattered by nucleus C

FB

B

FA nucleus

A

FN

5.

Particle moves slowest at B, because this is the distance of closest approach, the particle has been decelerating due to repulsive force. After B it accelerates away.

6.

Here the particle is furthest from the nucleus, the alpha particle has been accelerated between B and C and is therefore going fastest at C.

7.

The alpha particle path is symmetrical about the line from the nucleus to B. So the net change of momentum of the nucleus is along this direction.

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Path of alpha particle scattered by nucleus C

FB

B

FA nucleus

A

recoil

8.

Uphill, because the particle is approaching the nucleus but being pushed away from it.

9.

Along contour, because the particle is travelling at right angles to the direction of the force on it.

10.

Downhill, because the particle is travelling away from the nucleus and is being pushed away from it.

11.

Potential at B larger than potential at A by a factor of 2, because the distance is halved and the potential varies as 1/r.

External reference This activity is taken from Advancing Physics chapter 17, 80S

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TAP 522- 4: Rutherford scattering: Energy and closest approach Scattering of alpha particles Rutherford did not have a particle accelerator. Instead he used alpha particles, typically of energy 5 MeV, from radioactive decay. These questions are about how close an alpha particle can get to different nuclei. An alpha particle has charge 2e, where e = 1.60 × 10–19 C. A nucleus has charge Ze, where Z is the number of protons in the nucleus (and the number of electrons in the atom). The electrical potential energy of the two charges at a distance r is: electrical potential energy =

2e × Ze 4πε 0 r

where ε0 = 8.85 × 10–12 C2 J–1 m–1. The electrical potential energy in electron volts is obtained by dividing by 1.60 × 10–19 J eV–1

Calculating the potential energy 1.

Show that the units of energy from the expression electrical potential energy =

2e × Ze 4πε 0 r

are joules.

2.

Show that the energy in MeV is given by

2Ze × 10 − 6. 4πε 0r

Alpha scattering by gold This graph shows the energy in MeV of an alpha particle at distances r from a gold nucleus, Z = 79.

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INDEX

Approach of alpha particle to nucleus Z = 79 (gold)

25 20 15 10 5 0 0

2

4

6

8

10

distance from nucleus / 10–14 m

3.

Make an arithmetical check to show that at distance r = 1.0 × 10–14 m the electrical potential energy is between 20 MeV and 25 MeV, as shown by the graph.

4.

How does the electrical potential energy change if the distance r is doubled?

5.

From the graph, at what distance r will an alpha particle with initial kinetic energy 5 MeV colliding head-on with the nucleus, come to rest momentarily?

6.

From the graph, at what distance r will a 5 MeV alpha particle have lost half its initial kinetic energy?

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7.

From the graph, what energy would an alpha particle need to approach as close as 2.0 × 10–14 m in a head-on collision?

Alpha scattering by tin The next graph shows, on the same scale as before, the potential energy of an alpha particle near a nucleus of tin, Z = 50.

Approach of alpha particle to nucleus Z = 50 (tin)

25 20 15 10 5 0 0

2

4

6

8

10

distance from nucleus / 10–14 m

8.

Why are the values of the potential energy smaller at the same values of r in this second graph?

9.

At r = 5.0 × 10–14 m the electrical potential energies of an alpha particle are 4.55 MeV for gold, Z = 79 and 2.88 MeV for tin, Z = 50. Explain the ratio, 1.58, of the two energies.

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INDEX

10.

Approximately how close can a 5 MeV alpha particle get to a tin nucleus, in a head-on collision?

Alpha scattering by aluminium The next graph shows the potential energy of an alpha particle close to an aluminium nucleus, Z = 13.

Approach of alpha particle to nucleus Z = 13 (aluminium)

25 20 15 10 5 0 0

2

4

6

distance from nucleus /

11.

8 10–14

10 m

From the graph, how close could a 5 MeV alpha particle get to a nucleus of charge? Z = 13, in a head-on collision?

12.

The radius of an aluminium nucleus is approximately 3 × 10–15 m. Does a 5 MeV alpha particle get close to the nucleus, compared with the dimensions of the nucleus itself? Could the pattern of scattering be affected?

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Heavy ion colliders Recently, to investigate very high-energy collisions, accelerators have been used to make head-on collisions between lead nuclei travelling in opposite directions. 13.

How much kinetic energy is needed to get two lead nuclei, Z = 82, within 1.0 × 10–14 m of one another? (Assume that electrical forces are the only forces operating.)

Hints 1.

Treat units like algebraic quantities in the expression for potential energy.

2.

Remember that the charge e coulomb is also the conversion joule per electron volt.

3.

Substitute values in the equation for potential energy.

4.

Remember 1/r.

5.

Read approximately from the graph.

6.

Read approximately from the graph.

7.

Read approximately from the graph.

8.

Look at the equation for electrical potential energy.

9.

Try the ratio 79/50.

10.

Read approximately from the graph.

11.

Read approximately from the graph.

12.

Remember that 10–15 is 1/10 of 10–14.

13.

Substitute in the expression for electrical potential energy. Or start with the answer to question 3.

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INDEX

Practical advice The questions focus on the distance of closest approach of an alpha particle to a nucleus. The approach is through the shape of the 1/r curve of electric potential energy, and the way the curve varies with radius and charge.

Alternative approaches Students could explore the electric potential energy close to a nucleus, using a spreadsheet.

Social and human context With hindsight, Rutherford was very clever to have managed without a particle accelerator. But how could he have raised the money to build one without knowing what he would find?

Answers and worked solutions 1.

The units are: C×C 2

C J

2.

−1

m −1 × m

= J.

In the expression electrical potential energy =

2e × Ze 4πε 0 r

dividing by e gives

2Ze 4πε 0r for the energy in eV. Multiply by 10–6 to get the energy in MeV. 3.

Substituting values gives

EP =

2 × 79 × 1.6 × 10 −19 C 4π × 8.85 × 10 −12 C 2 J −1 m −1 × 1.0 × 10 −14 m

× 10 −6 = 22.7 MeV.

4.

Halves, because the potential energy is proportional to 1/r.

5.

About 4.6 × 10–14 m, where the graph reaches 5 MeV.

6.

Just less than 10.0 × 10–14 m.

7.

About 12 MeV.

8.

The charge on the nucleus is smaller, so the potential energy is smaller in the same ratio.

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INDEX

9.

The ratio of the charges, 79 / 50 = 1.58.

10.

About 3 × 10–14 m.

11.

About 0.75 × 10–14 m.

12.

The alpha particle approaches to about 2.5 times the radius of the nucleus. Attractive forces between nucleons might begin to be important, and modify the scattering.

13.

Inserting values:

EP =

82 × 82 × 1.6 × 10 −19 C 4π × 8.85 × 10 −12 C 2 J −1 m −1 × 1.0 × 10 −14 m

× 10 −6 = 967 MeV.

External reference This activity is taken from Advancing Physics chapter 17, 70S

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TAP 522-5: Deep inelastic scattering Deep inelastic scattering Medium energy: elastic scattering

electron proton

Quarks move rapidly inside proton. The interaction time is long enough for the proton to behave like a blur of charge.

Deep inelastic scattering High energy: deep inelastic scattering

electron scattered at large angle

u

electron

d Electron can hit one quark, and be scattered. Exchange of high energy photons leads to the creation of a jet of particles and antiparticles.

243

u ‘jet’ of particles, mainly mesons

INDEX

Quarks as relativistic stationary pancakes

proton as seen by observer not moving relative to it: rapidly moving spherical quarks fill a sphere

coming towards the electron

proton as seen by electron moving rapidly towards it: almost stationary pancake quarks filling a flat disk

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INDEX

Practical advice The diagram could be used as an OHT and discussed in class

External reference This activity is taken from Advancing Physics chapter 17, 150O

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INDEX

TAP 522-6: Electrons measure the size of nuclei Scattering by small particles Hold a glass plate smeared with a little milk, or dusted with lycopodium powder, in front of a point source of light and you may see rings of light round the source. The photons are diffracted by globules of fat in the milk or by the lycopodium spores. Similarly to diffraction by a small hole of diameter d, there is a first minimum intensity at an angle θ of order of magnitude given by sin θ = λ / d. (For circular objects or apertures a more exact expression is sin θ = 1.22 λ / d.) Angles and wavelengths 1.

Show that if you see a first dark ring at θ = 30°, the circular objects have diameter approximately twice the wavelength.

2.

Use the expression sin θ = 1.22 λ / d to find the angle of the first dark ring for particles four wavelengths in diameter.

Wavelengths for electrons The de Broglie wavelength λ of an electron of momentum p is given by λ = h / p, where h is the Planck constant, 6.6 × 10–34 J Hz–1. Since the rest energy of an electron is 0.5 MeV, at energies of hundreds of MeV, the rest energy can be ignored as part of the total energy E. In this case the momentum p is given to a good approximation by p = E / c. 3.

Calculate the energy in joules of an electron with energy 100 MeV (take 1 eV = 1.6 × 10–19 J).

4.

Use the value of the energy from question 3 to calculate the momentum of the electron.

5.

Use the value of the momentum from question 4 to calculate the de Broglie wavelength of 100 MeV electrons.

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INDEX

6.

The radius of a single proton or neutron is of the order 1.2 × 10–15 m. What approximately is the ratio of the wavelength of the electrons to the diameter of a proton or neutron?

7.

Using the relations p = E / c and λ = h / p show that the de Broglie wavelength is inversely proportional to the energy E.

8.

Use the result of question 7 and the answer to question 5 to show that the de Broglie wavelength for 400 MeV electrons is about 3.0 × 10–15 m.

Electron scattering by nuclei You have seen that electrons of a few hundred MeV have de Broglie wavelengths comparable to the diameter of a nucleus. Suppose that in an experiment a beam of 400 MeV electrons is scattered by carbon-12 nuclei. The angle θ at which the scattering is first a minimum is 42°, for which sin θ = 0.67. 9.

Calculate the ratio λ / d of the de Broglie wavelength to the diameter of a carbon-12 nucleus.

10.

Use the de Broglie wavelength of 400 MeV electrons from question 8 to show that the radius of a carbon-12 nucleus is about 2.7 × 10–15 m.

11.

You might expect the volume occupied by the 12 nucleons of carbon-12 to be 12 times the volume occupied by one nucleon. The radius of a nucleon is about 1.2 × 10–15 m. Show that the ratio of the volumes is about 12 (expect some rounding error in these figures).

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12.

A uranium-238 nucleus has a radius of about 7.4 × 10–15 m. What roughly would be a good energy of electrons to use to determine its radius by scattering?

Hints 1.

sin 30° = ½.

2.

Substitute in the expression for sin q.

3.

The conversion factor is equal to the magnitude of the charge on the electron.

4.

Use p = E/c.

5.

Use λ = h/p.

6.

Remember that the diameter = 2 x radius.

7.

Substitute p = E/c for p.

8.

Scale down the wavelength in proportion to the increase in energy.

9.

Obtain the ratio λ/d from the value of sin q.

10.

Remember the radius is half the diameter.

11.

The ratio of the volumes is the cube of the ratio of the radii.

12.

Choose a reasonable angle, say 30°.

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INDEX

Practical advice These questions lead students through a numerical example of the measurement of nuclear dimensions by electron scattering, using the diffraction of the electrons to obtain the scale. Many students will require help with the powers of ten involved, and the conversion between electron volts and energy in joules. The questions are intended primarily as ‘learning questions’ to be gone through slowly. Some students may profit from tackling them alone, but most will need to be taken through them, and have the general message pointed out for them.

Social and human context To measure these tiny distances needed a linear accelerator a few kilometres in length. Large energies to measure small dimensions are expensive.

Answers and worked solutions 1.

sin 30° = ½ = λ/d approximately.

2. sin θ = 1.22

λ d

1.22 4 = 0.305. =

The angle whose sin is 0.305 is 17.8°. 3. Energy = 100 MeV = 10 8 eV = 10 8 eV × (1.6 × 10 −19 J eV −1 ) = 1.6 × 10 −11 J.

4. p= =

E c 1.6 × 10−11 J 3.0 × 108 m s −1

= 0.53 × 10−19 kg m s −1.

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INDEX

5.

λ= =

h p 6.6 × 10 −34 J s 0.53 × 10 −19 J m −1 s

= 12 × 10 −15 m.

6.

Ratio of wavelength to radius = 12 x 10–15 m / 1.2 x 10–15 m = 10; ratio of wavelength to diameter = 5.

7.

Substitute p=

E c

in λ=

h p

gives λ=

8.

hc . E

From λ=

hc E

λ is inversely proportional to E. Since λ for 100 MeV is 12 x 10–15 m then λ for 400 MeV is 3 x 10–15 m. 9.

Since sin θ = 0.67, then 0.67 = 1.22 λ / d, and λ / d = 0.55.

10.

Since λ for 400 MeV is 3 x 10–15 m, and λ / d = 0.55, then d = 5.5 x 10–15 m and r = 2.7 x 10–15 m.

11.

(2.7/1.2)3 = 11.4. It would be closer to 12 but for rounding errors.

12.

Diameter of the uranium-238 nucleus = 15 x 10–15 m. Choosing λ = 2d gives λ = 30 x 10–15 m, about five times the wavelength of 400 MeV electrons (3 x 10–15 m). So scale down energy by a factor 5, to say 80 MeV.

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INDEX

External reference This activity is taken from Advancing Physics chapter 17, 90S

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INDEX

Episode 523: Preparation for nuclear stability topic Why are some nuclides stable while others are unstable (radioactive)? Starting from the pattern of stability, this section looks at forces in the nucleus and the idea of binding energy.

Episode 524: Stable nuclides Episode 525: Binding energy

Main aims Students will: 1. Sketch the N-Z graph for stable nuclei. 2. Describe the balance of forces that results in a stable nucleus. 3. Calculate mass defect and binding energy. 4. Relate nuclear fission and fusion to the graph of binding energy per nucleon.

Prior knowledge Students should know about the composition of nuclei in terms of protons and neutrons. They should be familiar with nuclear notation (e.g. 209 83 Bi )

Where this leads An understanding of nuclear stability leads on to a study of nuclear fusion and fission.

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Episode 524: Stable nuclides Hundreds of nuclides are known; this episode looks at the pattern of stability using an N-Z plot, and considers the need for an additional, attractive force to hold the nucleus together.

Summary Discussion: Which stable nuclides exist? (10 minutes) Student Activity: Using Excel to generate an N-Z plot. (20 minutes) Discussion: Interpreting the graph. (5 minutes) Worked example: Calculating the force between protons. (10 minutes) Discussion: The nature of the strong nuclear force. (5 minutes) Worked example + student example: Calculating nuclear densities. (15 minutes)

Discussion: Which stable nuclides exist Rehearse and extend your students’ existing knowledge. Nuclei are composed of N neutrons and Z protons – collectively A = N + Z nucleons. Explain (or revise) nuclide notation; for example: The heaviest stable element is bismuth-83, 209 83 Bi . How many protons and neutrons in this nucleus? (83 and 126) There are 90 naturally-occurring elements between hydrogen (Z = 1) to uranium (Z = 92). Is there anything odd about this? (Two are “missing”; these are purely artificial and man made, 99 43Tc 63 Pm (promethium). Both are radioactive with relatively short half lives on a (technitium) and 61 geological time scale, and thus would have decayed long ago.)

All elements beyond Z = 92 are man-made. So far (as of February 2004) the record is Z = 116.

Student activity: Using Excel to generate an N-Z plot Use a spreadsheet to generate a plot of N versus Z for stable nuclides. To save time (and transcription errors!) you could prepare an Excel spreadsheet with the data below already entered (columns for the element name, A, Z and, anticipating work to follow, the nucleus mass in atomic mass units u – see below). Representative stable nuclides 2 1H

3 2 He

6 3 Li

9 4 Be

11 5B

15 7N

19 9F

24 21 Mg

36 17 Cl

56 26 Fe

75 33 As

89 39Y

100 42 Mo

113 48 Cd

126 52Te

141 59 Pr

160 66 Dy

180 72 Hf

197 79 Au

212 83 Bi

238 92 U

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INDEX

TAP 524-1: Stability: Balanced numbers of neutrons and protons

Discussion: Interpreting the graph Describe the graph. (It is linear (N ∝ Z) up to Z ~ 20, then increasingly N > Z – there is a ‘neutron excess’. The neutron excess is crucial in explaining nuclear stability, and for setting up a chain reaction in the exploitation of nuclear energy. Think about the Coulomb repulsion between protons. What are the neutrons doing in there? (Neutrons must help overcome the strong repulsion between the protons, partly by ‘diluting them’, but also providing an attractive force to balance the electric repulsion. Hence the name strong nuclear force.

Worked example: Calculating the force between protons If Coulomb’s law has been covered, calculate the repulsive force between two protons that just ‘touch’, so the separation of their centres r is the diameter of a proton (1.4 × 10-15 m).

Fe =

1

q1 q 2

4πε 0 r 2

where q1 = q2 = +e = 1.6 × 10-19 C

Fe ~ 100 N

Discussion: The nature of the strong nuclear force Explain the need for the strong nuclear force to balance the Coulomb repulsion. This force must be attractive – it overcomes the coulomb repulsion; independent of electric charge – it acts between nn, pp and np; and very short range ~ 1fm = 10-15 m.

Worked example + student example: Calculating nuclear densities Calculate the density of a nucleon: nucleon mass ~ 1.7 × 10-27 kg; radius ~ 1.4 × 10-15 m; density

ρ ~ 1.4 × 1017 kg m-3

This is enormous compared to the density of everyday matter. Ask your students to repeat the calculation for He-4 (or give data for other nuclides): Nuclide mass ~ 6.6 × 10-27 kg; radius ~ 2 × 10-15m; density ~ 1.6 × 1017 kg m-3 So the density is roughly the same for all nuclei. This is summed up in the relationship: r = r0 A1/3

with r0 = 1.4 fm

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INDEX

TAP 524-1: Stability: Balanced numbers of neutrons and protons Stable and unstable nuclei: balance of numbers of protons and neutrons stable alpha decay

150

beta decay electron capture positron emission fission

100 N=Z

50

0 0

20

40

60

80

100

proton number Z

This plot of neutron number against proton number shows nuclei decay paths, and the trend of stable nuclei.

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INDEX

Practical advice The diagram could be used as an OHT and discussed in class

External reference This activity is taken from Advancing Physics chapter 18, 50O

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INDEX

Episode 525: Binding energy Summary Discussion: Introducing mass defect and atomic mass units. (10 minutes) Discussion: Mass defect and binding energy (10 minutes) Worked example: Calculating binding energy. (10 minutes) Student questions: Calculations. (20 minutes) Student activity: Spreadsheet calculations. (20 minutes) Student activity: Spreadsheet calculations of binding energy per nucleon. (20 minutes) Discussion: Fission and fusion linked to binding energy graph. (10 minutes)

Discussion: Introducing mass defect and atomic mass units Ask your students to consider whether the following data is self-consistent: proton mass, mp = 1.673 × 10-27 kg neutron mass, mn = 1.675 × 10-27 kg mass of a

4 2 He

nucleus = 6.643 × 10-27 kg

The mass of a 24 He nucleus is less than the sum of the masses of its parts; this is true for all nuclides. So much for conservation of mass. Introduce the atomic mass unit (amu, or u) as a convenient unit of nuclear mass. 1 amu or 1 u = 1/12 the mass of a neutral 12C atom (i.e. including its six electrons) = 1.66056 × 10-27 kg. Thus: mp = 1.0073 u mn = 1.0087 u me = 0.00055 u mass of a neutral

4 2 He

atom = 4.0026 u

Discussion: Mass defect and binding energy What has happened to the missing mass – or mass defect – between the whole and the sum of the parts? To separate the particles, they must be pulled apart against the attractive strong force. They thus have potential energy when they are separated. When the particles come together to form a nucleus, their potential energy decreases. So energy must be put in to separate the nucleons of a nucleus. This energy is known as the binding energy, a rather confusing term because students often think that this means that energy is required to bind nucleons together. As with chemical bonds, this is the opposite of the truth. Energy is needed to break bonds.

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INDEX

Einstein’s Special Theory of Relativity (1905) relates mass and energy via the equation E = mc2 (where c is the speed of light in a vacuum). In this case, we have: binding energy = mass defect × c2

or

ΔE = Δm × c2

(It is not advisable to talk about mass being ‘converted to energy’ or similar expressions. It is better to say that, in measuring an object’s mass, we are determining its energy. A helium nucleus has less mass than its constituent nucleons; in pulling them apart, we do work and so give them energy; hence their mass is greater.)

Worked example: Calculating binding energy Calculate the mass defect and binding energy for 24 He . (Mass defect = 0.053 × 10-27 kg; binding energy = 1.59 × 10-12 J = 9.94 MeV)

Student Questions TAP 525-1: Change in energy: Change in mass TAP 525-2: Finding binding energy TAP 525-3: Fusion in a kettle?

Student activity A data analysis exercise using Excel. This uses a spreadsheet to calculate binding energy for a number of nuclides. TAP 525-4: A binding energy calculator

Student activity Another spreadsheet activity, this time looking at the binding energy per nucleon. Note that it is desirable to plot this graph with a negative energy axis; this means that the lowest values are for the most stable nuclides. TAP 525-5: Binding energy of nuclei

Discussion Briefly discuss fission and fusion in terms of the graph. Although the fission ‘jump’ looks quite small compared to a typical fusion jump, the graph is plotting BE per nucleon. Many more nucleons are involved in the fission of heavy atoms than in the fusion of lighter ones. (This topic can be developed further when discussing nuclear power.) TAP 525-6: Binding energy per nucleon

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TAP 525-1: Change in energy: Change in mass Calculating using Erest = mc2 These questions show you how to calculate changes in energy from changes in mass, using Einstein’s relation Erest = mc2 linking the rest energy of a particle to its mass.

Transmutation of chemical elements The dream of the ancients was alchemy: turning base metals into gold. Although this is chemically impossible, at the end of the nineteenth century radioactivity was discovered by Henri Becquerel. When alpha and beta radiation are emitted atomic nuclei are ‘transmuted’ from one element to another. For example: 238 92 uranium

→ 24 alpha + 234 90 thorium.

In 1932 using protons (hydrogen nuclei) accelerated through a potential difference of 800 000 V, two English physicists, Cockcroft and Walton, carried out the first artificial transmutation: by bombarding lithium with the protons they produced two helium nuclei: 1 7 4 1H + 3Li → 2

He + 24He.

Change in mass Notice that in both these reactions the mass number and charge (proton number) are conserved. Energy, however, is only conserved if you take account of changes to the rest energy – in effect of changes to the masses – of the particles. In Cockcroft and Walton’s experiment, the masses of the particles are: •

H: 1.0073 atomic mass units

•

Li: 7.0160 atomic mass units

•

He: 4.0015 atomic mass units.

An atomic mass unit, symbol u, is equal to 1.6605 × 10–27 kg. 1

Show that the mass decreases in this reaction. 1 7 4 1H + 3Li → 2

He + 24He.

Calculate Δm in atomic mass units and in kilograms.

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INDEX

Change in energy 2

The energy of the protons was 800 000 electron volts (800 keV). The lithium was in solid form so the nuclei would only have been vibrating due to thermal energy, less than an electron volt. The reaction was captured in this photograph:

Two pairs of alpha particles, emerging in opposite directions, can be seen in the photograph. From the range of the tracks through the cloud chamber the energy of the alpha particles was measured to be 8.5 MeV each. Show that the total kinetic energy of the particles increases, and calculate ΔE in MeV and in joules.

3

If the increase in kinetic energy comes from the decrease in rest energy you should expect ΔE = Δmc2. Calculate the ratio of the change in kinetic energy to the change in mass ΔE/Δm in J kg–1.

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INDEX

4

Show that the value of the ratio ΔE/Δm is approximately consistent with the relationship ΔE = Δmc2.

The large value of c 2 (9 × 1016 J kg–1: use this value from now on in calculations) means that a small change in mass represents a vast change in rest energy. This relationship between mass and energy is why particle physicists measure masses in MeV / c2; any unit of energy divided by c2 is a unit of mass.

Creating massive particles Energy is ‘materialised’ in matter–antimatter production. A photon of electromagnetic radiation can produce an electron and a positron. In this case, the energy of the photon vanishes and the rest energy of the particles appears. (This reaction needs to take place near to the nucleus of a heavy atom to conserve momentum but this is not going to affect your calculations here.) In this bubble chamber photograph a photon enters from the bottom. It is uncharged and so produces no observable track. After some distance the photon disappears and produces the electron–positron pair. These two charged particles ionise the liquid in the chamber and bubbles form near the ions and are photographed.

In this case the chamber is filled with liquid hydrogen mixed with liquid neon. It is held under pressure which is released just as the particles enter the chamber to encourage bubbles to form and enlarge near the ions. 5

The bubble chamber is in a magnetic field, so charged particles bend due to the force Bqv on a moving charge. How does the photograph show that the two particles have opposite charges?

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INDEX

The mass of the electron is 5.5 × 10–4 u. What is the minimum energy photon that will produce an electron–positron pair? From what part of the electromagnetic spectrum is this? (Planck constant h = 6.63 × 10–34 J Hz–1.)

6

Nuclear binding energy If protons and neutrons (together known as nucleons) are bound together in a nucleus, the bound nucleus must have less energy than the nucleons of which it is made. That is, the rest energy of the nucleus must be less than the sum of the rest energies of its nucleons. In turn, this means that the mass of the nucleus must be less than the sum of the masses of its nucleons. The simplest compound nucleus is the deuteron, the nucleus of hydrogen-2. It consists of a proton and a neutron bound together by the strong nuclear force. The masses of these particles are: •

proton: 1.0073 u

•

neutron: 1.0087 u

•

deuteron: 2.0136 u.

7

Calculate the difference in mass between a deuteron and one proton and one neutron.

8

Calculate the binding energy of the deuteron in J and in MeV.

9

Calculate the binding energy per nucleon of the deuteron.

10

Express the difference in mass as a percentage of the sum of the masses of the proton and neutron.

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INDEX

Mass change in nuclear fission A possible reaction for the nuclear fission of uranium-235 is: 235 1 133 99 92 U + 0 n→ 51 Sb + 41 Nb

+ 4 10 n.

The masses of the particles are •

U-235 = 235.0439 u

•

Sb-133 = 132.9152 u

•

Nb-99 = 98.9116 u

•

neutron (n) = 1.0087 u.

11

Show that the energy change per atom of uranium is about 200 MeV and calculate Δm/m.

Summary Einstein's famous equation Erest = mc2 reveals a Universe that is not as simple as it seems at first sight. The mass of a particle is generally a very large part of its total energy. The existence of rest energy was not suspected until after Einstein had predicted it, because the change in mass is usually so small, because changes in energy are usually a small fraction of the rest energy. Only in nuclear reactions where Δm/m ~ 0.1% or more are you able to see the change in mass, accompanied by what appears to be a huge change in energy.

Hints 1

Compare masses of H plus Li with mass of two He nuclei.

2

Two 8.5 MeV alpha particles come out, but one 800 keV proton goes in.

3

Compare the answers to questions 1 and 2.

4

Don’t expect to get exactly the speed of light. Remember to take the square root of c2!

5

What is the difference between forces F and - F?

6

Start with the mass of an electron in atomic mass units. Convert to kilograms. Write down the mass of an electron–positron pair. Use Erest = mc2 to get the rest energy of the pair in joules. Then use E = hf.

7

Do this one in the same way as question 1.

8

Erest = mc2 again. But now use the electron charge to get to electron volts and MeV.

9

How many nucleons in a deuteron?

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INDEX

10

Best to take the difference as a fraction of the mass before.

11

Add up before and after masses in atomic mass units first. Don’t forget there’s one extra neutron to start with and four extra neutrons afterwards. Then convert mass changes first to joules and then to MeV.

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INDEX

Practical advice These questions practise the use of the relation between rest energy and mass in various contexts: nuclear transmutation, creation of particle–antiparticle pairs, nuclear binding and nuclear fission. You may need to select only certain groups of questions, depending on what the class has covered. Alternatively, the whole set could be used for revision. Some of the questions make extra demands in frequent changes of units, between atomic mass units, kilograms, joules and electron volts or MeV. You may well need to give extra help here. Note the consistent use of the term ‘rest energy’. The rest energy is treated as part of the total energy. It manifests itself in the mass of a particle. If mass is measured in kilograms and energy in joules, then the conversion is Erest = mc2. Remember that the mass, an invariant, is a physical property of a particle independent of frame of reference. The questions bring out the fact that the rest energy is a very large fraction of the total energy, in many cases.

Alternative approaches You may want to show that energies involved in everyday processes involve negligible changes in mass. The calculations of percentage change of mass in the questions here provide a starting point.

Social and human context It is not possible to ignore the consequences for war and peace of the possibility of tapping these very large sources of energy.

Answers and worked solutions 1.

Mass of H plus Li = 1.0073 u + 7.0160 u = 8.0233 u Mass of two He = 2 × 4.0015 u = 8.0030 u Difference Δm = 8.0030 u − 8.0233 u = −0.0203 u So we can find the mass difference in kg: Δm = −0.0203 u × 1.6605 × 10 −27 kg = −3.3708 × 10 −29 kg

2. Increase in energy: ΔE = 2 × 8.5 MeV − 0.8 MeV = 16.2 MeV

In joules: ΔE = (16.2 × 10 6 eV) × (1.6 × 10 −19 J eV −1 ) = 2.6 × 10 −12 J

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INDEX

3. 2.60 × 10 −12 J ΔE = = 7.7 × 1016 J kg −1 . − 29 Δm 3.37 × 10 kg

4.

If ΔE = Δmc2, then c2 = 7.7 × 1016J kg–1, so c = 2.8 × 108 m s–1.

5.

The force on a moving charged particle is Bqv. If the charge q changes sign, the direction of the force is reversed, so the curvature is opposite.

6.

The mass of an electron or positron is equal to: (5.5 × 10 −4 u) × (1.66 × 10 −27 kg) = 9.1 × 10 −31 kg.

From Erest = mc2, the rest energy of an electron–positron pair is: E rest = 2 × 9.1× 10 −31 kg × (3 × 10 8 m s −1 ) 2 = 1.6 × 10 −13 J.

If this energy is supplied by a photon of energy E = hf, then: f =

1.6 × 10 −13 J 6.63 × 10

−34

J Hz

−1

= 2.5 × 10 20 Hz.

This is the frequency of a gamma ray. 7.

The mass difference is: 2.0136 u − (1.0073 u + 1.0087 u) = −0.0024 u.

I

n kg the mass difference is: − 0.0024 u × (1.66 × 10 −27 kg) = −3.98 × 10 −30 kg.

8.

Bindingenergy = −3.98 × 10−30 kg × (3 × 108 m s−1)2 = –3.58 × 10−13 J =–

3.58 × 10−13 J

1.6 × 10−19 J eV−1 = –2.2 MeV.

9.

= –2.2 × 106 eV

The deuteron has two nucleons so the binding energy per nucleon is –2.2 MeV / 2 = –1.1 MeV.

10.

As a percentage the mass difference is equal to: 0.0024 u = 1.2 × 10 −3 × 100 = 0.1% (approximately) 1.0073 u + 1.0087 u

11.

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INDEX

Mass after = 132.9152 u + 98.9116 u + (4 ×1.0087 u) = 235.8616 u Mass difference = 236.0526 u − 235.8616 u = 0.191 u. Change in rest energy =

0.191 u × (1.66 × 10 -27 kg) × (3 × 10 8 m s −1 ) 2 1.6 ×10

−19

J eV

−1

The ratio is given by:

Δm / m = 0.191 u / 236 u = 8.1×10 −4 ~ 0.1%.

External reference This activity is taken from Advancing Physics chapter 18, 200S

267

= 1.78 × 10 8 eV = 178 MeV.

INDEX

TAP 525-2: Finding binding energy Binding energy of carbon-12 nucleus

Mass of carbon-12 atom

Mass of 6 electrons

1 atomic mass unit u = 1/12 of mass of C-12 atom 1 u = 1.66056 × 10–27 kg

mass of electron = 9.1095 × 10–31 kg = 0.000 549 u

mass of C-12 atom = 12.0 u

mass of 6 electrons = 0.0033 u

6 electrons 6 protons 6 neutrons

Calculate mass of all the protons and neutrons

Calculate mass of carbon-12 nucleus mass of carbon-12 nucleus = mass of carbon-12 atom – mass of 6 electrons

mass of proton = 1.67265 × 10–27 kg = 1.00728 u

mass of carbon-12 nucleus = (12.000 – 0.0033) u = 11.9967 u

Difference in mass

= – 0.0989 u = – 1.643 × 10–28 kg

Energy well for carbon-12 is 7.7 MeV per nucleon deep

7.7 MeV

Binding energy Erest = mc2

= 1.67495 × 10–27 kg = 1.00866 u

mass of 6 protons and 6 neutrons = 6 (1.007 28 + 1.008 66) u = 12.0956 u

= mass of carbon-12 nucleus – mass of protons and neutrons = (11.9967 – 12.0956) u = – 0.0989 u

in mass units:

mass of neutron

in energy units: = – 1.477 × 10–11 J = – 92.16 MeV

Binding energy per nucleon –92.16 MeV for 12 nucleons = – 7.7 MeV per nucleon

Binding energy of a nucleus is the difference between its mass and the sum of the masses of its neutrons and protons

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INDEX

Practical advice This diagram is reproduced here so that you can use it for discussion with your class.

External reference This activity is taken from Advancing Physics chapter 18, 60P

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INDEX

TAP 525-3: Fusion in a kettle? A change of scale When you are confident with basic calculations of fission and fusion energy changes, you should work through these questions that try to put the energies of these changes into a more human scale for you. You will also need to understand the conversion of atomic mass units to energy and the meaning of the term ‘electron volt’.

Try these One of the reactions that fuel the stars is the fusion of two protons to give deuterium. In turn the deuterium goes through a series of reactions, the end product being helium. This is also a process that releases energy. In this question you are asked to consider the energy that would be released if all the deuterium in the water contained in an electric kettle were to be converted by fusion into helium. The kettle contains 1 litre of water. The data you need are listed below. 1 atomic mass unit (u) = 931 MeV 1 eV = 1.6 × 10–19 J NA = 6.02 × 1023 mol–1

1

2

Particle

Mass / u

1 1H

1.007 825

2 1H

2.014 102

3 2 He

3.016 030

1 0n

1.008 665

3 2 Two deuterium nuclei 1 H can fuse to give one nucleus of helium 2 He with the ejection of one other particle. Write down the balanced equation that represents this reaction.

Calculate the mass change that occurs in this reaction.

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INDEX

3

Convert this energy into joules.

This gives you the energy released when two deuterium nuclei fuse. The next steps take you through the calculation of the total energy released if all the deuterium in the kettle water were to fuse to make helium-3. The ratio of deuterium atoms to hydrogen in water is roughly 1 to 7000. 4

What is the mass of 1 mole of water (H = 1 u; O = 16 u roughly)?

5

How many moles of water are contained in the litre?

6

How many molecules of water (H2O) are in the kettle?

7

How many molecules of deuterium oxide (D2O) are in the kettle?

8

Each heavy water molecule has two atoms of deuterium; what total energy is released if all the deuterium in the kettle is converted to helium-3?

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INDEX

Now to put this number in a new perspective. It requires 4200 J to increase the temperature of 1kg of water by 1K. 9

How many litres of water could be heated through 100 K by the fusion energy you calculated in question 8?

Hints 1

It is important to consider the atomic electrons in this equation. You begin with two, one for each hydrogen. How many electrons does an un-ionised atom of deuterium have? So what must one of the emitted particles be? This should lead you to the other particle.

2

The conversions you need are near the data table in the question.

4

The formula of water shows that there are two hydrogen atoms and one oxygen for each water molecule.

5

1 litre of water has a mass of 1 kg.

6

1 mole contains 6 x 1023 molecules of water.

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INDEX

Practical advice These questions can be modified in many ways, not least by changing the homely example of a kettle to perhaps a bath full of water or even to Lake Windermere or the local reservoir.

Social and human context The 6000 litres of heated water may not seem so significant until you realise that this has come from the fusion of deuterium which had an original volume of 0.15 cm3.

Answers and worked solutions 1. 2 2 3 1 H + 1 H→ 2 He

+ 01n

2.

Δm = (3.016 030 u + 1.008 665 u) – 2 × 2.014 102 u = – 0.0035 u

3.

0.003509 u × 931 × 106 eV u–1 × 1.6 × 10–19 J eV–1 = 5.23 × 10–13 J

4.

18 g

5.

1 litre of water has a mass of 1 kg. number of moles = 1000 g / 18 gmol–1 = 56 mol

6.

56 mol × 6.02 × 1023 mol–1 = 3 × 1025

7.

(3.4 ×1025)/7000= 4.9 × 1021

8.

energy released = 4.9 ×1021 × (5.23 ×10–13 J) = 2.49 × 109 J

9.

(2.49 ×109 J)/ (4200 J kg–1K–1 × 100K) = 6000 kg = 6000 litres

External reference This activity is taken from Advancing Physics chapter 18, 260S

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INDEX

TAP 525- 4: A binding energy calculator This model removes the drudgery from the calculation of binding energies. There are three sheets. Sheet 1 does the calculations.

4 4 The calculator numbers masses / amu electrons

5

0.0025

protons

5

5.0365

neutrons

7

7.0609

constitutents total

energy / J

12.0999

quoted mass of isotope

11.00931

difference in mass

-1.09059

energy released

-1.63E-10

Sheet 2 holds the common data – masses for neutrons, protons, electrons, the value of one atomic mass unit in kilograms and the speed of light

Masses of particles name symbol

mass / unified atomic

electron

0

-1e

0.0005

proton

1

1p

1.0073

neutron

1

0n

1.0087

Constants Speed of light 1 amu

c

3.00E+08 m/s 1.67E-27 kg

274

INDEX

Sheet 3 holds a sample of isotope data.

275

INDEX

Element H He Li Be Be C N O F Ne Na Mg Al Si P Si Cl Ar K Ca Sc Ti V Cr Mn Fe Ni Co Cu Zn Ga Ge As Br Se Kr Rb Sr Y Zr Nb Mo Ru Rh Pd Ag Cd In Sn Sb In Te Xe Cs Ba

Z

A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 28 27 29 30 31 32 33 35 34 36 37 38 39 40 41 42 44 45 46 47 48 49 50 51 53 52 54 55 56

1 4 7 9 11 12 14 16 19 20 23 24 27 28 31 32 35 38 39 40 45 47 51 52 55 56 58 59 63 64 69 74 75 79 80 82 85 88 89 90 93 98 102 103 106 107 114 115 118 121 127 130 132 133 138

Mass / amu 1.00783 4.00260 7.01600 9.01218 11.00931 12.00000 14.00307 15.99491 18.99840 19.99244 22.98980 23.98504 26.98153 27.97693 30.97376 31.97207 34.96885 37.96272 38.96371 39.96259 44.95592 46.95180 50.94400 51.94050 54.93810 55.93490 57.93530 58.93320 62.92980 63.92910 68.92570 73.92190 74.92160 78.91830 79.91650 81.91350 84.91170 87.90560 88.90540 89.90430 92.90600 97.90550 101.90370 102.90480 105.90320 106.90509 113.90360 114.90410 117.90180 120.90380 126.90040 129.90670 131.90420 132.90510 137 90500

276

INDEX

What to do: Enter the values for the chosen isotope into the pale yellow boxes in sheet 1, following the tips in the comment boxes, and the binding energy is calculated and displayed in the pale blue box. You will need to double click on the boxes and have a computer running Excel

277

INDEX

Practical advice This is provided as a constructed calculator, together with some useful data.

External reference This activity is taken from Advancing Physics chapter 18, File 30T

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INDEX

TAP 525-5: Binding energy of nuclei Looking for patterns You will use the data in a spreadsheet to calculate the binding energy of a set of nuclei. You will then produce a plot to show how the binding energy per nucleon varies with mass of the nucleus.

You will need 9

computer running a spreadsheet

9

data provided in spreadsheet format

Building the spreadsheet Take a look at the four columns in the spreadsheet data. The first is simply the name of some of the stable elements. This is followed by a column showing the atomic number (Z, the number of protons in the nucleus) and a column giving the mass number (A, the total number of nucleons, i.e. protons plus neutrons). Finally there is a column giving the actual atomic mass. The units of this column are atomic mass units, which are defined as exactly one-twelfth of the mass of a carbon-12 atom. The atomic mass unit (u) is also called the unified atomic mass constant, and has a value of 1.660 5402 × 10–27 kg. Use this information to calculate the binding energy of each nucleus. The binding energy is simply the difference in energy between a nucleus and its constituent parts. This energy change can be measured as a change in the mass of the nucleus. A useful shortcut is that a mass difference of 1 atomic mass unit is equivalent to 931 MeV (million electron volts) of energy. To find the binding energy you will need to subtract the mass of the constituents from the atomic mass. The constituents are Z protons, (A – Z) neutrons and Z electrons (electrons are included in the atomic mass). The masses of these in atomic mass units are:

•

mass of neutron = 1.008 665 u

•

mass of proton = 1.007 277 u

•

mass of electron = 0.000 548 u

Create new columns in the spreadsheet giving the number of neutrons and the mass of the constituents. Now calculate the binding energy of the entire nucleus and the binding energy per nucleon. Plot this last quantity against mass number (not atomic number). Double click on the chart below, you will need a computer running Excel.

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INDEX

Element H He Li Be Be C N O F Ne Na Mg Al Si P Si Cl Ar K Ca Sc Ti V Cr Mn Fe Ni Co Cu Zn Ga Ge As Br Se Kr Rb Sr Y Zr Nb Mo Ru Rh Pd Ag Cd In Sn Sb In Te Xe Cs Ba

Z 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 28 27 29 30 31 32 33 35 34 36 37 38 39 40 41 42 44 45 46 47 48 49 50 51 53 52 54 55 56

A 1 4 7 9 11 12 14 16 19 20 23 24 27 28 31 32 35 38 39 40 45 47 51 52 55 56 58 59 63 64 69 74 75 79 80 82 85 88 89 90 93 98 102 103 106 107 114 115 118 121 127 130 132 133 138

Mass / u 1.00783 4.00260 7.01600 9.01218 11.00931 12.00000 14.00307 15.99491 18.99840 19.99244 22.98980 23.98504 26.98153 27.97693 30.97376 31.97207 34.96885 37.96272 38.96371 39.96259 44.95592 46.95180 50.94400 51.94050 54.93810 55.93490 57.93530 58.93320 62.92980 63.92910 68.92570 73.92190 74.92160 78.91830 79.91650 81.91350 84.91170 87.90560 88.90540 89.90430 92.90600 97.90550 101.90370 102.90480 105.90320 106.90509 113.90360 114.90410 117.90180 120.90380 126.90040 129.90670 131.90420 132.90510 137 90500

particle neutron proton electron

280

mass /u 1.008 665 1.007 277 0.000 548

INDEX

There are four columns in the spreadsheet data.

•

The name of some of the stable elements.

•

The atomic number (Z, the number of protons in the nucleus).

•

The mass number (A, the total number of nucleons: protons plus neutrons).

•

The actual atomic mass. The units of this column are atomic mass units, which are defined as exactly one-twelfth of the mass of a carbon-12 atom. The atomic mass unit is also called the unified atomic mass constant, and has a value of 1.660 5402 x 10–27 kg.

You will have 1. A spreadsheet giving the binding energy of a selection of nuclei. 2. A graph of binding energy per nucleon against mass number.

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INDEX

Practical advice Only a selection of stable nuclei have been included, and the data have been pre-sorted so they are in mass number order rather than atomic number order, and should therefore produce a graph very readily. Students need to be encouraged to change the default settings in their spreadsheet to make the graph clearer and more easily read - an example from Excel is included here. There are some obvious spikes in the graph, which students should be encouraged to think about. This chart is a springboard for discussing why binding energies are negative, why fission and fusion release energy and why certain nuclei are more stable than others. The chart given here indicates some of the key features.

Alternative approaches Use the chart given and ask students to investigate different parts of it - the long slow slope showing where fission releases energy, the steeper slope where fusion releases energy and the spikes at 4He, 12C and 16O. These are particularly important for stellar fusion.

Social and human context It has often been claimed that our Universe is a fluke because the values of certain fundamental constants are closely tuned to values that produce a Universe we can live in. One of these claims is that the fusion of helium in stars to produce carbon and hence all the heavier elements of which we are made requires a lucky coincidence of energy levels between 4He, 8Be (which is unstable and forms for a short time) and 12C. However, a glance at the chart shows that elements such as 12C and 16O are very close to being clusters of helium nuclei so it is, perhaps, no surprise that the relevant energy levels are close to coincidence. A good reference on this, and other aspects of basic laws, is: Dreams of a Final Theory by Steven Weinberg (published by Vintage).

External reference This activity is taken from Advancing Physics chapter 18, 140s

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TAP 525- 6: Binding energy per nucleon The graph below shows the binding energy per nucleon against nucleon number. Elements with a high binding energy per nucleon are very difficult to break up. Iron 56 has the highest binding energy per nucleon of any element and this which explains why there is so much of it in the universe.

9

Energy released by fusion

Binding energy per nucleon (MeV)

8 7 6 5

Energy released by fission

4 3 2 1 0

20

40

60

80

100 120

140 160 180

200

220 240 Mass number (A)

The part of the curve to the left shows that two light elements can produce energy by fusion while the part of the curve to the right shows that a heavy element can produce energy by fission. Therefore if a reaction takes place where the products are closer to the base then the original nucleus (nuclei) then energy is given out. For helium the binding energy per nucleon is 28.3/4 = 7.1 MeV. The helium nucleus has a high binding energy per nucleon and is more stable than some of the other nuclei close to it in the periodic table.

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INDEX

Some of the binding energies per nucleon for some common elements are shown in the following table.

Element

Mass of nucleons (u) 2.01594 4.03188 7.05649 9.07243 56.44913 107.86187 128.02684 207.67109 211.70297 236.90849 239.93448

Deuterium Helium 4 Lithium 7 Beryllium 9 Iron 56 Silver 107 Iodine 127 Lead 206 Polonium 210 Uranium 235 Uranium 238

Nuclear Mass (u) 2.01355 4.00151 7.01336 9.00999 55.92069 106.87934 126.87544 205.92952 209.93683 234.99351 238.00037

Binding Energy (MeV) 2.23 28.29 40.15 58.13 492.24 915.23 1072.53 1622.27 1645.16 1783.80 1801.63

Binding Energy per Nucleon (MeV) 1.12 7.07 5.74 6.46 8.79 8.55 8.45 7.88 7.83 7.59 7.57

A very useful web site containing a huge nuclear database is to be found at: http://nucleardata.nuclear.lu.se It may be more helpful to consider the binding energy per nucleon diagram in the form shown in Figure 2 where reactions tend to move the nuclei towards the valley at the bottom of the curve. (In this case note that the binding energies per nucleon are given as negative values).

0

-2 -3

Energy released by fusion

Binding energy per nucleon (MeV)

-1

-4 -5 -6 4

Energy released by fission

He

238

-7 -8

0

12 16

85

O

20

U

C 56

40

Fe

60

Br 148

80

100 120

La

140 160 180

200

220 240 Mass number (A)

Figure 2

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Episode 526: Preparation for nuclear fission topic This topic does not lend itself to much practical work; there are a few analogue demonstrations which can be valuable. Students may have strong feelings about nuclear technology – power stations and bombs. You could make this a focus of your coverage of this area: the physics ideas won’t tell you what’s right or wrong, but they should allow you to make more informed judgments on questions such as nuclear waste disposal or the use of depleted uranium in weaponry.

Episode 527: Nuclear transmutation Episode 528: Controlling fission

Main aims Students will: 1. Use balanced equations to represent transmutation, fission and fusion events. 2. Calculate mass and energy changes in such events. 3. Understand how a chain reaction can arise in fissile material. 4. Explain how a controlled chain reaction is managed in a nuclear power reactor.

Prior knowledge Students should already be familiar with standard notation for nuclides and with balanced nuclear equations. They should know how to calculate energy changes from changes in nuclide masses.

Where this leads If your students are to study an astronomy topic, this work forms a basis for understanding nuclear processes in stars. Mass-energy calculations using E = mc2, as well as the ability to balance nuclear equations, will also be useful in any study of particle physics.

285

neutron uranium

INDEX

Episode 527: Nuclear transmutation Students need to move beyond the idea that nuclear changes are represented solely by alpha, beta and gamma decay. There are other decay processes, and there are other events that occur when a nucleus absorbs a particle and becomes unstable.

Summary Discussion: Transmutation of elements. (15 minutes) Student questions: Balancing equations. (30 minutes) Discussion: Induced fission. (10 minutes) Demonstration: The nucleus as a liquid drop. (10 minutes) Discussion: Fission products and radioactive waste. (10 minutes) Worked example: A fission reaction. (10 minutes) Discussion and demonstrations: Controlled chain reactions. (15 minutes) Discussion: The possibility of fission. (10 minutes) Student questions: Fission calculations. (20 minutes)

Discussion: Transmutation of elements Start by rehearsing some assumed knowledge. What is the nucleus made of? (Protons and neutrons, collectively know as nucleons.) What two natural processes change one element into another? (α and β decay). This is transmutation. Using a Periodic Table, explain that α decay moves two places down the periodic table. What about β- decay? (Moves one place up the periodic table.) Introduce the idea of β+ decay. (Moves one place down the periodic table.)

β+ decay

β− decay

A, Z-1

A, Z

A, Z+1

Nucleon number

β-

β+

α A-4, Z-2 α decay 286

Proton number

INDEX

Write general equations for these processes. There is another way in which an element may be transmuted; for example, the production of radioactive 14C used in radio-carbon dating in the atmosphere by the neutrons in cosmic rays. 14 1 14 1 7 N + 0 n→ 6 C +1H

The first artificial transmutation was achieved by Rutherford by bombarding nitrogen with α particles. (This experiment was also important in demonstrating that protons are found inside nuclei.) Ask your students to complete the following nuclear equation that summarizes Rutherford’s transmutation of nitrogen into oxygen: He+147N → O +11H

They should get 24 He+147N →178O +11H Cockroft and Walton were the first to ‘split’ the atom, by bombarding lithium with protons from their accelerator. 1 7 8 1 H + 3 Li → 4 Be

→2 24 He

Student questions: Balancing equations Students can practise balancing equations. TAP 527-1: Isotope production

Discussion: Induced fission In the examples above, small parts are ‘chipped off’ nuclei. The behaviour of the heaviest natural element, uranium, is different. It breaks up into two large chunks – into two elements nearer to the middle of the periodic table – so-called induced fission. The two lighter elements are referred to as fission fragments. How do the two common isotopes of uranium than 235 92U .) It is the

235 92U

not the

238 92U

235 92U

and

238 92U

differ? ( 238 92U has three more neutrons

that fissions. It absorbs a neutron, then splits into fission

fragments, i.e. any two smaller nuclei that can be made from the 235 nucleons of the

235 92U

.

TAP 527-2: Nuclear fission

Demonstration: The nucleus as a liquid drop In many ways, nuclei behave like a drop of liquid. Show a water filled balloon - a good model for a nucleus. After the absorption of the neutron, the nucleus of 238 92U wobbles. As soon as the electric charge distribution departs from the spherical (pinch the balloon into a dumbbell like shape) the

287

INDEX

uranium nucleus

neutron

fission fragments

The nuclear chain reaction

neutrons (resourcefulphysics.org)

mutual coulomb repulsion between the two ends drives the fission process. An alternative is to grease a plate and put a large drop of water on it. Wobble the plate about and watch the drop split.

Discussion: Fission products and radioactive waste Most of the energy released is in the form of the kinetic energy of the fission fragments. Because they have a relatively high fraction of neutrons, they are unstable, and decay with short half-lives. They form the ‘high-level’ radioactive waste that cannot be simply disposed of; it has to be stored somewhere for a minimum of 20 half lives. By what factor will the activity fall after 20 half lives? (1/220 is about 10-6, or one-millionth) 137

Cs has a half life of 30.23 years: 20 half lives = 605 years

90

Sr has a half life of 28.1 years: 20 half lives = 562 years

Think about the consequences if waste disposal has to be engineered to remain intact for many centuries. (Which engineering structures have existed for the last 600 years?)

Worked example: A fission reaction Here is the nuclear equation for a typical fission process: 1 235 236 138 95 0 n + 92 U → 92 U → 53 I + 39Y

+?

What is required to balance the equation? (3 neutrons) Why are there some neutrons left over? (Relate this to the N - Z curve. The heaviest elements have the largest neutron excess to remain stable. The two lighter fission fragments have a higher fractional neutron excess; hence some are ‘left over’.) These ‘left over’ neutrons are the vital key to unlock nuclear power using fission.

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INDEX

Discussion + demonstrations: Controlled chain reactions If at least one surplus neutron can induce fission in another 235U nucleus and so on, then a self sustaining release of nuclear energy is possible. For a power station a controlled chain reaction is needed. Should each fission result in more than one further fission, then the chain reaction is said to diverge. In a bomb the aim is to get the chain reaction to diverge as fast as possible. Blow up two balloons; let one fly off; release the other slowly, to illustrate the difference between uncontrolled and controlled energy release. There are a number of analogues of chain reactions that can be demonstrated at this point, using matches or lines of dominoes. TAP 527-3: Chain reactions TAP 527-4: Fission analogues

Discussion: The possibility of fission What are the chances that a neutron will strike another nucleus? First recall that atoms are mostly empty space. The nuclei of two adjacent uranium atoms are typically 10,000 nuclear diameters apart. Emphasise this by picking a pupil in the middle of the class, and estimating her/his width (0.3 m?). Where will the next ‘pupil nuclei’ be situated? (3 km away.) A fast-moving neutron will travel a long way before it strikes another nucleus. In fact, most neutrons are absorbed by 238U nuclei, which are much more common than 235U, and quite good at absorbing fast neutrons. Instead of fissioning they transmute into 239Pu which is fissile, the favourite explosive material for making nuclear bombs. Pure natural uranium is incapable of sustaining a fission reaction – less than one fission neutron succeeds in inducing a further fission. Ask your students how this problem might be overcome in order to have a controlled chain reaction. (The answer is the introduction to the next episode.)

Student questions: Fission calculations Calculations of energy released in fission events. TAP 527-5: Fission – practice questions

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TAP 527-1: Isotope production The production of radio-nuclides by nuclear transmutation is now big business. They are for use in medicine (diagnostic and therapy) and industry (imaging, tracers, process monitoring etc) and are made by neutron bombardment in nuclear reactors and by proton, deuteron ( 12 H ) or α particle bombardment by accelerators. Complete these equations that represent examples of these processes.

31 15 P

32 + ?→15 P + ?+ γ

2 17 15 1H + 7N → 8O

+?

4 16 18 2 He + 8 O → 9 F

4 121 2 He + 51Sb

+ ?+ ?

→ ?+ 01n

Find out the half-lives of the isotopes produced in these processes Find out a typical use of these products.

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INDEX

Practical advice These questions are to help your students write decay equations and carry out research. Half lives can be found at http://www.kayelaby.npl.co.uk/atomic_and_nuclear_physics/4_6/4_6_1.html

Answers

31 15

Isotope

Half Life

Emitted particle(s)

Energy / Mev

32 15 P

14.3 days

beta

1.178

15 8O

126 seconds

positron

1.7

18 9F

112 minutes

positron

0.6

123 53 I

13.2 hours

gamma

0.159

32 P →15 P + 01n + γ

Phosphorus-32 is commonly the highest energy radionuclide encountered in a research setting, and thus requires special caution. Exposure should be avoided and handling should be limited as much as possible. http://www.unh.edu/ehs/radsafety/Fact_Sheets/P-32.html A solution of phosphate, containing radioactive phosphorus-32, is injected into the root system of a plant. Since phosphorus-32 behaves identically to that of phosphorus-31, the more common and non-radioactive form of the element, it is used by the plant in the same way. A Geiger counter is then used to detect the movement of the radioactive phosphorus-32 throughout the plant. This information helps scientists understand the detailed mechanism of how plants utilized phosphorus to grow and reproduce. http://www.chem.duke.edu/~jds/cruise_chem/nuclear/agriculture.html

2 17 15 1 1 H + 7 N → 8 O+0 n

Air tagged with O15 has been applied to the study in dogs for (a) the kinetics of the transfer of oxygen from pulmonary gases to blood, (b) the rate of incorporation of oxygen into water during metabolism, c) the rate of exchange of plasma water with tissue water http://ajplegacy.physiology.org/cgi/content/abstract/201/3/582

4 2

He+ 168O→189 F + 11H + 01n

F-18 is for many reasons a useful radioisotope for bio-medical studies. Physically, the longer half-life (112 min) allows more time for relatively complex synthetic manipulations and for biological studies. In addition, the lowest positron energy, and thus its shortest positron range, allows for the

291

INDEX

sharpest imaging with a high-resolution PET. http://laxmi.nuc.ucla.edu:8248/M248_98/synprob/part4/org_chem_f.html

4 121 123 2 He + 51Sb→ 53 I

+ 201n

The usefulness of iodine-123 whole-body scans in evaluating thyroid cancer http://tech.snmjournals.org/cgi/content/short/27/4/279 Clinical Pharmacology: Sodium Iodide is readily absorbed from the upper gastrointestinal tract. Following absorption, the iodide is distributed primarily within the extracellular fluid of the body. It is concentrated and organically bound by the thyroid and concentrated by the stomach, choroid plexus, and salivary glands. It is also promptly excreted by the kidneys. The normal range of urinary excretion in 24 hours is reported to be 37-75% of the administered dose varying with thyroid and renal function. The iodide-concentrating mechanism of the thyroid, variously termed the iodide “trap” or “pump” accounts for an iodide concentration some 25 times that of the plasma level, but may increase to as much as 500 times under certain conditions. http://www.nuclearonline.org/PI/Nycomed%20I-123%20Na%20I%20capsules.pdf

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INDEX

TAP 527- 2: Nuclear fission This sequence shows how to calculate the energy changes as fission occurs.

Nuclear fission of uranium-235

neutron comes towards U-235 nucleus

U-236 nucleus oscillates like liquid drop

U-236 breaks into two pieces

electrical repulsion

neutron is captured

neutron

U-235 nucleus U-236 nucleus in excited state

293

electrical repulsion

two or three neutrons set free

INDEX

Nuclear fission of uranium-235 Fission takes nucleons down the binding energy valley 0

curve of binding energy as number of nucleons changes

U-235 breaks into two unequal fragments of variable size

–7.6

235 about 0.9 MeV

133

Coulomb slope

100

–8.8

nucleon number A

Nuclear fission of uranium-235 Energy from fission energy per U-235 fission event = 202 MeV

fissile nucleus about 0.9 MeV per nucleon

energy per nucleon = 202 MeV/235 nucleons = 0.86 MeV per nucleon

fission products

294

INDEX

295

INDEX

Practical advice These diagrams are reproduced here so that you can use them for discussion with your class.

External reference This activity is taken from Advancing Physics chapter 18, 110O

296

INDEX

TAP 527- 3: Chain reactions Chain reaction and critical mass Critical chain reaction escapes

The chain reaction is self-sustaining at a steady rate if on average one neutron from a fission produces a further fission. escapes

absorbed

absorbed neutron from fission

absorbed

escapes

Some neutrons escape from the surface of the reactor. Other neutrons are absorbed without causing fission. Sub-critical mass

all chains die out as neutrons are absorbed or escape

Critical mass

Super-critical mass

one new fission follows each fission, on average. Reaction goes at steady rate

Rate of escape of neutrons ∝ surface area Rate of production of neutrons ∝ volume

several new fissions follow each fission: reaction grows rapidly

ratio

297

volume increases with size surface area

INDEX

Practical advice These diagrams are reproduced here so that you can use them for discussion with your class.

External reference This activity is taken from Advancing Physics chapter 18, 120O

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INDEX

TAP 527- 4: Fission analogues 1. Controlled energy release Air filled balloons can be used to show the difference between the controlled release of energy and an uncontrolled explosion, to mimic the controlled chain reaction or the diverging chain reaction (used in a nuclear bomb). Inflate two balloons and tie the necks. On one stick a piece of sellotape about 5 cm long. When holding the balloon by the neck, have the sellotape facing upwards. Take a good-sized pin (a ray tracing optics pin is ideal). Burst one balloon with the pin – the uncontrolled energy release. To ham up the demo, now claim that by pushing the pin very carefully into the balloon it is possible not to burst the balloon and thus achieve a controlled energy release. Slowly push the pin into the sellotaped portion of the balloon. Stop when the pin is still sticking out of the balloon. Then remove the pin slowly. Keep checking during the rest of the lesson to monitor the slow deflation of the balloon.

2. Diverging chain reaction Set up dominoes on edge in the form of a triangle, so that when a domino at the apex is pushed over, it hits two dominoes, they hit 3, which hit 4, and so on.

Alternatively: Drill a piece of wood with a triangular array of small holes to hold matches, ‘live’ end upper most. Arrange the spacing so that one lighted match can ignite its neighbours in the next row. Ten rows are ample. Set the board upright, with the apex of the triangle at the bottom. Light the match at the apex, and see the diverging chain reaction.

!

Safety assessment! Try it by yourself first to see what to expect. Have a fire extinguisher handy (it helps to hype up the demo). Beware any heat or smoke alarms that might be activated.

Extinguish by blowing out – ask if anybody in the audience has a birthday today, or close to the actual date, to volunteer. For a controlled chain reaction: use a simple line of dominoes or matches.

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INDEX

Practical advice These analogues are provided so you can make a choice of demonstration(s) or activities.

300

INDEX

TAP 527- 5: Fission – practice questions What these are for These questions will give you some simple practice in handling the ideas and calculations that physicists meet in nuclear fission.

Try these The process of fission in one type of nuclear reactor proceeds as follows: a nucleus of uranium 235 92 U captures a single neutron. The resulting nucleus is unstable and splits into two or more 90 144 fragments. These fragments could typically be a pair of nuclei, 36 Kr and 56 Ba for example. Neutrons are also ejected as a result of the fission. It is these neutrons that go on to cause subsequent fission events and maintain the chain reaction. 1.

Write down two balanced equations (the first to the unstable uranium; the second to the final products) that represent this fission process.

2.

Calculate the total mass of the original uranium isotope and the neutron. The table gives the atomic masses (in atomic mass units) of the particles found in this question. (1 atomic mass unit (u) ≡ 931 MeV.)

Particle

Mass (u)

1 0n

1.008 665

90 36 Kr

89.919 528

92 36 Kr

91.926 153

96 37 Rb

95.934 284

138 55 Cs

137.911 011

138 56 Ba

137.905 241

144 56 Ba

143.922 941

235 92 U

235.043 923

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INDEX

3.

Calculate the total mass of the four products.

4.

Calculate the change in mass. Does this represent energy gained or lost by the system?

5.

Convert the mass change into the energy released (in MeV) in the fission event.

6.

These particular barium and krypton isotopes are not the only products possible in nuclear fission. Repeat the calculation steps 1–5 with the following possible products caesium-138 and rubidium-96.

Hints 1.

There are two equations, the first for the absorption of the neutron; the second for the splitting of the unstable nucleus formed in the absorption. Write down all the original nucleons on the left-hand side of the first equation (do not forget the original neutron). Put all the products on the right-hand side. Check that all protons, neutrons and electrons balance. Energy is also an output of the reaction, call it Q.

2.

Add the atomic mass unit values for the uranium and the neutron together.

3.

Add the atomic mass unit values for the barium, krypton and two neutrons together.

5.

Use ΔE = Δmc2 to carry out this conversion. c2 = 9 x 1016 J kg–1.

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INDEX

Practical advice This question set provides repetitive practice in handling nuclear mass changes and conversions between mass and energy. It is suitable for students meeting these ideas for the first time. There is an energy release / nucleon perspective here – a useful teaching point when students have completed this task

Answers and worked solutions 1.

235 92 U

2.

m = 235.043923 u – 1.008665 u = 236.052 588 u

3.

m = 89.919528 u + 143.922941 u + 1.008665 u + 1.008665 u =235.859 799 u

4.

Δm =236.052 588 u – 235.859799 u = 0.192 789 u; energy lost

5.

ΔE = 0.192 789 u × 931.3 MeV u–1 =179.49 MeV

6.

m = 137.905241 u + 95.934284 u + 1.008665 u + 1.008665 u =235.856855 u

+

1 0n

=

236 92 U

=

90 36 Kr

+

144 56 Ba

+2 01 n + Q

Δm = 236.052 588 u – 235.856855 u = 0.195733 u ΔE = 0.195733 u × 931.3 MeV u–1 = 182.2 MeV

External reference This activity is taken from Advancing Physics chapter 18, 250S

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Episode 528: Controlling fission In this episode, you can look at the different features of the core of a nuclear reactor, and explain its operation using your students’ knowledge of nuclear physics.

Summary Discussion: The construction of a nuclear reactor. (10 minutes) Discussion: Moderation (10 minutes) Discussion: Enrichment. (10 minutes) Discussion: Critical mass. (10 minutes) Discussion: Control rods and coolant. (10 minutes) Student questions: Power reactors. (30 minutes) Discussion: Nuclear fusion. (15 minutes) Student questions: Fusion calculations. (30 minutes) charge face

boron control rod hot gas graphite moderator reactor core

fuel element channel

heat exchanger concrete steel

cold gas

Discussion: The construction of a nuclear reactor

304

(resourcefulphysics.org)

INDEX

Look at a diagram or animation of a nuclear reactor. Check what your students already know about the reactor’s construction. TAP 528-1: Nuclear fission reactor

Discussion: Moderation How can we make it more likely that a neutron will collide with a 235U nucleus? There are two ways, both used in nuclear power reactors: Slow down the fast neutrons to increase their chance of being captured by a fissile 235U nucleus. This process is called moderation. Concentrate the 235U compared to the 238U. This process is called enrichment. The speed of the fast fission neutrons is slowed down (‘moderated’) by allowing them to collide with a suitable moderator nucleus. Conservation of momentum tells us that the speed of a light neutron colliding with a massive nucleus will be little affected. We need a material with relatively light nuclei to absorb momentum and energy from the neutron. Look at the periodic table for some ideas: Hydrogen – i.e. protons. Virtually the same mass (great), but gaseous (not very dense) and explosive. Hydrogen in water maybe? Yes, pressurised water reactors use water as the moderator (as well as the coolant), but the protons are attached to the rest of the water molecule and have an effective mass of 18 times that of a free proton. Helium – inert (good) but gaseous, so not dense enough. Lithium – too rare (expensive), melting point too low anyway. Beryllium – possible but expensive. Boron absorbs neutrons. Carbon – mass equivalent to 12 protons, solid (good), flammable (bad). Used in the first generation of UK ‘Magnox’ reactors. So there are a number of possibilities, each with a balance of advantages and disadvantages.

Discussion: Enrichment Nuclear power stations use uranium enriched to typically 2.5% - a factor of 2.5/0.7 = 3.6 times the proportion found in natural uranium. Ask your students how much 238U must be discarded to produce 1 tonne of enriched uranium, i.e. with the fraction of 235U increased from 0.7% to 2.5%. (You need 3.6 tonnes of natural uranium, so you discard 2.6 tonnes of 238U.) Bombs require 90% enrichment. Power station enrichment can be easily extended to get pure fissile 235U. Herein lies an easy route to the proliferation of nuclear weapons by countries that have nuclear power programs.

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Discussion: Critical mass Extend your earlier discussion of chain reactions to introduce the idea of critical mass. At least one of the fission neutrons must induce a further fission to allow for a chain reaction. Some may simply escape from the fuel assembly; others may be absorbed by the 238U, by structural materials used in the construction, by the coolant, by the fission fragments etc. Fewer will escape if there is a smaller surface area to volume ratio. For enriched uranium, the critical mass is roughly the size of a grapefruit. Picture bringing two half-grapefruit together to cause an explosion. Why would the critical mass be different for shapes other than a sphere? (A sphere has the lowest area to volume ratio. Other shapes with the same mass would have greater areas, so more neutrons would escape, making a chain reaction less likely.)

Discussion: Control rods and coolant The chain reaction in a nuclear power stations must be controlled, which means that the number of neutrons must be continuously regulated to stop the chain reaction diverging or closing down. To do this control rods are moved into or out of the reactor core. They are made from a substance that absorbs neutrons (e.g. boron). A coolant carries energy away from the core. What are the desirable properties of the coolant? (It must not absorb neutrons; it must have high thermal conductivity, high specific heat capacity and high boiling point.)

Student questions These questions compare Magnox and PWR reactors. TAP 528-2: Fission in a nuclear reactor – how the mass changes

0

(resourcefulphysics.org)

-2 -3

Energy released by fusion

Binding energy per nucleon (MeV)

-1

-4 -5 -6 4

Energy released by fission

He

238

-7 -8

0

12 16

85

O

20

U

C 56

40

Fe

60

Br

306 148

80

100 120

La

140 160 180

200

220 240

Mass number (A)

INDEX

Discussion: Nuclear fusion You can now look at the process of nuclear fusion. (This will have been touched on when considering the graph of binding energy per nucleon.) Students should be able to calculate the energy changes from values of nuclide mass. Emphasise that the energy released per nucleon in fusion is larger than for fission. TAP 528-3: Fusion

Student questions: Fusion calculations Calculating the energy released in fusion reactions. TAP 528-4: Fusion questions TAP 525-3: Fusion in a kettle?

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TAP 528-1: Nuclear fission reactor A Magnox reactor charge face

boron control rod hot gas graphite moderator reactor core

fuel element channel

heat exchanger concrete steel

cold gas

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Practical advice This diagram is reproduced here so that you can use it for discussion with your class.

External reference This activity is taken from Resourceful Physics

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TAP 528-2: Fission in a nuclear reactor – how the mass changes Some rather harder questions These extended questions will test your ability to deal with calculations involving the physics of nuclear fission. Use the following conversions and values for some of the questions:

•

1 eV = 1.6 × 10–19 J

•

1 atomic mass unit = 1.66 × 10–27 kg

•

c = 3 × 108 m s–1

Particle

Mass (u)

235 92 U

235.043 94

1 1H

1.007 825

3 2 He

3.016 030

1 0n

1.008 665

Try these Magnox power stations produce about 20 TW h of electrical energy in the UK every year by fission of uranium. (This energy supplies roughly the electrical needs of Greater London.) 1

The overall efficiency of the process that converts the energy for heating released in the fission to the final electrical product is 40%. How much energy, in joules, is produced each second in the company’s reactors?

2

Each fission releases about 200 MeV of energy. How many atoms of 235 92 U need to fission in each second to produce the heating energy you calculated in question 1?

3

What was the mass of these atoms before they underwent fission?

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4

What is the total mass change due to fission in Magnox reactors each second?

In the pressurised water reactor (PWR) the fuel rods do not contain pure 235 92 U . The uranium comes from mined ore that contains a mixture of

238 92 U

and 235 92 U . The fuel delivered to the reactor contains

0.7% of 235 92 U . The fuel rod stays in the reactor for about 3 years and is then removed to allow reprocessing. This time consider just one reactor with an output of 1 GW. 5

Calculate the number of uranium nuclei disintegrating every second.

6

Calculate the mass of

7

Estimate the mass of

8

Estimate the total mass of both uranium isotopes required in the core for a 3 year cycle.

9

Is your estimate in question 8 likely to be an upper or a lower limit?

235 92 U

235 92

that undergoes fission every second.

U required in the core for a 3 year cycle.

Hints 1

Remember the meaning of the term watt-hour. It corresponds to the amount of energy delivered at a rate of 1 joule per second for 1 hour. Do not forget to include the efficiency in your calculation.

2

Convert to J from MeV.

3

Use the nucleon number of the uranium and the conversion from atomic mass units to kilograms.

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4

Use ΔE = Δmc2 to calculate this.

7

The information indicates that one-third of the fuel needs to be removed for reprocessing every year. Your answer to question 6 can be multiplied up to give the fuel usage in 1 year. This is one-third of the total.

8

The answer to question 7 represents the fuel used, and this is 3% of all the uranium (both isotopes). Hence the total mass.

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Practical advice These questions revise basic conversions between electron volts and joules and atomic mass units and kilograms. Students will need to be familiar with gigawatts (GW) and terawatts (TW) in powers of 10. The questions could be extended either verbally or in writing to ask students about the volume of uranium inside the core and about the equivalent volumes of coal or oil that might be required in a conventional power station. For example, 1 megatonne of coal is equivalent to 29 x 1015 J, 1 megatonne of oil is equivalent to 42 x 1015 J.

Social and human context The questions provide an opportunity for debate about fission power generation.

Answers and worked solutions 1.

energy = ((20 × 1012 Wh × 3600 J W h–1) / (3.16 × 107)) × (100 / 40) = 5.7 × 1019 J every second.

2.

number of atoms = (5.7 × 1019 J) / (200 × 106 eV × 1.6 × 10–19 JeV–1) = 1.8 × 1026 atoms.

3.

mass per second = 1.8 × 1026 s–1 × 235.04394 u × 1.66 × 10–27 kg u–1 = 7.0 × 10–5 kg s–1

4.

mass change = (5.7 × 1019 J s–1)/ ((3 × 108)2 kg s–1) = about 5 g

5.

disintegrations per second = ((1 × 1019 J s–1)/ (200 × 106 MeV × 1.6 × 10–19 J eV–1)) × (100/40) = 7.8 × 1019s–1

6.

mass per second = 7.8 × 1026 s–1 × 235.04394 u × 1.66 × 10–27 kg u–1 = 3.0 × 10–5 kg s–1

7.

mass = 3.0 × 10–5 kg s–1 × 3 years × 3.2 × 107 sy–1 = 2900 kg

8.

mass = 2800 kg × (100/0.7) = 400 000 kg

9.

Lower limit.

External reference This activity is taken from Advancing Physics chapter 18, 270S

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TAP 528-3: Fusion

Fusion in the Sun and on Earth Fusion in the Sun: three-stage process proton

proton

deuterium H-2

positron

proton

deuterium H-2

helium He-3

helium He-3

helium He-3

helium He-4

two protons

γ

neutrino

Two protons fuse, converting one to The deuterium H-2 captures a neutron, to form deuterium H-2. another proton, to form He-3.

Two He-3 nuclei fuse, giving He-4 and freeing two protons.

Fusion on Earth: two-stage process deuterium H-2

helium He-4

tritium H-3

neutron

neutron

lithium Li-6

helium He-4

tritium H-3

Deuterium and tritium are heated to very high temperature. Neutrons from their fusion then fuse with lithium in a ‘blanket’ around the hot gases. Tritium is renewed.

Here you can compare what happens in the Sun with reactions on Earth.

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Practical advice This diagram is reproduced here so that you can use it for discussion with your class.

External reference This activity is taken from Advancing Physics chapter 18, 140O

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TAP 528- 4: Fusion questions Nuclear fusion is the process in which nuclei combine to give heavier elements. In one fusion reaction, two atoms of deuterium (hydrogen-2) fuse together to give one atom of a helium isotope (helium-3) together with one other particle. 1.

Write out a balanced equation for this fusion process and say what the fourth particle is?

2.

Calculate the energy release in this equation. Values you need are in the table.

Particle

Mass (u)

1 0n

1.008 665

2 1H

2.014 102

3 1H

3.016 050

3 2 He

3.016 030

Another possible fusion process is represented by: 2 1H

+

1 0n

=

3 1H

(the formation of hydrogen-3, tritium, by a nucleus of deuterium absorbing a neutron). This equation is certainly balanced. But can it occur in practice? 3.

Calculate the change in mass in this reaction.

4.

Is the reaction possible or not?

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Practical advice These questions revise basic conversions between electron volts and joules and atomic mass units and kilograms. Students will need to be familiar with gigawatts (GW) and terawatts (TW) in powers of 10. The questions could be extended either verbally or in writing to ask students about the volume of deuterium inside the core and about the equivalent volumes of coal or oil that might be required in a conventional power station. For example, 1 megatonne of coal is equivalent to 29 x 1015 J, 1 megatonne of oil is equivalent to 42 x 1015 J.

Social and human context The questions provide an opportunity for debate about fusion power generation.

Answers and worked solutions 1.

2 1H

2.

mass of two hydrogen-2 = 2 × 2.014102 u = 4.028204 u

+

2 1H

=

3 2 He

+

1 0n

mass of helium-3 plus neutron = 3.016030 u + 1.008665 u = 4.024695 u

Δm = 4.028204 u – 4.024695 u = 0.003509 u ΔE = 0.003509 u × 931.3 MeV u–1 = 3.27 MeV 3.

mass of hydrogen-2 plus neutron = 2.014102 u + 1.008665 u = 3.022767

Δm =3.022767 – 3.016050 u = 0.006 717 u.

4.

The reaction can occur with a release ΔE = 0.006 717 u × 931.3 MeV u–1 = 6.26 MeV

External reference This activity is taken from Advancing Physics chapter 18, 250S

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Episode 529: Preparation for X- ray and neutron diffraction topic Students can apply their understanding of diffraction to X-ray and neutron diffraction studies of the structure of matter. This topic could extend a study of diffraction of waves, or be part of a study of material structures, or of atomic physics.

Episode 530: X-ray diffraction Episode 531: Neutron diffraction

Main aims Students will: 1. Relate diffraction effects to the spacing of gratings etc. and to the wavelength of the radiation involved. 2. State some uses of X-ray and neutron diffraction. 3. Apply Bragg’s law.

Prior knowledge Students should be familiar with diffraction due to a standard (transmission) grating (d sin θ = nλ), de Broglie’s formula λ = h/p. It will also help if they are aware that, at (absolute) temperature T, typical energy ~ kT.

Where this leads This topic could lead into a study of the structures and properties of solid materials.

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Episode 530: X- ray diffraction Most schools will not have an X-ray set, but there are plenty of effective analogue demonstrations using other types of waves (laser light, microwaves, and water waves). Note that students who are also studying chemistry may already have come across Bragg’s law.

Summary Discussion + student questions: Students’ knowledge of X-rays. (30 minutes) Discussion: Diffraction and the limits of resolution. (20 minutes) Demonstrations: Diffraction of laser light; crystal models. (20 minutes) Discussion – optional: Deriving Bragg’s law. (20 minutes) Demonstration: Crystal spacing by X-rays. (30 minutes) Demonstrations: Various analogues of X-ray diffraction. (30 minutes) Student activity: Chemical composition by X-ray analysis. (20 minutes)

Discussion + student questions: Students’ knowledge of X-rays Rehearse students’ assumed knowledge of X-rays. If a typical wavelength is λ = 1 nm, what is the frequency of such an X-ray? (f = c/λ = 3×108 / 10-9 = 3×1017 Hz) Students can learn about how X-rays were discovered, and how they are used. TAP 530-1: What are X-rays?

Discussion: Diffraction and the limits of resolution The shortest wavelength of visible light ~ 450 nm (1 nm = 10-9 m) sets a limit for the smallest thing that can be ‘seen’ using visible light. This limit comes about because of diffraction effects, when the wavelength is comparable to physical dimensions. To investigate matter on a smaller scale requires that we ‘look at it’ using shorter wavelengths. X-ray wavelengths are < 1 nm. The wavelengths of X-rays are comparable to the atomic spacing in solid matter. Hence X-rays will be diffracted by planes of atoms in crystalline solids. Show some X-ray diffraction patterns. Emphasise the idea that, the narrower the spacing, the greater the diffraction. This means that diffraction patterns can be used to determine the arrangement of atoms within a solid, and their separations. You could also point out that a single crystal gives a pattern of discrete dots; a polycrystalline material or powder gives rings (because all orientations are present), and an amorphous material gives blurred rings or dots.

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Demonstrations: Diffraction of laser light; crystal models Shine a laser beam at normal incidence onto a grating and note the separation of the diffracted beams. Now rotate the grating about a vertical axis. Observe that the separations of the diffracted beams increase as the effective slit width decreases. (Alternatively, use gratings with different spacing.)

Safety 



Provided the laser is class 2 (less than 1 mW of visible light), the warning ‘Do not stare down the beam' is sufficient. Avoid specular reflections.

If you have crystal models handy (ask your chemistry department), look through in different directions. Many ‘planes’ of atoms reveal themselves, each with its own separation d.

Discussion – optional: Deriving Bragg’s law You may have to derive Bragg’s law. Beware of potential confusion: students will have met the formula for diffraction by a (transmission) grating. In diffraction from crystals the angle is defined differently, and the crystal is acting as a reflection grating. Furthermore, the theory uses reflection rather than diffraction! X ray beam

θ

θ

A

}

B

C

Crystal atoms

d

TAP 530-2: Bragg reflection

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Demonstration: Crystal spacing by X-rays There are problems with using X-rays in school. However, some schools do have X-ray sets. A good alternative is to arrange a visit to a university department (Physics, Chemistry or Materials Science) to see one in use, and to learn about contemporary applications. You could arrange to show a determination of the crystal plane spacing in alkali halides.

Demonstrations: Various analogues of X-ray diffraction Here are some further analogues, for you to choose from (perhaps dependent on the equipment available). Using laser light and diffraction gratings: Two ‘crossed’ diffraction gratings represent the atomic planes, and give an array of diffracted spots. The fact that many solids are polycrystalline and are made up of many small crystallites orientated randomly can be simulated by slowly rotating the crossed gratings. The array of diffracted spots rotates too. What would the diffraction pattern look like if the crossed gratings were rotated quite quickly to simulate all the possible orientations? (A series of rings.) Rig up two crossed gratings with an electric motor to spin them round while diffracting the laser beam. This always makes a visual impact. LINKS TAP 530-3: Simulating X-ray diffraction Finding the structure of DNA is perhaps the best-known use of X-ray diffraction. The iconic ‘X’ shaped diffraction pattern from the helix can be simulated by diffracting a laser beam off a fine bolt thread. Using bolts of different pitches alters the angle of the ‘X’. (See School Science Review vol 85 (312) pp 18-19.) Use a ripple tank TAP 530-4: Diffraction with water waves Try using microwaves and /or a ripple tank with an array of pins LINKS TAP 530-5: Diffraction by crystals

Student activity: Chemical composition by X-ray analysis An interesting use of the Bragg equation is to find the wavelength of X-rays emitted by a substance or object, so that information about its chemical composition can be found.

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TAP 350-6: Where has it been?

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TAP 530- 1: What are X-rays? When Wilhelm Rőntgen first discovered X-rays in 1895, no one had any idea what this strange phenomenon was - hence the name X-rays (although they were referred to as Rőntgen rays for a while). Visits to Egypt to see the pyramids was very much in vogue for cultured people at this time, so it is perhaps not surprising that X-rays were soon used to `see inside' Egyptian mummies. Since then we have found out a great deal more about what X-rays are, how they are generated, their uses and their dangers. Try this short quiz to see just how much you know about X-rays.

Questions 1.

X-rays are part of the electromagnetic spectrum, but where are they positioned when the spectrum is listed in increasing wavelengths?

2.

X-rays are frequently used to diagnose broken or cracked bones. How is this possible?

3.

There are a recommended maximum number of X-ray examinations that a person should not exceed in a year. Why is it necessary to limit the number?

4.

At the dentist, a lead apron is often put onto the patient having a dental X-ray. Why is this apron made of lead?

5.

Women having an X-ray are always asked if they might be pregnant. Why should the radiographer be interested in this?

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Practical advice `What are X-rays?' is mainly a recap of work done at pre-16 level to remind students of the nature of X-rays, their uses and dangers.

Answers 1.

X-rays are listed after gamma rays and before ultraviolet in the electromagnetic spectrum (in increasing wavelength). (Their wavelengths are typically 10-10 m, compared with a bit less than 10-6 m for visible light.)

2.

Bones show up on an X-ray photograph because the X-radiation passes through soft tissue but is stopped (absorbed) by the bones (bones are much denser). The photographic film is blackened by the X-rays going through the soft tissue and through the crack in the bone, while the bones show up as white where little X-radiation has gone through. So a better name for the ‘photograph’ is a shadowgraph

3.

When the X-rays pass through the body they affect the body tissues. In fact the X-rays have so much energy that they can kill cells on the way through or at least damage them. The more X-rays you have over a short period of time, the more damage is done to your body tissue from which it does not have enough time to recover.

4.

The lead apron is to prevent X-rays entering the patient's body. The high density of lead and its high Z number prevents X-rays passing through it. (There are more dense metals - for example gold, but that would be a little expensive in an apron!)

5.

The danger of administering an X-ray to a pregnant woman is that the growing embryo or foetus may be damaged by the X-rays passing through it. This is particularly important because the rate of cell division in a developing baby is extremely high, and if cells are damaged or mutated by X-rays then this might cause the baby to be deformed.

External reference This activity is taken from Salters Horners Advanced Physics, section DUTP, additional activity 3

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TAP 530- 2: Bragg reflection The diagram below shows X-rays being reflected from a crystal. Each layer of atoms acts like a mirror and reflects X-rays strongly at an angle of reflection that equals the angle of incidence. The diagram shows reflection from successive layers

If the path difference between the beams from successive layers of atoms is a whole number of wavelengths, then there is constructive interference. The path difference is the distance AB + BC as above: (AB + BC) = 2dsinθ The reflected beam has maximum intensity when 2dsinθ = nλ

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Practical advice Some background information: von Laue's experiment In 1912, Max von Laue realised that X-rays were not charged particles as they were not deflected by electric or magnetic fields, but no-one had succeeded in demonstrating that X-rays were waves by creating interference until von Laue realised that it was the extremely small wavelength which was causing the problem. He used the spaces between atoms in a crystal to diffract the X-rays and thus produced interference. In the same year, following von Laue's discovery, William and Lawrence Bragg (father and son) varied the experiment and used a crystal to reflect X-rays and also obtained diffraction patterns. It may be useful to remind pupils of: TAP 322-5: Using a CD as a reflection grating

External reference This activity is taken from Salters Horners Advanced Physics, section DUTP, additional activity 8

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TAP 530- 3: Simulating X-ray diffraction Introduction: Basic X-ray diffraction involves the transmission of X-rays through a thin sample. This can be simulated using a laser through Lycopodium powder/talcum powder or tree pollen sandwiched between two microscope slides

You will need: 9

low-power laser

9

screen

9

pair of microscope slides (2)

9

stand to support slides, e.g. retort stand

9

powder (talc, Lycopodium) or tree pollen

Safety Provided the laser is class 2 (less than 1 mW for visible light), the warning ‘Do not stare down the beam' is sufficient. Avoid specular reflections.

Allergic reaction Class members may be allergic to pollen (e.g. Lycopodium), check before use.

What to do: Make a film of powder sandwiched between microscope slides. For best results the film should be very thin. Blowing away the excess usually achieves this.

You have seen: •

The pattern produced by Lycopodium powder will be a ring pattern due to the random orientation of the scattering centres which causes the maxima to occur as a cone of a particular angle.

•

Talcum powder consists of larger particles so, although the pattern is due to diffraction, the pattern does not show the ring shape but simply a random arrangement of spots.

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Practical advice Do not breath in the dust particularly Lycopodium powder.

External reference This activity is taken from Salters Horners Advanced Physics, section DUTP, activity 15

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TAP 530- 4: Diffraction with water waves Crystal diffraction can be simulated using water waves.

You will need: 9

ripple tank (levelled)

9

plane wave generator

9

power pack 6 V

9

rheostat to control speed of motor

9

desk lamp for illumination

9

barriers to provide at least 3 ‘holes’

9

hand held stroboscopes

What to do: 1.

First set up the ripple tank with about 1 cm depth of water.

2.

Set up a line of small barriers 5 cm from the vibrator, as shown. There should be a gap of 2 to 3 cm between each.

3.

Start the motor at a low speed (4 rev/second).

Ask: ‘Can you see semicircular ripples emerging from the gaps? Further out, can you see waves moving out in slanting directions, as well as a wave moving straight ahead?’ Students should observe the diffraction pattern carefully, with and without stroboscopes. Keeping the barriers arrangement the same, gradually increase the motor speed.

Ask, ‘How does the diffraction pattern change?

!

Safety Beware water on the laboratory floor. Make sure you have a sponge and bucket handy to mop up spills immediately. Place the power supply for the lamp on a bench, not on the floor by the tank

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Practical advice: You should refer back to the work in Episode 322: Diffraction gratings and Episode 323: Diffraction and particularly: TAP 323-1: Ripple tank diffraction The pattern produced with multiple gaps is less clear than the double gaps experiment but, with care in aligning the gaps, it is visible. It will help if students have first seen diagrams of sets of semicircles to represent a snapshot of waves proceeding from several gaps. Avoid very high motor speeds, which cause unwanted vibration of the barriers. The spacing of nodal lines will decrease as the wave frequency increases The following Applet can be set for a triple slit, (on the set up menu pick triple slit): http://www.falstad.com/ripple/index.html this may be simpler and easier to see than on the ripple tank

!

Safety Beware water on the laboratory floor. Make sure you have a sponge and bucket handy to mop up spills immediately. Place the power supply for the lamp on a bench, not on the floor by the tank

Technical notes You need 2 large barriers and 6 small barriers The hand stroboscope is a disc of hardboard or card with a simple pivot at its centre, so that the disc can be kept spinning by hand. The disc has a finger hole, off-centre, to enable the user to keep it spinning. It has narrow slits on its face, near the rim. The slits are evenly spaced and 12 slits are best This simple stroboscope enables students to 'freeze' repetitive motions – or to slow them down for closer study. For example, continuous ripples are easier to see by using a stroboscope, especially those ripples with higher frequencies. By viewing a vibrating object through the slits, students can calculate the frequency of a vibration. The stroboscopes are less likely to judder while rotating if the bearing is not too tight and the handle is held loosely.

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External references This activity is taken from Salters Horners Advanced Physics, section DUTP, activity 17 and http://www.practicalphysics.org/

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TAP 530- 5: Diffraction by crystals Introduction: This effect of scattering from successive layers can be shown with the ripple tank and a `bed of nails' - optical pins in a 1 cm array glued into a matrix board and placed in a ripple tank. Microwaves can also be used to show the effect of adding layers

You will need: 9

ripple tank (levelled)

9

plane wave generator

9

power pack 6 V

9

rheostat to control speed of motor

9

desk lamp for illumination, a special ripple tank lamp is better: the filament can be set parallel to the ripples.

9

bed of nails

9

microwave transmitter and receiver

9

wax lenses to concentrate beams

9

polystyrene ceiling tiles

What to do: This effect of scattering from successive layers can be shown with the ripple tank and a `bed of nails' i.e. optical pins in a 1 cm array glued into a matrix board and placed in a ripple tank. Plane waves should be incident on the nails at an angle, with the diffraction pattern appearing at the same angle on the opposite side of the bed of nails. The precise depth of water at which this works best depends on the thickness of the perspex and the nail length used. You will need to experiment. With microwaves, first show that a single polystyrene tile used as a mirror produces a moderate reflected signal irrespective of the angle of incidence. If a second tile is mounted parallel to the first and separated from it by a few cm, then there is a reasonable reflected signal at an angle that satisfies the Bragg equation, e.g. if d = 5 cm then n = 1, λ = 3 cm leads to sinθ = 0:3, y = 17:58O. The signal is noticeably weaker at other angles. Add further regularly-spaced tiles.

You have seen: •

A regular array produces a diffraction pattern.

•

Reflection from surfaces in parallel satisfies the Bragg equation.

•

Adding further regularly-spaced tiles makes the reflected signal sharper.

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Practical advice: The microwave equipment can be set up without the wax lenses, however, they will improve results if set up correctly.

Technician’s note: To make the bed of nails, use a piece of plain matrix board about 10 cm to 15 cm with optical pins glued in at approx. 1 cm intervals to create a square matrix.

External reference This activity is taken from Salters Horners Advanced Physics, section DUTP, activity 17

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TAP 350- 6: Where has it been? The Coppergate Helmet was subjected to X-ray diffraction analysis (XRD) in order to determine what it was made from and the likely conditions in which it was buried. Study the relevant X-ray diffraction spectra and write a short report on the deductions that can be made.

X-ray diffraction spectra Nowadays most X-ray diffraction analysis is done with a goniometer, a device in which the variation of intensity of the diffracted X-radiation is measured by scanning an ionisation detector through a range of angles - usually 0O (straight through) to 100O - but photographic detection is still also used. The goniometer spectrum for iron (II) carbonate is shown in Figure (a) together with the photographic film equivalent in Figure (b). (The XRD spectra in this activity were kindly made available by English Heritage.)

Note that in Figure (b) the dark lines indicate a high intensity of diffracted X-radiation. Sometimes you will find white lines on a black background - it depends on whether the negative or the positive print is used. Be aware that diffraction information can be displayed in different ways. The records shown here involved scanning from just over 90O to 0O, but you may come across others which show spectra fro + 90O to 0O to -90O or more, symmetrical about 0O.

The Coppergate Helmet As you might expect, all the metals on the helmet corroded to some extent while it was buried. The corrosion products revealed during XRD analysis provided evidence about the original materials

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and the conditions where the helmet had been buried. The diagram below shows the XRD spectra of some common corrosion products.

Bornite (Cu5FeS4) This is a corrosion product of iron and copper alloy artefacts when buried on waterlogged sites.

Chalcopyrite (CuFeS2) This is also a corrosion product of iron and copper alloy artefacts when buried on waterlogged sites.

Siderite (FeCO3) This is a corrosion product of iron. It is found where there is, or has been at some time, a high concentration of carbonates in the water. Typical sources of these are plaster and mortar from earlier Roman buildings that have long since been demolished.

Vivianite (Fe3 (PO4)2 8H2O) This is a corrosion product of iron which is found where the artefact has been buried alongside phosphates from bone and excreted waste. The XRD spectra in below are from two different parts of the Coppergate Helmet.

Use the XRD spectra for corrosion products and the two spectra A and B above to report

•

on what those parts of the helmet were made of,

•

together with the conditions in which it had been buried.

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Practical advice Archaeologists examining the Coppergate Helmet used X-ray diffraction (XRD) to investigate the conditions in which it had been buried. They used XRD to examine the corrosion products in small samples of the helmet, and compared their spectra with those of corrosion products produced in known conditions. Students are provided with X-ray spectra from two samples of the helmet, and the spectra of four corrosion products together with information about the conditions that give rise to such products. Their task is to identify the corrosion products found in the helmet and hence to deduce the nature of the original materials and the conditions in which they had been buried.

Answers X-ray diffraction spectrum A on the Activity Sheet shows that the corrosion sample is made up of siderite and vivianite. This indicates that the material that had corroded was made of iron. It also shows that, at some time, it was buried alongside organic phosphates probably produced from bone or excreted waste. In addition, the ground water around it had, at some time, a high concentration of carbonates, possibly as a result of demolition of earlier Roman buildings that had been built with plaster and mortar. XRD spectrum B shows that the corrosion sample is made up of bornite and chalcopyrite. This indicates that the material that had corroded was an iron and copper alloy, and had been buried in a waterlogged area.

External reference This activity is taken from Salters Horners Advanced Physics, section DUTP, activity 18

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Episode 531: Neutron diffraction This episode takes a very brief look at the topic of neutron diffraction.

Summary Discussion + worked example: de Broglie wavelength of a neutron. (15 minutes) Discussion: Sources and uses of neutrons. (15 minutes)

Discussion + worked example: de Broglie wavelength of a neutron Neutrons are very penetrating, typically 10 cm in lead. Why might this be? (Neutrons have no electric charge, so they do not interact with the atomic electrons.) However, neutrons do have a magnetic moment that allows them to interact with magnetic nuclei. Neutron diffraction was proposed in 1934, to exploit de Broglie’s hypothesis about the wave nature of matter. Use de Broglie’s equation λ = h/p to calculate the momentum and KE of a neutron whose wavelength is comparable to atomic spacing, say 1.8 × 10-10 m. p = h/λ = 6.6 × 10-34 / 1.8 × 10-10 = 3.7 × 10-24 kg m s-1 Ek = p2/2m = [3.7 × 10-24]2 / (2 × 1.67 × 10-27) = 4.0 × 10-21 J = 1/40 eV If we equate this energy to kT, we get: T = 4.0 × 10-21 / 1.38 x 10-23 = 290 K

which is roughly ‘ambient temperature’. (It is useful to be able to recall that 1/40 eV is the typical energy of a particle at room temperature.) The neutrons released in nuclear fission are ‘fast’ neutrons, i.e. much more energetic than this. Will graphite moderator

thermal neutron

fast neutrons their wavelengths be greater or less than atomic dimensions? (Less) So we need to slow down the fast neutrons to ‘thermalise’ them. How is this achieved in a nuclear power reactor? (By including a moderator.)

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Discussion: Sources and uses of neutrons Neutrons – where do we get them from? (Nuclear reactors; spallation sources, which use accelerators to crash ions into heavy nuclei.)

What can they tell us? For elastic interactions where the neutron energy is not changed, the Bragg equation allows the determination of the separation of atomic planes. In an inelastic process, the neutron gives energy to, or receives energy from, the vibrating ions that make up the crystal lattice. This leads to information about the forces between atoms in the crystal lattice. The interaction of magnetic moments gives information about magnetic properties of materials – e.g. the position of the domain boundaries, their size and orientation, the magnetic moments of the atoms etc.

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Episode 532: Preparation for the particle physics topic It is important to avoid turning this entire topic into ‘stamp collecting’. The underlying themes to develop here are that discoveries in nuclear physics, accelerating during the middle of the twentieth century, resulted in a need to classify the particles discovered. The key to this classification was the application of conservation laws, some of quantities already well established (charge, baryon number), some modified from earlier versions (mass/energy) and some new (lepton number) or even bizarre (strangeness is the one met here, but there are also parity, isospin and hypercharge). The other aspects developed are: antiparticles, the explanation of particle interactions in terms of exchange particles (vector bosons), and the development of the standard model of three generations of quarks and leptons to explain and simplify the ‘particle zoo.’

Episode 533: The particle zoo Episode 534: Antiparticles and the lepton family Episode 535: Particle reactions Episode 536: Vector bosons and Feynman diagrams

Advance warning The first topic makes use of activities based on card sorting and arranging. The baryon, meson and lepton cards suggested here have been constructed for this project, but other excellent examples have been published, and it is well worth your considering the following: A resource for particle physics teaching in schools, Ken Zetie, Physics Education, March 2003, 38, p 107 with sets of online resources Playing with Particles – a teaching approach for A-level (post-16) particle physics, David Brown, School Science Review, December 2002, 84 (307), pp 118-123 For more wide-ranging resources on the teaching of particle physics at this level, see the TRUMP Particle Physics project from York University (www.york.ac.uk/org/seg/trump/parthome.htm).

Main aims Students will: Use the terms baryon, meson, hadron and lepton. Know that, for each particle, there exists an antiparticle. Apply conservation rules to particle interactions, including particle annihilation and creation. Know that interactions between fundamental particles (quarks and leptons) are due to the exchange of other particles (vector bosons), and that these are (virtual) photons for the electromagnetic interaction and W+, W- and Z0 particles for the weak interaction. Construct and interpret Feynman diagrams showing interactions.

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Prior knowledge Students should know about the constituent particles of the atom (protons, neutrons, electrons). They should be familiar with the ideas of conservation of mass, energy and electric charge.

Where this leads The first topic deals with the fundamental particles that students have already met, and how their number proliferated in the second half of the twentieth century. Classification of the different sorts of fundamental particles then leads to use of conserved quantities (some rather bizarre) to establish rules for particle reactions. Students will learn that particles can be classified as hadrons – baryons and mesons – and leptons, each with its anti-particle, and they should know that interactions between these particles can be described in terms of transfer of other particles known as vector bosons. Once the ground rules for particle interactions have been established, students can go on to learn that there is a finer level of structure to mesons and baryons.

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Episode 533: The particle zoo Protons, neutrons and electrons are familiar particles of matter. However, students are likely to have heard of other particles, and this episode introduces some of these.

Summary Discussion: Establishing prior knowledge. (15 minutes) Discussion: Conservation of charge and mass. (20 minutes) Discussion: Units of mass. (10 minutes) Student activity: Classification of hadrons. (30 minutes) Student activity: Research task. (30 minutes)

Discussion: Establishing prior knowledge Brainstorm to review prior knowledge. What’s in an atom, and how do we know? Establish with the students that the atom contains: -

•

electrons (discovered by J J Thomson in 1897, Cavendish Lab, Cambridge, though he called them negative corpuscles, the name electron was ‘coined’ by G. Johnstone Stoney in 1891 )

•

nuclei (Rutherford, Geiger and Marsden, 1911)

•

it was soon realised that nuclei contained multiples of the nuclei of hydrogen atoms (i.e. protons; Rutherford suggested the name proton in 1920)

•

that the simple picture was completed with Chadwick’s discovery of the neutron (predicted by Rutherford) in 1932.

At this point, you may find it useful to ask how students think how a neutral particle such as a neutron may be detected. You can tell them that sub atomic particle detection in 1932 relied on ionisation (as in a GM tube, or in a photographic film), and that this only detects charged particles. They should realise that the neutron must do something to some other matter that then produces some ionising radiation. (For Chadwick, the ionising radiation consisted of protons emitted from paraffin wax when neutrons hit the nuclei of hydrogen atoms. The neutrons came from beryllium that had been bombarded with alpha particles. [J Chadwick, The Existence of a Neuron, Proceedings Royal Society of London, 1932]

Discussion: Conservation of charge and mass Use alpha and beta decay to show conservation of charge and baryon number (see below). Use the equations for one example of each, e.g.:

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220 216 4 86 Rn→ 84 Po + 2 He

14 14 0 6 C → 7 N + −1 e

to show conservation rules in action: the fact that the bottom numbers add to give the same total each side shows conservation of charge, while the fact that the top numbers do likewise shows that the number of nucleons (neutrons and protons) is also conserved. Introduce the term baryon (heavy particle) for these two particles, so that these processes also conserve baryon number. It is worthwhile giving each student an example to do, as this is confidence building, and it is a simple skill that is often examined.

Discussion: Units of mass It may help your students, before embarking on a discussion of the various families of particles, if they have an idea of the different units used for mass. All those powers of 10 are very inconvenient; hence, it is easier to work in energy units, particularly the MeV (mega electron volt). Although it is useful to have met E=mc2, this is not essential. All that is needed here is that students appreciate that mass and energy can be inter-converted in nuclear reactions (such as fission and fusion, which students should be familiar with), and that the energy that would be released if the entire mass of a particle were transferred into energy is used as a measure of its mass in particle physics. Rather than define an electron-volt in terms of charge and pd, you may prefer to state that it is a convenient ‘atomic size’ energy unit where 1 MeV = 1.6 × 10-13 J equivalent to 1.783 × 10-30 kg. It is useful to have a poster in your lab/ teaching room with different mass units on it. TAP 533-1: Mass units for particle physics

Student activity: Classification of hadrons Hadrons, is a collective term for both mesons and baryons; mesons are less massive than baryons. This activity is a card-sorting exercise. Helpfully, this slows down the pace compared with teacher exposition or question and answer. Introduce the activity by explaining that, following the discovery of the neutron, many other particles were discovered, including some strange ones: these seemed to be created in pairs, encouraging the physicist Murray Gell-Mann to allocate ‘strangeness numbers’ to some of these particles. In these reactions, strangeness was conserved. This is a kinaesthetic exercise using cards for a collection of hadrons. It is best to print up several sets of the two hadron card sheets, on card rather than paper. There is a similar sheet of lepton cards – they can be added to or simplified as you wish. It is best to use a different colour for the lepton cards. You may consider laminating them.

N

N P

ν e

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e

ν e

P

INDEX

For the first exercise, use the first sheet of hadron cards (omitting some mesons and all anti-baryons), cut into individual cards. Ask your students to sort particles by mass and to note the charge and baryon number. Mesons are those hadrons, mostly of lower mass, which have baryon number of 0. Later, the cards can be used, together with genuine particle reactions, to check on conserved quantities. TAP 533-2: Hadron cards TAP 533-3: Lepton Cards

Student activity: Research task Conclude this episode with some questions for students to research, leading to the next episode:

•

What is a positron?

•

Who suggested it must exist?

•

Who discovered it?

•

What is a muon?

•

When was it discovered?

•

Who said, ‘Who ordered that?’

•

What is a neutrino?

•

Who suggested it must exist?

•

Who discovered it?

TAP 533-4: Positron, muon, neutrino

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TAP 533- 1: Mass units for particle physics u = 1.661 × 10-27 kg

Atomic mass unit: MeV/c2

1 MeV/c2 = 1.783 × 10-30 kg

Some rest masses mass (kg)

mass (u)

mass (MeV/c2)

electron

9.109 × 10-31

5.486 × 10-4

0.5110

proton

1.673 × 10-27

1.009

938.3

-27

1.007

939.6

neutron

1.675 × 10

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TAP 533- 2: Hadron cards neutron

proton

n Mass 2 (MeV/c ) 939.6

Mass 2 (MeV/c ) 140

Charge (e) 0

Charge (e) -1

p

Baryon number +1

Lepton number 0

Strange -ness 0

Mass 2 (MeV/c ) 938.3

Charge (e) +1

pi minus

K zero

π

K0

-

Baryon number 0

Lepton number 0

Strange -ness 0

Mass 2 (MeV/c ) 498

Charge (e) 0

K plus

Mass 2 (MeV/c ) 1116

Charge (e) 0

Baryon number 0

Lepton number 0

Strange -ness +1

Mass 2 (MeV/c ) 1198

Charge (e) -1

sigma zero

Λ

Σ0 Lepton number 0

Strange -ness -1

Mass 2 (MeV/c ) 1192

Charge (e) 0

pi plus Baryon number 0

Lepton number 0

Strange -ness 0

Mass 2 (MeV/c ) 494

Charge (e) -1

sigma plus

Σ Mass 2 (MeV/c ) 1189

Charge (e) +1

Lepton number 0

Strange -ness -1

Mass 2 (MeV/c ) 1232

Charge (e) 0

Lepton number 0

Strange -ness -3

Mass 2 (MeV/c ) 1232

Charge (e) +1

Charge (e) -1

Baryon number +1

Ξ

π0

Baryon number +1

Δ Charge (e) -1

Baryon number +1

pi zero Lepton number 0

Strange -ness -2

Mass 2 (MeV/c ) 106

Charge (e) 0

delta minus Mass 2 (MeV/c ) 1232

Baryon number 0

xi minus

-

Mass 2 (MeV/c ) 1321

Strange -ness -1

Lepton number 0

Strange -ness -1

Lepton number 0

Strange -ness 0

Lepton number 0

Strange -ness 0

Baryon number 0

Lepton number 0

Strange -ness 0

Lepton number 0

Strange -ness -2

xi zero

-

Baryon number +1

Lepton number 0

Δ+

Ω

Baryon number +1

Baryon number +1

delta plus

-

Charge (e) -1

Strange -ness -1

Δ0

omega minus Mass 2 (MeV/c ) 1673

Lepton number 0

delta zero

+

Baryon number +1

Strange -ness +1

K-

π Charge (e) +1

Lepton number 0

K minus

+

Mass 2 (MeV/c ) 140

Baryon number +1

lambda Baryon number +1

Strange -ness 0

Σ-

K Charge (e) +1

Baryon number 0

Lepton number 0

sigma minus

+

Mass 2 (MeV/c ) 494

Baryon number +1

Ξ0 Lepton number 0

Strange -ness 0

Mass 2 (MeV/c ) 1315

345

Charge (e) 0

Baryon number +1

INDEX

Mass 2 (MeV/c ) 549

Charge (e) 0

eta zero

phi zero

η

φ0

0

Baryon number 0

Lepton number 0

Strange -ness 0

Mass 2 (MeV/c ) 1020

Charge (e) 0

Baryon number 0

anti-sigma minus

Σ Mass 2 (MeV/c ) 1198

Charge (e) +1

Λ

Baryon number -1

Lepton number 0

Strange -ness +1

Mass 2 (MeV/c ) 1116

J/psi Mass 2 (MeV/c ) 3097

Charge (e) 0

Charge (e) 0

Mass 2 (MeV/c ) 498

Mass 2 (MeV/c ) 770

Charge (e) 0

Charge (e) -1

Baryon number 0

Lepton number 0

Strange -ness 0

Mass 2 (MeV/c ) 1232

Charge (e) +1

Baryon number -1

Strange -ness 0

Mass 2 (MeV/c ) 938.3

Charge (e) -1

Lepton number 0

Strange -ness 0

Σ0

Baryon number 0

Lepton number 0

Strange -ness -1

Mass 2 (MeV/c ) 1192

Charge (e) 0

Baryon number -1

rho minus

anti-xi zero

ρ

Ξ0

-

Baryon number 0

Baryon number -1

Lepton number 0

Strange -ness 0

Mass 2 (MeV/c ) 1315

Charge (e) 0

Lepton number 0

Strange -ness +1

Lepton number 0

Strange -ness +2

Lepton number 0

Strange -ness +2

Ξ-

Strange -ness +3

Mass 2 (MeV/c ) 1321

Charge (e) +1

Baryon number -1

anti-sigma plus

Σ+

Lepton number 0

Strange -ness o

Mass 2 (MeV/c ) 1189

Charge (e) -1

Baryon number -1

Lepton number 0

Strange -ness +1

Lepton number 0

Strange -ness 0

rho plus

+

Baryon number -1

Baryon number -1

Lepton number 0

anti-xi minus

0

Baryon number -1

Δ Charge (e) -1

Baryon number -1

anti-sigma zero

anti-delta plus Mass 2 (MeV/c ) 1232

Strange -ness 0

K0

Δ Charge (e) 0

Lepton number 0

anti-K zero

anti-delta zero Mass 2 (MeV/c ) 1232

Baryon number -1

p Lepton number 0

Ω-

Charge (e) +1

Strange -ness +1

anti-proton

anti-omega minus Mass 2 (MeV/c ) 1673

Lepton number 0

Δ-

n Charge (e) 0

Baryon number -1

anti-delta minus 0

anti-neutron Mass 2 (MeV/c ) 939.6

Strange -ness 0

anti-lambda

-

J/ψ

Lepton number 0

ρ+

Lepton number 0

Strange -ness 0

Mass 2 (MeV/c ) 770

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Charge (e) +1

Baryon number 0

INDEX

TAP 533-3: Lepton Cards electron

electron neutrino

e-

Mass 2 (MeV/c ) 0.511

Charge (e) -1

Baryon number 0

νe Lepton number +1

Strange -ness 0

Mass 2 (MeV/c ) 0

positron Charge (e) -1

Baryon number 0

Strange -ness 0

Mass 2 (MeV/c ) 0

Charge (e) 0

Lepton number -1

Strange -ness 0

Mass 2 (MeV/c ) 106

muon neutrino Mass 2 (MeV/c ) 0

Charge (e) 0

νμ

Baryon number 0

Baryon number 0

Lepton number -1

Strange -ness 0

Lepton number +1

Strange -ness 0

μ-

μ

Baryon number 0

Strange -ness 0

muon

+

Charge (e) +1

Lepton number +1

νe Lepton number -1

anti-muon Mass 2 (MeV/c ) 106

Baryon number 0

electron anti-neutrino

e+

Mass 2 (MeV/c ) 0.511

Charge (e) 0

Charge (e) -1

Baryon number 0

muon anti-neutrino Lepton number +1

Strange -ness 0

Mass 2 (MeV/c ) 0

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Charge (e) 0

νμ

Baryon number 0

Lepton number -1

Strange -ness 0

INDEX

TAP 533- 4: Positron, muon, neutrino Practical advice: This is just an outline. Your students will probably be much more forthcoming. They should be encouraged to put ideas into their own words rather than copy chunks from websites. (Some chunks are below.) It is good practice for them to fully reference their sources. It is possible that there is some conflict with different web sources; an example is given with the muon answers. Episode 534 gives the basic outline that all students should have.

Outline answers What is a positron? Anderson saw an example of anti-electrons, “positrons” in photographs of cosmic rays passing through a cloud chamber. (Pictures are provided at the site below) The discovery of the positron http://www.physics.ubc.ca/~waltham/p400/presentation/asgeirsson.pdf [Accessed 14 October 2005]

Who discovered it? A direct quote from Anderson’s 1933 paper: “On August 2 1932 during the course of photographing cosmic-ray tracks produced in a vertical Wilson chamber (magnetic field 15,000 gauss) designed in the summer of 1930 by Prof R A Millikan and the writer the track shown in fig 1was obtained which seemed to be interpretable only on the basis of a particle carrying a positive charge but having the same mass of the same order of magnitude as that normally possessed by a free electron.” Anderson Carl D, The positive electron, Physical Review, Vol 43 pp 491-494 15 March 1933 http://www.hep.man.ac.uk/babarph/babarphysics/positron.html puts the discovery as 1933 but as Anderson’s paper shows the observations were made in 1932

Who suggested it must exist? “In the 1920's, Paul Adrien Maurice Dirac, a British physicist, developed a quantum-mechanical wave equation which was able to explain the origin of the electrons magnetic moment and spin. There was only one problem, the relativistic equation would only be satisfied if there were solutions for it which corresponded to negative energy states, and in the case of the electron, an electron with a positive charge.” The above is quote from: http://www.upei.ca/~phys221/mmh/

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Another suitable source is: http://www.infoplease.com/ce6/sci/A0804257.html “The existence of antiparticles for electrons was predicted in 1928 by P. A. M. Dirac's relativistic quantum theory of the electron. According to the theory both positive and negative values are possible for the total relativistic energy of a free electron. In 1932, Carl D. Anderson, while studying cosmic rays discovered the predicted positron, the first known antiparticle.” Or “Paul Dirac: Relativity and antiparticles Paul Dirac set himself the task of repairing the big problem with Schrödinger’s wave theory: that it didn’t agree with the theory of relativity. What he found in 1928 was totally unexpected. If relativity was included the theory necessarily predicted that every particle must have a ‘mirror image’ particle, exactly the same but opposite in all properties except mass. These pairs of particles could annihilate one another. This interpretation wasn’t clear at first to Dirac, but it gradually became a second cornerstone of quantum physics. If particles can be created or destroyed they have to come in such pairs. Otherwise quantities like electric charge won’t be conserved. Soon afterwards, in 1933, the creation of pairs of electrons and antielectrons was observed by Carl Anderson. By the way, as a bonus, Dirac’s theory explained the existence of spin too.” Advancing Physics chapter 18 40T Anderson was given 1936 Nobel Prize for Physics for this discovery

What is a muon? The muon is a lepton that decays to form an electron or positron. It carries a charge of the same size as that on an electron but is much more massive, more than 2000 times that of an electron

When was it discovered? “The muon was discovered in 1937 by J. C. Street and E. C. Stevenson in a cloud chamber. The discovery was published in "New Evidence for the Existence of a Particle Intermediate Between the Proton and Electron", Phys. Rev. 52, 1003 (1937).” http://hyperphysics.phy-astr.gsu.edu/hbase/particles/muonhist.html However “Muons were discovered 60 years ago, by Carl Anderson and Seth Neddermeyer at Caltech. Muons live for about 2.2 microseconds, and often survive to ground level, before changing into electrons and invisible neutrinos. In 1947, ten years after the muon discovery, Cecil Powell's group at Bristol University discovered that the muons are produced by other particles - pions - which live for only a few hundredths of a microsecond. In this image, pions fly out from a collision in the streamer chamber.” http://www.cs.wisc.edu/~kaxiras/decay.html But muons had been predicted by Yukaya:

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Yukaya Hideka, On the Interaction of Elementary Particles 1, Proceedings of the Physico-Mathematical Society of Japan (3) 17, 48, pp 139-148 (1935). (Read 17 Nov 1934) “It seems natural to modify the theory of Heisenberg and Fermi in the following way. The transition of a heavy particle from neutron state to proton state is not always accompanied by the emission of light particles. The transition is sometimes taken up by another heavy particle”.

Who said, ‘Who ordered that?’ The muon was discovered accidentally in 1936-1937 in the hunt for a particle predicted by Yukaya. At the time it seemed to serve no purpose in particle physics. I I Rabi asked, "Who ordered that?", when he was told of it. http://www.everything2.com/index.pl?node_id=92185&displaytype=printable&lastnode_id=92185

What is a neutrino? “Three types of neutrinos are known; there is strong evidence that no additional neutrinos exist, unless their properties are unexpectedly very different from the known types. Each type or "flavour" of neutrino is related to a charged particle (which gives the corresponding neutrino its name). Hence, the "electron neutrino" is associated with the electron, and two other neutrinos are associated with heavier versions of the electron called the muon and the tau (elementary particles are frequently labelled with Greek letters, which confuses the layman)”. “Neutrinos are similar to the more familiar electron, with one crucial difference: neutrinos do not carry electric charge”.

Who suggested it must exist? “1931 - A hypothetical particle is predicted by the theorist Wolfgang Pauli. He based his prediction on the fact that energy and momentum did not appear to be conserved in certain radioactive decays. Pauli suggested that this missing energy might be carried off, unseen, by a neutral particle which was escaping detection. 1934 - Enrico Fermi develops a comprehensive theory of radioactive decays, including Pauli's hypothetical particle, which Fermi coins the neutrino (Italian: "little neutral one"). With inclusion of the neutrino, Fermi's theory accurately explains many experimentally observed results” Who discovered it? 1959 - Discovery of a particle fitting the expected characteristics of the neutrino is announced by Clyde Cowan and Fred Reines. Source for the above: http://www.ps.uci.edu/~superk/neutrino.html

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Episode 534: Antiparticles and the lepton family The purpose of this episode is to introduce the lepton family, and also to bring in the idea of anti-particles, which annihilate when they meet particles. If you have had students research the questions at the end of Episode 533, time must now be allowed for them to feed back what they have found. You may need to supplement their finding with some of the details mentioned below.

μ

Summary Student presentations: Information about leptons. (15 minutes)

e

Discussion: PET scans. (10 minutes) Student activity: Examining particle tracks. (10 minutes) Discussion: Summarising the main points. (5 minutes)

ν

e e+ e+ e+

ν μο

Student activities: Readings. (20 minutes)

μ

Student presentations: Information about leptons Your students should present their findings in response to the questions posed at the end of Episode 533. Important points to establish:

Positron e+ Dirac’s theoretical prediction of the antiparticle to the electron is too difficult to elaborate here, but Carl Anderson’s discovery of it in cosmic rays – he discovered the muon a few years later in the same way – is worth describing. The original cloud chamber track of Carl Anderson’s positron is shown in this famous photograph (Projecting this using a digital projector makes for more dramatic discussion.):

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TAP 534-1: Anderson’s positron photograph The particle is moving up the photograph. It has been slowed down by passing through the lead plate across the centre, and the curvature of the path is caused by a magnetic field. At this stage, it’s enough to say that the particle is curved more when it is slower because the particle spends longer in the magnetic field. Anderson could deduce, from the direction and magnitude of the curvature and the length of the particle track, that the particle was positive and had a mass not more than twice that of an electron. The positron was the first anti-particle discovered: since then it has been found that every particle has its antiparticle.

Muon The muon quote (‘Who ordered that?’) was from physicist Isadore Rabi – it’s whimsically supposed to be the sort of thing you say in a Chinese restaurant when you get some strange dish you don’t recognise. The muon was a problem because it had exactly the mass predicted for Yukawa’s meson (episode 534), but it didn’t undergo strong nuclear interactions at all, which the meson had to do (that was its job, after all!). It turned out to be a heavy type of electron. Like the electron, it has an anti-particle (the anti-muon, μ+) which is positively charged.

Neutrinos ν The problems with beta decay are worth describing in detail. Reactions such as carbon-14 → nitrogen-14 + β- were expected to produce beta particles with identical kinetic energy: this is what happens in alpha decay. This does not happen in beta decay; sometimes a lot of the energy seems to be missing. In 1930 Wolfgang Pauli suggested, in a famous letter to fellow physicists starting ‘Dear Radioactive Ladies and Gentlemen’, in which he wrote ‘I’ve done something terrible: I have predicted an undetectable particle’. He suggested that the ‘lost’ energy was carried away by a new particle,

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which must be chargeless and have virtually no mass. Enrico Fermi developed the theory of this new particle, which he called a neutrino, but it wasn’t until 1951 that Reines and Cowan discovered it at the Savannah River nuclear reactor. Current (2005) thought is that the mass of the electron neutrino is in the range 0 < mass < 3 eV/c2 (compare with the electron, me = 0.511 MeV/c2). Like the electron and the muon, neutrinos have antiparticles. Furthermore, there are different neutrinos associated with the electron and the muon. Because these light particles do not experience the strong force of hadrons, they form a different category of particle and given the name leptons. There are now a total of 12 leptons: the electron, the muon, and a super-heavy version called the tau (τ); a neutrino for each of these three; and six antiparticles for these six particles. The six leptons each have a lepton number of +1, while the six anti-leptons each have a lepton number of -1.

Discussion: PET scans Take a look at PET scans and how they are made. When a positron meets an electron, they annihilate to produce a pair of gamma ray photons, each of energy 511 keV. (Both the electron and the positron have a mass of 511 keV/c2.) This principle is used in medicine, in Positron-Electron Tomography (PET) scans. A radiochemical emitting positrons is injected into the body. When the chemical reaches the organ of interest, positrons emitted very soon meet electrons and annihilate. The scan reveals exactly where the radiochemical is by looking for a pair of gamma photons travelling in opposite directions. You may like to ask students how they think radiochemicals which emit positrons are made: they can be led to realise that the unstable nuclei lose positive charge when a positron is emitted, and so have too many protons. This suggests that you have to fire protons into the nucleus, which is one way this is actually done. TAP 534-2: Making PET scans

Student activity: Examining particle tracks Examining particle tracks: This can be done as class discussion with a digital projector, as suggested for the Anderson photograph of the positron track shown above, or students can work individually or in pairs, using printed copies of the images or looking at them on computer screens. If you adopt the latter approach, give a little more time for the activity: it could be a homework activity. The questions positron are intended for students late in the Post-16 level nucleus course, so you should concentrate on simple patterns: Gamma photons are not very ionising, so tend not to leave tracks in bubble chambers or cloud chambers. gamma ray electron 353

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Gamma photons of enough energy (2 × 511 keV) can produce an electron-positron pair (provided they are near nuclei at the time: don’t emphasise this point) Particles and anti-particles – protons and antiprotons in this case – can annihilate with production of energy, and also ‘new’ mass in the case of big particles In a magnetic field, charged particles follow curved paths, with opposite charges curving in opposite directions

TAP 519-3: Particle tracks TAP 534-3: Annihilation and pair production: Bubble chamber pictures

Discussion: Summarising the main points Establish the main points of this episode: The electron is one of a small family of fundamental particles called leptons, which are quite different from the nuclear particles (hadrons) of Episode 533. Particle have anti-particles, with opposite value of charge, lepton number and (by implication) baryon number and strangeness as well).

Student activities: Readings Here are a number of supplementary readings that you may care to use to broaden students’ background knowledge: The discovery of beta decay. TAP 534-5: The discovery of beta decay Three poems about particles. TAP 534-6: Three poems about particles Tracking particles. TAP 519-3: Particle tracks 354

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TAP 534- 1: Anderson’s positron photograph

“On August 2 1932 during the course of photographing cosmic-ray tracks produced in a vertical Wilson chamber (magnetic field 15,000 gauss) designed in the summer of 1930 by Prof R A Millikan and the writer the track shown in fig 1was obtained which seemed to be interpretable only on the basis of a particle carrying a positive charge but having the same mass of the same order of magnitude as that normally possessed by a free electron”.

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External reference The original reference is: Anderson Carl D, The positive electron, Physical Review, Vol 43 pp 491-494 15 March 1933

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TAP 534- 2: Making PET scans Making PET scans A pair of gamma rays are emitted in opposite directions as a result of electron/positron annihilation inside the patient

Scintillator – captures gamma ray photon and emits lower energy photons into photomultiplier tubes

signal processing

γ Photomultiplier – incoming photon creates a cascade of electrons, giving an electrical pulse output

One pair of detectors will respond almost simultaneously. This near coincidence shows that the two gamma rays came from a common source. The tiny time difference between the two signals is then used to work out where they came from along the line between the detectors.

γ

Scintillators are arranged in a grid on the inside surface of the scanner. In any short period of time many detectors will respond to gamma rays from many different annihilations inside the body. A computer produces a slice-by-slice map of activity in the brain.

γ

γ

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Practical advice This diagram is reproduced here so that you can discuss it with your class.

External reference This activity is taken from Advancing Physics chapter 17, 20O

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TAP 534- 3: Annihilation and pair production: bubble chamber pictures Pair production: The original image

A processed image, with some tracks removed, and (below) the tracks coloured by curvature:

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Both images can be interpreted by assuming two photons to be entering from the top of picture, leaving no track. One (top) has created a positron / electron pair and a ‘knock on’ electron from within an atom. The other has simply produced a positron / electron pair. To think about:

•

is there a minimum photon energy required to produce a positron / electron pair?

The pair produced in the lower event carry considerably more kinetic energy than the upper pair:

•

what feature of the picture shows this?

•

what is the physical reason for this?

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This image above shows pair production of electrons and positrons as a stream of photons enters a bubble chamber. The photons leave no track, so that the particles appear to come from nowhere. How many of these events can you identify?

Proton–antiproton annihilation

Here an antiproton (coming in from the bottom left) strikes a proton. Mutual annihilation leads to four pairs of pions (π+ and π–). These curve in opposite directions in the magnetic field. To think about:

•

the antiproton is being deflected slightly to the right. In which plane is the magnetic field?

•

can you identify the π+ and π– particle tracks?

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Practical advice These pictures may be used to illustrate annihilation and pair production. They might be printed out or made into OHP transparencies, with suitable captions. Use the questions as you see fit.

Answers and worked solutions Pair production The minimum photon energy required to produce a positron / electron pair is 2mc2. The feature that shows higher speed is less curvature. Initial photon had greater energy.

Proton–antiproton annihilation The magnetic field is directed “into the page” (NB anti-proton has a negative charge) The π+ and π– particle tracks are red and green respectively (the π - will deflect the same way as the anti-proton)

External reference This activity is taken from Advancing Physics chapter 17, display material 10S

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TAP 534- 5: The discovery of beta decay This reading is about an important discovery made in the early days of the study of radioactivity. It involves many of the most famous nuclear scientists of the day and led to an amazingly bold prediction of the existence of a small neutral particle, eventually called the neutrino. The theory also meant that physicists had discovered a hitherto unknown force in nature – the weak force. Radioactivity – the emission of mysterious ‘rays’ from certain rare elements – was discovered in 1896 by the French physicist Henri Becquerel. The emissions were soon classified in terms of how penetrating they were as alpha, beta and gamma rays. The alpha rays – soon found to be massive particles, helium nuclei – left a particular source with a definite kinetic energy that was a characteristic identifier of the source. In the early days it was assumed that beta rays, identified as high-speed electrons, also left the source with a characteristic energy. Indeed early experiments carried out between 1907 and 1914 seemed to confirm this assumption. But they were wrong: the detection relied on photographic techniques that were too insensitive to reveal that some beta particles from a particular source had less energy than others.

Constant energy produces a line spectrum Early experimenters were looking for patterns in the outputs of the various radioactive elements that were being discovered. It was possible to identify elements from the spectrum of the light they gave out when heated. It should be possible to do much the same for radioactive sources using these newer emissions. What they were beginning to realise at this time was that a good emitter like radium was changing into other radioactive elements with each emission, and it would be nice to be able to identify these from the emission they gave out. To start with it had been thought that the new elements were all forms of the original source, and they were named radium B, radium C, radium D, uranium X, etc. The idea that elements could change was against all the beliefs about atoms and elements so carefully built up in the nineteenth century. The energy of beta particles was found by making them move through a strong magnetic field before landing on a photographic plate. The slower less energetic ones were deviated more. These experiments showed that many beta emitters produced definite lines on the plate, where betas with a particular speed congregated. But the line spectra produced were ‘messy’. For some emitters it was hard to detect the betas with a discrete and definite energy against the noisy background in the exposed plates. In Berlin, Lise Meitner and Otto Hahn were two of several scientists across Europe investigating this problem. Remember that nobody realised that all these effects were due to changes in the nucleus – the nucleus had not then been discovered. The worst of all the emitters investigated by Meitner and Hahn was named radium E, which we now know is a radioactive isotope of bismuth (210Bi). In 1911 Otto Hahn (with Lise Meitner, one of the discoverers of nuclear fission – but that was yet to come) wrote to Ernest Rutherford: RaE is the worst of all. We can only obtain a fairly broad band. We formerly thought that it was as narrow as the other bands [as found in other emitters], but that is not true. It looks as if secondary … effects had a maximum influence on rays of a medium velocity like RaE. Hahn was thinking that there might be some secondary process inside the atom which somehow altered the energy distribution and smoothed things out, especially effective with low-energy (soft) beta rays. Then later: The trouble with the soft rays [from RaE] is very great and we do not feel sure that we can overpass the difficulties to obtain good lines. But even the great Rutherford at this time was not too worried, and replied:

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The continuous β-ray spectrum observed for uranium X and radium E may be ultimately resolved for a number of lines. The problem was that there were two effects happening at the same time. Electrons from beta decay do always produce a continuous spectrum – the vital fact that was to lead to so much. But several beta emitting nuclei also emit gamma rays, which like alpha particles have single definite energies. These gamma rays can then give their energy to an electron in the outer part of the atom, knocking it out of the atom so that it looks like a beta particle from the nucleus, but one with a single definite energy. It was too attractive to physicists looking for a pattern to put more emphasis on the nice line spectra than on the vague and inexplicable continuous one. And when such lines occurred they were much easier to see on the photographic plates used than was the vaguely darkened background due to the continuous spectrum. Remember also that sharp line spectra were what physicists expected to see; that was what they were used to, and what theory told them ought to happen. So they saw what they expected.

Why is a continuous energy spectrum a problem? So what was so important about the continuous spectrum anyway? It is always annoying for a scientist or a detective to discover that the interesting clues they are looking for are in fact far less important than the messy stuff that just seems to get in the way. But as the evidence for a continuous range of energies being carried away from a radioactive source grew, the very small numbers of what were to be called ‘nuclear physicists’ began to get more and more worried. The decisive evidence was coming not from hard-to-read photographic plates but from the new generation of detectors using an electrical discharge produced by the ionising radiations. In April 1914 a young 23-year-old physicist called James Chadwick was given a scholarship to do research and chose to join Hans Geiger in Berlin. Using a primitive form of what was later to be called a Geiger counter Chadwick obtained clear evidence for the continuous spectrum of beta particle energies – and not just with radium E. Chadwick was to become famous in 1932 for discovering the neutron, but choosing to go to Berlin in 1914 was not one of his better ideas. He spent the war years in an internment camp.

Is the law of conservation of energy true? This was the question raised by the continuous spectrum of beta particle energies. To avoid answering the question with a ‘NO!’, physicists had to postulate a completely new particle that had no charge and negligible or zero mass. It was not shown experimentally to exist until 1956, and required a new fundamental force in nature to explain it. The problem was simple: if some beta particles leave a nucleus with less energy than others, what has happened to the ‘missing’ energy? It was seriously considered in the 1920s that maybe the law of conservation of energy just didn’t apply to the strange world of the quantum and the relativistic equivalence of mass and energy. In 1930 the distinguished physicist Niels Bohr said in a lecture: At the present stage of atomic theory we have no argument, either empirical or theoretical, for upholding the energy principle in β-ray disintegrations, and are even led to complications and difficulties in trying to do so. Then the rather eccentric Austrian physicist Wolfgang Pauli came to the rescue with a very brave idea indeed. In December 1930 he wrote a letter to a meeting of physicists at Tubingen University: Dear radioactive ladies and gentlemen, I have come upon a desperate way out regarding … [some fairly obscure data], as well as to the continuous β-spectrum, in order to save …. The energy law. To wit, the possibility that there could exist in the nucleus electrically neutral particles, which I shall call neutrons, which have spin ½ and satisfy the exclusion principle and which are further distinct from light-quanta in that they do not

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move with light velocity. The mass of the neutrons should be of the same order of magnitude as the electron mass and in any case not larger than 0.01 times the proton mass. … The continuous β-spectrum would then become understandable from the assumption that in β-decay a neutron is emitted along with the electron, in such a way that the sum of the energies of the neutron and the electron is constant. Some theoretical calculations pointed to Pauli’s neutral particle having zero rest mass. The word neutron seemed to too big for such a tiny, if important, object. The Italian physicist Enrico Fermi started calling it the little neutral one – or neutrino. This became the accepted name when in 1932 Chadwick discovered a much larger neutral component of the nucleus that better deserved to be called the neutron.

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Practical advice This reading is designed as an extra to support the episode. It adds some personal context to the hard to accept discovery of the continuous energy spectrum associated with beta decay. It is interesting that one of the first challenges thrown up by the comparatively simple observation of a continuous beta energy spectrum should throw doubt on one of the most fundamental laws in physics, the conservation of energy.

Social and human context The problem of the continuous beta spectrum shows how even the greatest physicists of the day were confused by what was being discovered, and found it hard to accept. When the observations were accepted they seemed to shake one of the most basic laws of physics.

External reference This activity is taken from Advancing Physics chapter 17, 10T

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TAP 534- 6: Three poems about particles For many centuries, scientific ideas have inspired poets with raw material. These poems all deal with elementary particles, portraying everyday experiences and commenting on the scientific imagination.

Cosmic Gall Neutrinos, they are very small. They have no charge and have no mass And do not interact at all. The earth is just a silly ball To them, through which they simply pass, Like dust maids down a drafty hall Or photons through a sheet of glass. They snub the most exquisite gas, Ignore the most substantial wall, Cold-shoulder steel and sounding brass, Insult the stallion in his stall, And, scorning barriers of class, Infiltrate you and me! Like tall And painless guillotines they fall Down through our heads into the grass. At night, they enter at Nepal And pierce the lover and his lass From underneath the bed – you call It wonderful; I call it crass. by John Updike From: John Updike 1964 Telephone Poles and Other Poems (London: Andre Deutsch)

Nomad The particle scientist is more or less happy. He has no home. All his ladders

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go straight down and claim the nameless. by Anthony Piccione From: Anthony Piccione 1977 Anchor Dragging Poems (BOA Editions)

Little Cosmic Dust Poem Out of the debris of dying stars, this rain of particles that waters the waste with brightness; the sea-wave of atoms hurrying home, collapse of the giant, unstable guest who cannot stay; the sun’s heart reddens and expands, his might aspiration is lasting, as the shell of his substance one day will be with frost. In the radiant fields of Orion great hordes of stars are forming, just as we see every night, fiery and faithful to the end. Out of the cold and fleeing dust that is never and always, the silence and the waste to come – this arm, this hand, my voice, your face, this love. by John Haines From Bonnie Bilyeu Gordon 1985 Songs From Unsung Worlds (Boston, MA: Birkhauser)

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Practical advice Each in their own way, these poets reflect on the success of the scientific imagination. Is their aim to grasp the incredible ideas better through the language of poetry? Certainly they link astronomical and nuclear scales to the human scale of reality.

External references This activity is taken from Advancing Physics chapter 17, 20T These poems, and many others inspired by science, are published in: Carey J (ed.) 1995 The Faber Book of Science (London: Faber & Faber) Ferris T (ed.) 1991 The World Treasury of Physics, Astronomy and Mathematics (Boston, MA: Little, Brown)

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Episode 535: Particle reactions This episode considers both hadrons and leptons in particle reactions. Students must take account of both conservation of lepton number and conservation of baryon number.

Summary Student activity: Applying conservation rules. (20 minutes) Discussion: Identifying conservation rules. (10 minutes) Student questions: Questions on conservation rules. (30 minutes)

Student activity: Applying conservation rules Students should first check on the conservation of (electric) charge, baryon number, lepton number and strangeness in real reactions. They should also note that the mass/energy of products should be less/equal to the mass/energy of reactants. Use the first sheet of hadron cards from the previous episode and the four leptons from the ‘lepton cards’ document to decide which particle is needed to complete several reactions. TAP 535-1: Applying conservation of baryon and lepton number You should expect some ambiguity as to which neutrino or antineutrino is involved: after all, this ambiguity was not resolved until recently. Ask students what they would expect, from symmetry, in each case. Now add the second sheet of cards from the ‘hadron cards’ document, containing all the anti-baryons and some more mesons. TAP 535-2: Particle card student activities A quick sort of all cards should reveal that all baryons and leptons have their anti-particles, all with obvious names except electron/positron. Tell students that all mesons – those with baryon number and lepton number of zero – have antiparticles, but that some are their own anti-particles; they can then sort out which is which. Students are now able to check whether reactions can proceed according to the conservation rules met so far.

Discussion: Identifying conservation rules Invite students to sum up what they now know about particles and particle reactions. Look for the following points.

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Baryon number is always conserved (Electric) charge is always conserved Lepton number is always conserved Mass on the left hand side of the equation must be bigger than the mass on the right hand side. Strangeness may be conserved, but not always. (In weak interactions it can change by 1).

Student questions: Questions on conservation rules Here are some suggestions for questions that could be given as student exercises at this point. TAP 535-3: Things that don’t change Questions about creation and annihilation (but note that this question uses E = mc2; however, you may find that more mathematical students will accept mass values quoted as e.g. 939.6 MeV/c2 once they have done a calculation of this sort. TAP 535-4: Creation and annihilation Questions about creation from annihilation. (To reduce the demand, you may wish to delete all the text from ‘Exotic forms of matter can occur fleetingly…’ and dropping questions 7 to 11.) TAP 535-5: Creation from annihilation

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TAP 535- 1: Applying conservation of baryon and lepton number Use the hadron and lepton cards to decide which particle is needed to complete each of the following reactions. Assume that electric charge, baryon number and lepton number are all conserved in each reaction, and that the mass of the reactants cannot be less than the mass of the products. 1. n

→

2. …… 3. π+ 4. p

p +

→ →

→

n

μ+ n

e-

+

+ +

+

……

……

+

e-

……

νe

+

e+

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Answers and worked solutions 1. n

→

p

e-

+

+

νe-bar

allow any antineutrino. 2. νe

+

n

→

p

+

e-

allow any antineutrino, and of course the scattering e3. π+

+

→

n

→

μ+

+

n

+

e- is also a correct answer!

νμ

allow any antineutrino 4. p

→

n

+

νe

+ e+

allow μ+

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TAP 535- 2: Particle card student activities 1.

Sorting particles and anti-particles. Apart from the positron, all the anti-particles of baryons and leptons have obvious names. Not so for mesons! The mesons have baryon number 0 and lepton number 0. Some of them are anti-particles of others, but three of the mesons in these cards are their own anti-particles. Can you identify all these?

2.

Decay reactions. (a)

Use the particle cards to check that the following decay reactions are all possible. The rules are:

•

Mass/energy is conserved. In practice, this means that the mass on the left hand side of the equation must be more than the mass on the right hand side if the reaction is to go.

•

(Electric) charge is conserved

•

Baryon number is conserved

•

Lepton number is conserved

•

Strangeness may be conserved, or may change by 1 (The symbol γ refers to a gamma photon.)

Κ+ → μ+ + νμ Λ → p + π− μ- → e- + νe-bar + νμ Ω− → Ξ0 + π− Σ0 → Λ + γ (b)

Here are some impossible decay reactions. For each one, use the particle cards to find why the reaction is not possible. n → p + e- + νe

Δ+ → π+ + π0 Ξ0 → p + π0 Σ+ → p + K0

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Answers and worked solutions 1.

The particle – antiparticle pairs are: (π+, π−), (K+, K-), (K0, K0-bar), (ρ+, ρ -) and the particles which are their own antiparticles are: π0, η0, φ0 and J/ψ0.

2. (b) n → p + e- + νe

lepton number not conserved

Δ+ → π+ + π0

baryon number not conserved

Ξ0 → p + π0

charge not conserved

Σ+ → p + K0

mass/energy not conserved

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TAP 535- 3: Things that don’t change Checking on conserved quantities Try these questions, which require you to notice quantities that are conserved. Change, and no change Here is the equation for the reaction in which a neutron decays to a proton p, an electron e and an antineutrino ν : 1 0n

→ 11p +

0 ? e

0

+ 0 ν.

1.

How does the equation show that the total number of nucleons remains the same?

2.

How does the equation show that the total electric charge remains the same?

3.

Which symbols show that the electron and antineutrino are not nucleons, but leptons with zero nucleon number?

4.

An electron is a lepton. An antineutrino is an antilepton. Explain how in this reaction the lepton number is conserved, even though it creates two leptons.

Here is the equation for an electron interacting with a proton to produce a neutron and a neutrino: 0 ? e

+ 11p → 01n + 00 ν.

5.

How does the equation show that the total number of nucleons remains the same?

6.

How does the equation show that the total electric charge remains the same?

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7.

How does the equation show that the total lepton number remains the same? Why is this neutrino not an antineutrino?

56 49 The stable iron nucleus is 26 Fe ; ‘lightweight’ iron 26 Fe decays emitting a positron and a neutrino. 62 ‘Heavyweight’ iron 26 Fe decays emitting an electron and an antineutrino. Here are the reactions: 49 26 Fe

→

49 25 Mn

+

0 1e

+

0 0ν

62 26 Fe

→

62 27 Co

+

0 -1e

+

0 0 ν.

8.

In the first equation, what values show that the nucleon number stays the same?

9.

In the second equation, what values show that the nucleon number stays the same?

10.

How can you see from the first equation that charge is conserved?

11.

How can you see from the second equation that charge is conserved?

12.

In the first equation, the positive electron (positron) is an antilepton. How does the equation show that total lepton number is unchanged even though two are produced?

13.

In the second equation, why is an antineutrino emitted?

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14.

In both equations, how does the reaction change the chemical element involved?

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Practical advice These are intended as simple practice questions, on conservation principles for particles. The most difficult point is that an antielectron plus a neutrino or an electron plus an antineutrino do not change the total number of leptons (a lepton and an antilepton in each case).

Social and human context Particle physics involves creating meaningful categories, such as ‘lepton’ and ‘nucleon’, to form families. These come from thought, not ‘directly’ from experiment.

Answers and worked solutions 1.

The upper numbers (nucleon number) add up to the same on the left and right: 1 = 1 + 0 + 0.

2.

The lower numbers (charge number) add up to the same on the left and right: 0 = 1 + (– 1) + 0.

3.

The upper (nucleon) number is zero for both (even though an electron has some mass).

4.

The reaction produces one lepton and one antilepton. The electron has lepton number 1. The antineutrino has lepton number – 1. Together, they add to zero. The total lepton number is zero before and after.

5.

The upper numbers (nucleon number) add up to the same on the left and right: 0 + 1 = 1 + 0.

6.

The lower numbers (charge number) add up to the same on the left and right: – 1 + 1 = 0 + 0.

7.

An electron (a lepton) vanishes on the left. A neutrino (also a lepton) is created on the right. The total lepton number remains equal to 1. If the neutrino were an antineutrino the lepton number would go from + 1 to – 1, and not be conserved.

8.

The total nucleon number 49 remains constant: 49 = 49 + 0 + 0.

9.

The total nucleon number 62 remains constant: 62 = 62 + 0 + 0.

10.

The electric charge on the nucleus is decreased, the charge being carried away by the positron: 26 = 25 + 1 + 0.

11.

The positive electric charge on the nucleus increases, a negative charge being carried away by the electron: 26 = 27 + (– 1) + 0.

12.

A lepton–antilepton pair is produced, so the lepton number does not change.

13.

The antineutrino ‘cancels’ the increase in lepton number due to producing an electron.

14.

The charge on the nucleus changes. So the number of electrons around the nucleus changes, altering the chemical properties of the atom

External reference This activity is taken from Advancing Physics chapter 17, 10S

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TAP 535- 4: Creation and annihilation Using Erest = m c 2 These questions give practice in using Erest = m c 2 to calculate photon energies and masses of particles created or annihilated.

Energy and mass The energy of a particle at rest is all due to its mass. This energy is called the rest energy. The rest energy in joules of a particle with mass m measured in kilograms is given by Erest = m c 2, where c is the speed of light in metres per second. An energy in joules can be converted to electron volts by dividing by 1.60 × 10–19 J eV–1. The mass of an electron or positron is 9.11 × 10–31 kg. The speed of light is 3.00 × 108 m s–1. 1.

Show that the rest energy of an electron is 8.2 × 10–14 J.

2.

Use the answer to question 1 to show that the rest energy of an electron is 0.51 MeV.

3.

Write down the rest energy of a positron (antielectron).

4.

An electron and a positron which meet annihilate one another. By how much does the rest energy decrease in total? Express the answer in MeV.

5.

The annihilation of an electron and a positron at rest produces a pair of identical gamma ray photons travelling in opposite directions. Write down in MeV the energy you expect each photon to have

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6.

A single photon passing near a nucleus can create an electron–positron pair. Their rest energy comes from the energy of the photon. Write down the smallest photon energy that can produce one such pair.

7.

Cosmic rays can send high-energy photons through the atmosphere. What approximately is the maximum number of electron–positron pairs that a 10 GeV photon can create?

8.

24 22 The isotope 12 Mg is stable. The light isotope 12 Mg emits positrons and gamma rays including a photon of energy 1.28 MeV. How can decays of this nucleus result in both annihilation and creation of electron–positron pairs?

9.

A photon can create particle–antiparticle pairs of greater mass than electrons and positrons. Approximately what energy must a photon have to create a proton–antiproton pair? (The mass of a proton is 2000 times the mass of an electron).

10.

Why do the photons from the annihilation of an electron–positron pair not themselves go on to create new electron–positron pairs?

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Practical advice These are intended as simple practice calculations on energy changes in particle–antiparticle creation and annihilation.

Alternative approaches You might start in the context of PET scans, which use positron annihilation.

Social and human context The existence of the creation and annihilation of matter makes for a profound shift in the way the world has to be imagined. Its permanence is undermined.

Answers and worked solutions 1. Erest = mc2

(

Erest = 9.11×10−31 kg× 3.00×108 ms–1

)

2

= 8.2 ×10−14 J. 2.

Energy in eV is energy in joules divided by 1.60 × 10–19 J eV–1: 8.2 × 10 −14 J 1.60 × 10 −19 J eV −1

= 5.1 × 10 5 eV = 0.51 MeV

3.

0.51 MeV since electrons and positrons have identical mass.

4.

1.02 MeV. Two particles, each of rest energy 0.51 MeV, are annihilated.

5.

0.51 MeV each. The two photons share the total 1.02 MeV.

6.

1.02 MeV. The photon has to create the rest energy of two particles, each 0.51 MeV.

7.

Approximately 104 pairs. Each pair requires approximately 1 MeV = 1 × 106 eV and 10 9 GeV = 10 × 10 eV is available.

8.

The 1.28 MeV photon has enough energy to create an electron–positron pair. The positron from the decay can annihilate with any nearby electron.

9.

Approximately 2 GeV. The rest energy of a proton–antiproton pair is 2000 times greater than the roughly 1 MeV rest energy of an electron–positron pair.

10.

Their 0.51 MeV energy is too small (minimum 1.02 MeV).

External references This activity is taken from Advancing Physics chapter 17, 30S

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TAP 535- 5: Creation from annihilation Colliding electrons and positrons The passage below is from The New Physics, edited by Paul Davies. Read the passage, in which some phrases or words are highlighted in bold. Write a fuller explanation of what is meant by each highlighted word or phrase. Questions to guide you are offered, but you can go further if you wish.

Electron–positron annihilation Electrons colliding with positrons provide one of the most exciting ways of learning about bizarre varieties of matter…. The key feature is that positrons are the antiparticles of electrons. When matter and antimatter meet, they can mutually annihilate. The energy in their masses has been unlocked: Erest = mc2 at work. What is the point of this? Destroying the electrons and positrons is just the start. The aim is to watch what happens when their energy ‘recongeals’ into new forms of matter and antimatter. It can return whence it came, into electron and positron, but more interestingly it may produce new forms of matter with their corresponding antimatter. The hunt is on for those occasions when new forms of matter, not previously seen on Earth, emerge from the encounter. Exotic forms of matter can occur fleetingly in the heat of stars, and when we temporarily simulate that heat on Earth, so that we can capture these new varieties in earthbound laboratories. This continuous destruction of matter and antimatter was common in the brief heat of the primordial Big Bang. By annihilating electrons and positrons in the laboratory we are reproducing conditions similar to those that occurred a split second after the Big Bang. We can create matter and antimatter, built for example of quarks and antiquarks, to order. 1.

Positrons: what are they?

2.

Antiparticles: what’s the relation between a particle and antiparticle?

3.

Antimatter: what is antimatter? Where can it be found?

4.

Mutually annihilate: what happens? What vanishes and what does not vanish?

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5.

Erest = mc2: give an example of how to use this equation.

6.

It can return whence it came, into electron and positron: what is happening here?

7.

Emerge from the encounter: what emerges?

8.

The heat of stars: why does it matter how hot the stars are?

9.

Temporarily simulate that heat: how is this temporary simulation done? By heating stuff up?

10.

Primordial Big Bang: what is this?

11.

By annihilating electrons and positrons in the laboratory: how?

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Practical advice This passage comes from a book intended for a wide audience, but an audience with considerable knowledge of physics. Suggest to students that they should write enough to help a friend who is not studying physics at A-level at least to make sense of the passage. Remind them that this often means saying the same thing in one or two other ways. Examples always help, too, so encourage that.

Alternative approaches You could choose a different passage, for example from a text about medical applications of particle physics.

Social and human context New ideas often need new words. Learning physics is sometimes like learning a new language.

Example answers 1.

Positrons: are antiparticles of electrons, they have the same mass but positive charge.

2.

Antiparticles include the fact that antiparticles have all properties except mass opposite to those of their particle counterparts.

3.

Antimatter: is made of antiparticles. No bulk antimatter is known in the Universe, but antiparticles can be produced in accelerators.

4.

Mutually annihilate: the opposite properties of particle and antiparticle all cancel, and the rest energy (mass) goes into kinetic energy of a photon (which has zero mass).

5.

Erest = mc2: you could show that the mass of an electron corresponds to 0.5 MeV approximately. Or give an example from nuclear reactions.

6.

It can return whence it came, into electron and positron: a photon (near a nucleus) can create an electron–positron pair.

7.

Emerge from the encounter: pairs of particles and antiparticles emerge.

8.

The heat of stars: because the stars are hot, particles have a lot of random thermal energy per particle. This may be enough to cause an event in a collision.

9.

Temporarily simulate that heat: accelerated particles are collided, so that their kinetic energy is available in the collision event.

10.

Primordial Big Bang: the hypothetical origin of the Universe, when space and time were created and a hot fireball of space, time and matter exploded.

11.

By annihilating electrons and positrons in the laboratory: the best way is to make accelerated particles collide head on, to get the maximum energy into the products of the collision.

External references This activity is taken from Advancing Physics chapter 17, 50C

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Episode 536: Vector bosons and Feynman diagrams You need to check your own specification here for details of what students will need to do in examinations, and to look at past papers: although Feynman diagrams give clarity to particle interactions, they are not required by all specifications. e

Summary

-

e

-

Demonstration: Exchange particles. (5 minutes)

virtual photon

Discussion: Interactions of different types. (15 minutes) Demonstration: Model Feynman diagrams.

e

(15 minutes)

-

e

-

Discussion: Rules for Feynman diagrams. (10 minutes) Student activity: Constructing Feynman diagrams. (20 minutes)

Demonstration: Exchange particles Yukawa’s theory of an exchange particle to explain repulsive and attractive forces in nuclei is worth demonstrating with two students and a football or other large object. If two students throw (gently) a heavy object such as a schoolbag or football to each other, each will report feeling an outwards force both on throwing and on catching (Why? Conservation of momentum). If the rules are changed so that, instead of throwing, each student pulls the object from the other’s hands in turn, then each will report feeling an inwards force both on gaining and on losing the particle.

Discussion: Interactions of different types This crude model in the demonstration above will illustrate the idea of an exchange particle originated by Yukawa, who suggested that a nuclear exchange particle (it turned out to be the pion) could explain the strong interaction between protons and neutrons. In the last episode, it will be clear that a similar fundamental exchange works at a level that is more fundamental than mesons and baryons. The electromagnetic interaction, which consists of just the well-known attractions and repulsions of static electricity (pre-16 level), is a different interaction, much weaker than the strong interaction. Here the exchange particle is the photon. The weak interactions, which are harder to classify, and are similar in strength to the electromagnetic interactions, are associated with changes in the nature of particles.

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Demonstration: Model Feynman diagrams Feynman diagrams can be introduced via a physical model that can be twisted to show different interactions. The key aspects – direction of time, transfer of the force-carrying boson, difference between particles and anti-particles – can be quickly illustrated for an electromagnetic interaction. As an example of these points (including the last), you may wish to use a simple physical model. It is quick and easy to use cheap coat hangers linked by their hooks, with triangles of card attached midway across the ‘shoulder’ of each. The supporting ‘shoulders’ of the coat hangers are the interacting particles, while the interlocked hooks constitute the vector boson. With one twist each time, it is possible to go from electron-electron interaction to electron positron interaction to positron-positron interaction to electron-positron annihilation. For these electromagnetic interactions, the particle exchanged is a photon. For the weak interaction, there are three particles, depending on the changes in charge taking place. If you deal with quark interactions later, the exchange particle is the gluon. Some teacher notes: TAP 536-1: Feynman diagrams TAP 536-2: Coat-hanger Feynman diagrams

Discussion: Rules for Feynman diagrams If your specification requires Feynman diagrams, you will need to emphasise the rules for drawing them. These are not consistent from source to source! In this episode, the following conventions are followed. Time goes vertically up the diagram (many sources have time horizontal). Side-to-side displacement in the diagrams has no meaning other than to show separate particles. If two paths are heading outwards, it does not imply that particles are repelling each other. Particles are shown by ‘normal’ arrow-heads, while anti-particles are shown by reversed arrow-heads (remember that the direction of time is upwards), so a collision between a proton and an anti-proton can be represented as:

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From any vertex, such as the collision point of the proton and anti-proton, a boson can be drawn. This can be a photon (wavy line), a weak interaction boson (a dotted line) A Feynman diagram – certainly the simple ones in this episode – can be pivoted about any of the vertices to produce another valid diagram.

Student activity: Constructing Feynman diagrams Students are supplied with cards from which they can construct Feynman diagrams. They use the different ‘left-hand sides’ of the diagrams with the single vector boson and the appropriate ‘right-hand side’ to produce the different possible weak interactions, and then to label the boson with W+, W- or Z0 as appropriate. TAP 536-3: Feynman diagram student activities TAP 536-4: Feynman weak interaction cards

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TAP 536- 1: Feynman diagrams Teacher notes A Feynman diagram is not a picture in space of the paths of actual particles. It is a picture of the structure of one of the terms that contributes to the total “quantum amplitude” for a process. A ‘process’ here is an event in which definite particles enter and definite particles leave. A ‘quantum amplitude’ is a measure of the likelihood of the process happening. For example, scattering of a pair of electrons has two electrons enter and two leave. Here are some of the terms that Feynman diagrams keep track of and all of which have to be taken into account to obtain the final amplitude: indistinguishable

No photon exchanged. The amplitude for these terms just contains an expression for an electron propagating from one place to another. Here we have included terms obtained by interchanging the indistinguishable electrons. Both have to be counted in. indistinguishable

One photon exchanged. The expression for the amplitude is now a product of four expressions for electrons going from place to place, one expression for a virtual photon going from place to place, and two couplings of electron to photon.

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A further term (omitting ones which involve interchanging electrons). Here an electron emits a virtual photon and absorbs it again.

e+ e–

Photon emits an electron-positron pair, which then recombines. This term has four couplings of electron (or positron) with a photon. Each introduces a factor of the electronic charge into the product giving the amplitude. Doing the adding up The expression for each diagram has to be integrated over all space-time positions for the particles and where they interact. They are also integrated over all possible energies and momenta. This gives a total quantum amplitude (‘arrow’) for each diagram. Finally, the arrows for all computed diagrams are added to give a final arrow.

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Practical advice We strongly recommend that you read Feynman’s ‘QED: The Strange Theory of Light and Matter’, particularly chapters 3 and 4. It gives a very clear picture of the nature of Feynman diagrams, and the calculations made using them, without technical detail. A careful reading will amplify several of the teaching notes above

External reference This activity is taken from Advancing Physics chapter 17, further teaching notes

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TAP536- 2: Coat-hanger Feynman diagrams One way of showing Feynman diagrams is by using coat-hangers and cardboard triangles.

Electron-electron interaction

Electron-positron interaction

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Positron-positron interaction

Electron-positron annihilation

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TAP 536 - 3: Feynman diagram student activities This is the Feynman diagram for an electron interacting with another electron by the electromagnetic interaction.

In a Feynman diagram, the electromagnetic interaction is shown by the exchange of a photon. The ‘left hand’ electron emits a photon, and changes direction. The ‘right hand’ electron absorbs the photon, and also changes direction. In a Feynman diagram,

•

Time goes upwards. (This is not consistent. Some particle physicists prefer to have time going sideways.)

•

Particles are shown by arrows going upwards

Antiparticles are shown by arrows going downwards, so the electromagnetic interaction between two positrons (the anti-particles of electrons) is

The outwards movement after emitting or absorbing a photon just shows a change in direction. It does not mean that they particles are repelling each other. As an example, the attraction between an electron and a positron is

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The photon (wavy line) in the electromagnetic interaction is called a vector boson.

•

Vector – carries the interaction

•

Bosons – the type of particles that carry interactions.

In the weak interaction, there are three vector bosons: the W+, W- and Z0 bosons. On Feynman diagrams, these are shown by dotted lines. The first two weak interaction bosons are charged, and are emitted when the ‘left hand’ particle changes its charge. If a proton turns into a neutron, for example, it emits a W+.

You have the following cards Four ‘left hand’ cards, similar to the electron in the third example above One ‘middle’ card to serve as the vector boson (dotted) Four ‘right hand’ cards, similar to the similar to the positron in the third example above. Three vector boson labels: W+, W- and Z0 to put on top of the ‘middle’ card. 1.

Use these cards to construct four possible weak interactions. Make sure that your reaction conserves charge as well as baryon number and lepton number.

2.

Label each diagram with one of these titles:

3.

•

Beta-minus decay

•

Beta-plus decay

•

Electron capture

•

Neutrino-neutron collision For each one, write the equation.

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Answers and worked solutions 2.

Beta-minus decay

Beta-plus decay

Electron capture

Neutrino-neutron collision

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It is worth emphasising how the diagrams show conservation of baryon number (upward arrow head before and after the vertex on the baryon path), conservation of lepton number (upward arrow head before and after the vertex for the electron capture and the neutrino-neutron collision, or no lepton becoming and upward and a downward arrow after for the two beta decays) and conservation of charge, including the vector boson, reading from left to right. 3.

n → p + e- + νebar p → n + e++ νe p + e- → n + νe

νe + n → n + νe

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TAP 536 - 4: Feynman weak interaction cards

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Episode 537: Preparation for the deep scattering and quarks topics Episodes 538 and 539 of this topic give clear evidence for the size of the nucleus, and for the fact that nucleons are not fundamental particles but contain different parts. This leads onto Gell-Mann & Zweig’s quark model.

Episode 538: Electron scattering Episode 539: Deep inelastic scattering Episode 540: Quarks and the standard model

Main aims Students will: 1. Know that Rutherford’s experiment, using alpha particles, cannot probe the nucleus because the alpha particles will interact with the nucleus by the strong nuclear force. 2. Know that electrons, being leptons, do not ‘feel’ the strong nuclear force, and so can probe the nucleus. 3. Use electron wavelength and scattering data to calculate the size of the nucleus. 4. Understand that the complex scattering from a nucleus reveals that nucleons are not simple points, but are themselves composed of smaller particles. 5. Describe how hadrons are made from two or three quarks. 6. Deduce the properties of a hadron from the properties of its constituent quarks. 7. Draw Feynman diagrams involving quarks and gluons.

Prior knowledge Rutherford’s experiment, diffraction, and the quantum nature of electrons (‘wave-particle duality’). If you have not covered diffraction already, you will have to modify the suggested approach to take account of this.

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Episode 538: Electron scattering This could follow work on Rutherford scattering, where alpha particles are used to identify the nucleus as the region of the atoms containing all the positive charge and most of the mass. Calculation of closest approach of alpha particles gives an estimate of nuclear size, but more direct experimental evidence is given by scattering of electrons.

Summary Student experiment: An optical analogue. (10 minutes) Discussion: Diffraction graphs. (10 minutes) Worked example: Calculating nuclear diameter. (10 minutes) Student questions: Further examples. (20 minutes)

Student experiment: An optical analogue When waves pass through a collection of spheres, which ‘look’ like circles to the oncoming waves, they diffract around them in the same way as through holes of the same size. This means that they produce a diffraction pattern with a minimum at roughly the angle given by the single-slit diffraction equation sinθ = λ/d (it should really be sinθ = 1.22λ/d for a circle; you could use that form, but the familiar equation is close enough here). Look at a point source of light (a small circuitry lamp is fine) through a pinhole made in aluminium foil – the smaller the hole the better, so use a fine sewing needle or else pull thin copper wire until it snaps, and push the work-hardened tip through the aluminium foil. A diffraction pattern consisting of a circle with a surrounding ring (more than one if you’re lucky) should be visible. If you look at the point source through a microscope slide dusted with lycopodium powder (cornflour will work, but is not as good, because the particle sizes vary more) you see a similar pattern. You may like to refer to the halo seen around the Moon in the winter, which is due to similar scattering by ice crystals in the upper atmosphere. The point here is that waves encountering round (spherical) objects will give a diffraction pattern, involving a central maximum, and then a minimum before the secondary maximum of the surrounding ring.

Discussion: Diffraction graphs The graph below compares a typical electron diffraction pattern with the Rutherford alpha -scattering pattern and the single slit pattern. Show this graph, and identify the differences.

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TAP 538-1: Rutherford alpha -scattering pattern and the single slit pattern The electron scattering clearly shows features of both Rutherford scattering (due to electrostatic interaction: the fact that the force is attractive rather than repulsive is not important) and diffraction. Concentrating on the minimum, it suggests that the nucleus is behaving as a circular object of diameter d.

Worked example: Calculating nuclear diameter The nuclear diameter d can be calculated from the electron scattering data. If your students can cope with the maths, you can work through a sample calculation. (The virtue of the approach is not that students should be able to reproduce the mathematics, but that they should be able to see that a systematic application of the mathematics related to diffraction and the wave nature of the electron can give the size of the nucleus.) Electron beam energy = 100 MeV = 100 × 106 × 1.6 × 10-19 J = 1.6 × 10-11 J Momentum p = E/c = 1.6 × 10-11/3.0 × 108 = 5.3 × 10-20 kg m s-1 de Broglie relationship λ = h/p = 6.6 × 10-34/5.3 × 10-20 = 1.2 × 10-14 m The first diffraction minimum occurs at about 22°, so using the single slit diffraction equation sinθ = λ/d we have d = λ/sinθ = 1.2 × 10-14 m /sin (22°) = 3.3 × 10-14 m

Student questions: Further examples A structured question will allow your students to follow through this logic for themselves. It should be covered here if you did not do so in Episode 522.

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TAP 522-6: Electrons measure the size of nuclei.

TAP 538-1: Rutherford alpha -scattering pattern and the single slit pattern

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Practical advice The graph is reproduced here for use in the classroom.

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Episode 539: Deep inelastic scattering At higher electron energies, two things happen:

•

The electrons penetrate deeper into the nucleus and scatter off sub-units within protons and neutrons.

•

The electrons lose energy (they undergo inelastic collisions); this energy is ‘converted’ into mass as a jet of pions is produced.

Hence higher energy electrons give us information about the structure of nucleons.

Summary Student experiment: Analogue of electron scattering by quarks. (20 minutes) Discussion: Deductions from electron scattering. (10 minutes)

Student experiment: Analogue of electron scattering by quarks Magnets are concealed in a box. These represent ‘charges’ which are probed using a freely-suspended magnet. The important point here is that each of the ‘charges’ represented by the magnet poles will affect the trajectory of the target ‘charge’ of the moving magnet pole in a more complex way than a single one would. TAP 539-1: Probing arrangements

Discussion: Deductions from electron scattering Students should appreciate that the necessarily complex analysis of particle paths from deep electron scattering indicates that the neutron and proton are not simple point charges but contain simpler structure within them. Electrons must be given very large kinetic energies to penetrate nuclei. Those energies can be sufficiently great that some of the energy can be converted (via E = mc2) into the mass of new particles. This results in the electrons losing energy – this is why the scattering is inelastic – to produce pions. Jets of pions are typical of the ‘events’ seen in particle accelerators colliding protons and anti-protons. At this stage you could ask the open question: if, as indicated by deep inelastic scattering, neutrons and protons are each made of three particles called quarks, what’s the smallest number of quarks you need? Obviously you could have any number of quarks, but there must be more than one, or else neutrons and protons would not be different. The simplest model would have two different types.

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Assuming there are just two types of quark, then possibilities could be AAA and BBB (chargeless A and +e/3 for B) or AAB and BBA, which is actually correct, with A being d (-e/3) and B being u (+2e/3). The latter (correct) version also explains other particles.

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TAP 539-1: Probing arrangements Trajectories show structure Probing unknown regions of space by tracking probes can show what lies in the volume of space. If a charged probe is used then expect to be able to tell where the charged particles are in that space (or perhaps were – the nature of the interactions will depend on the mass of the target charges, as well as the probes). A further potential difficulty is that the charges within the probed volume may not themselves be stationary, even before the probe arrives. And often you can usually only see where the probes enter and leave the volume, not track them through the volume. However, there are usually few alternatives available. Highly inferential knowledge gained in precisely this way forms a major part of what is known about the subatomic world. Here the north-seeking poles of magnets are used, to model repulsive interactions.

You will need: 9

target box

9

three bar magnets

9

cylindrical magnet mounted on 1 m of dowel

9

retort stand, boss and clamp

A model of high-energy collisions Set up the probe to swing freely. Set up a number of magnets in the target volume of space:

height of magnets

15 cm 15 cm

Then challenge a partner to discern what is in that space – using the probe! Keep it to fairly simple distributions.

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Suspend the probe so that the centre of the swing is over the centre of the box. Place the aiming card over the centre of the box. Then probe. Make a good model – don’t cheat!

In probing you might try: 1.

Altering the initial kinetic energy of the probe.

2.

Altering the aiming error of the probe.

Suggest reasons for your choice of what is under the box. Change roles and try again.

You have 1.

Used a model of high-energy probing of a volume of space.

2.

Seen the effect of probe energy on the ability to resolve what is in the space.

3.

Found some of the difficulties in interpreting trajectories to infer the existence of particles.

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Practical advice The apparatus is simple enough, and variations may suit local conditions. The box will need to be of a suitable inside height to allow the magnets to stand inside the box, just reaching to the top.

External reference This activity is taken from Advancing Physics chapter 17, 70E

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Episode 540: Quarks and the standard model The quark model, justified by the results of deep inelastic electron scattering, creates relative order out of the chaos of particle classification. Quarks have three ‘colour’ charges, and the rule for stability is that combinations must be colourless.

Summary Discussion: Rules for quarks. (15 minutes) Student activity: Making non-strange hadrons with quark triangles. (15 minutes) Student questions: Making strange hadrons with quark triangles. (15 minutes) Discussion: Gluons and the force between quarks. (15 minutes) Student activity: Constructing Feynman diagrams. (20 minutes) Discussion: Summing up. (10 minutes)

Discussion: Rules for quarks Students need to be informed of the rules for combining quarks. Where the electromagnetic interaction is due to the property of electric charge, which can be positive or negative, quark interaction is due to a property which can have three different states. This has been called ‘colour’ charge, because the three primary colours red, green and blue add to give white, a colourless combination. These three-quark combinations are the baryons. Anti-quarks have an anti-colour; you may wish to think of the complementary colours to the three primary colours, so anti-red is cyan, anti-green is magenta and anti-blue is yellow. Three anti-quark combinations are the anti-baryons. By combining a quark with an anti-quark of the appropriate anti-colour, a two-quark hadron can be produced. These two-quark combinations are the mesons.

Student activity: Making non-strange hadrons with quark triangles The quark triangle are constructed so that they can be fitted together in threes, with the 120° vertices together, with the combination red, blue and green giving a baryon. A similar arrangement with anti-red, anti-blue and anti-green gives an anti-baryon. By taking one quark and fitting an anti-quark of the appropriate anti-colour alongside it so that the two long sides coincide, a meson can be constructed. TAP 540-1: Quark triangles TAP 540-2: Quark models

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The document contains two versions of each anti-quark: one version uses the same colours as the related quarks, but with the white and coloured regions of the triangles reversed. The second version has the same pattern of colours on the triangles as the related quarks, but the complementary colours are used instead of red, blue and green: thus anti-red is cyan, anti-blue is yellow and anti-green is magenta. You can choose whichever form you prefer! Using only u (+2/3 e) and d quarks (- 1/3 e) and their anti-quarks, u bar and d bar, in all possible colours, students can quickly use the three-colour rule to construct four possible baryons (n, p and the unstable Δ− and Δ++) together with their anti-particles. Other non-strange baryons are high energy states of these four, e.g. the Δ and Δparticles are uud and udd respectively, and can decay to the proton and neutron by emitting the extra energy as a gamma photon Δ+ → p + γ

Δ → n + γ

They should also be able to construct four different mesons, all pions. Two of these are, in fact, the same particle (u + u bar = d + d bar = pi zero). Issue the strange quarks, s and s bar, at this point, with the explanation that they are heavier versions of d and d bar, and that they have strangeness of -1 and +1 respectively. This allows the construction of all the particles in the baryons decuplet and meson octet.

Student questions: Gluons and the force between quarks These questions use the u and d quark triangles and their anti-quark triangles. TAP 540-3: Putting quarks together Further questions take a similar approach to explain the baryon decuplet and meson octet. These questions can be used to structure the student activity, or else as homework to consolidate the learning afterwards. TAP 540-4: Strange quarks

Discussion: Gluons and the force between quarks Just as pions (Yukawa’s mesons) are the particles that bind baryons, the quarks in a baryon are bound by exchanging a particle. The particles here are gluons, and we envisage the transfer as exchanging the colour of the two quarks concerned: e.g. a red quark will change into a blue quark by emitting a red-antiblue gluon, which is then absorbed by a blue quark that becomes red.

Student activity: Constructing Feynman diagrams

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If your specification requires it, the Feynman diagrams used for the weak interaction can now be extended to the strong interaction, and the ‘n’ and ‘p’ of the weak interactions replaced by‘d’ and ‘u’ quarks respectively. TAP 540-5: Feynman diagrams with quarks

Discussion: Summing up This is a conclusion to the entire particle physics sequence. It should be realised that normal matter we see around us consists of two quarks and two leptons, and their anti-particles. Larger mass versions, which are not stable, occur in two generations. The muon and strange quark have already been met, and, together with the muon neutrino and the heavier version of the u quark, the charm quark (c), form the second generation. There is only one more generation, so all matter, whether stable or not, can be described in terms of 6 quarks and 6 leptons.

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TAP 540- 1: Quark triangles TAP 540-2: Quark models

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External reference This activity is taken from Advancing Physics chapter 17, 60P

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TAP 540- 2: Quark models Hadrons and quarks This simple physical model allows construction of baryons (hadrons containing three quarks) using the basic up (u) and down (d) quarks or antiquarks (the same letters with bars above them – here called ubar or dbar quarks) to form equilateral triangles containing red, green and blue. The antiquarks have the same colours in opposite arrangement on the models. Quarks and antiquarks can combine to give mesons (hadrons containing two quarks). Quarks aren’t really 'coloured' of course. But each has a property that is an analogue of colour with a rule that says that to make a valid particle these properties must combine to give a colourless (white) result. No one can see these colours – or even investigate the properties of single quarks because they can’t exist on their own. The colours have to be worked out 'backwards' – from seeing what combinations actually exist in nature to make hadrons. Historically, the names baryons, mesons and leptons are descriptions of observed particles in terms of their masses. Baryons are heavy, leptons are light, and mesons are in-between. It was later realised that leptons (electrons and neutrinos) were fundamental particles, while both baryons and mesons were combinations of smaller particles — quarks. This led to the introduction of the term hadrons (‘dense particles’) for both baryons and mesons.

You need 9

Triangular models of the six possible u and d quarks and their six possible antiquarks. Each triangle labels the quark (u or d), whether it is a quark or an antiquark, gives its charge and its colour.

TAP 540-1: Quark triangles

Rules of the game: building up hadrons from quarks When quarks combine to form stable matter they have to satisfy two basic conditions:

•

the net electric charge must be zero or a whole number of the electronic charge e;

•

the final colour must be 'white'

Physicists knew that three primary colours make white: red, blue and green. Each of the primary colours has it complementary colour, which when added also makes white: yellow and blue make white because yellow is a combination of red and green. Yellow is sometimes called antiblue for this reason. The other two complementary colours are magenta (antigreen) and cyan (antired). This means that to make a stable combination of quarks we have one of three possibilities. 1.

Three quarks of different primary colours, one red, one blue and one green.

2.

Three antiquarks of different complementary colours, one antired, one antiblue and one antigreen. (The antiquarks each have a bar above the letter u or d)

3.

A quark and an antiquark of the primary colour and its complement: e.g. red and antired.

The three-quark and three-antiquark combinations each form triangles with circles at their centres. The quark–antiquark combinations are made by placing the long edges together.

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What to do 1.

Produce three-quark combinations (baryons) with charge 0 (a neutron) and with charge +e (a proton).

2.

What are the charges of the antiparticles to these two baryons?

3.

Are there any other different combinations of three quarks? What is the resulting charge of each hadron?

4.

Now produce quark–antiquark combinations (mesons) with charge 0, +e and –e.

5.

Can you make or think of any other different combinations of quark and antiquark that make viable hadrons? Do they actually exist – or have they been discovered in experiment?

You have seen that 1.

Deep inelastic scattering experiments showed that the proton and neutron have a substructure – they are made of smaller particles. With this simple model it can be seen that different combinations of two quark ‘building blocks,’ with charges of +2/3 e and –1/3 e, give the expected proton and neutron.

2.

The quark–antiquark combinations give the pions (π -mesons). Two of the possible combinations give the neutral π0, while the other two combinations give the π + and π – particles.

3.

Other combinations of three quarks are possible: the ddd combination, with charge –e and uuu with charge +2e are predicted. These particles are the unstable Δ− and Δ++.

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Practical advice The six possible u and d quarks and their six possible antiquarks are best printed on card and laminated. Teachers may choose between two versions of the antiquarks: in one, each antiquark has the same colours as the quark but in the opposite sections of the triangle, whereas in the other, the same section is coloured but with the complementary colour, cyan for red, magenta for green, and yellow for blue. These could be backed if wished (each antiquark on the reverse of its quark), which does emphasise their complementary nature, but it will restrict the production of mesons, so extra sets will be needed. Note that there is a choice of two antiquark styles as detailed in the requirements section above. Have handy a table of discovered particles so that students may check what it is they have constructed. Although this model can readily be extended to strange and charmed hadrons, this is likely to extend it beyond its value, and to give undue emphasis to a ‘stamp-collecting’ approach to particle physics. A more useful extension could be to explain the creation of pions in high energy interactions via the creation (ΔE = Δmc2) of quark–antiquark pairs, e.g. in the decay Δ++ → p + π+ where (u+u+u) becomes (u+u+d) + (dbar+u). This would involve starting with a hadron, and adding an extra (quark + its antiquark) pair to make new combinations using the colour combination rules.

External resources Useful additional information can be gleaned from PPARC materials. http://www.pparc.ac.uk/ Frank Close 1983 The Cosmic Onion (London: Heinemann)

External reference This activity is taken from Advancing Physics chapter 17, 60P

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TAP 540- 3: Putting quarks together Three quarks for Muster Mark The American physicist Murray Gell-Mann gave the name ‘quarks’ to the particles he proposed as the basic building bricks of other particles. The name refers to a line in the novel Finnegans Wake by James Joyce, who was famed for his word play. The line is: ‘Three quarks for Muster Mark’. His colleague George Zwieg wanted to call the particles ‘aces’, but Gell-Mann’s choice won out. In spite of the rhyme suggested by Joyce’s line, the word ‘quark’ is generally pronounced ‘quork’. The word is also German slang for ‘nonsense’ and the trade name for a type of yoghurt! These questions ask about how quarks go together to make other particles.

Two kinds of quark The simplest particles, including all the ones that everyday matter is made of, are built from two kinds (‘flavours’) of quark: ‘up’ and ‘down’. The most peculiar thing about them is that their electric charges come in multiples of 1/3 of the charge on an electron. On a scale where the charge on an electron is –1e, with e = 1.6 × 10–19 C, the charges on the quarks are:

•

Up quark u: charge + 2/3 e.

•

Down quark d: charge – 1/3 e.

Making massive particles Relatively massive particles like the proton and neutron are made of combinations of three quarks. 1.

What is the charge on the combination uuu?

2.

What is the charge on the combination uud?

3.

What is the charge on the combination udd?

4.

What is the charge on the combination ddd?

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There are four compound particles here. 5.

Which combination has the right charge to be a proton?

6.

Which combination has the right charge to be a neutron?

7.

There is a particle called the Δ– which has a charge of –1e. Which quark combination could be the Δ–?

8.

There is a particle called the Δ++ which has a charge of + 2e. Which quark combination could be the Δ–?

9.

A neutron can be changed to a proton if one quark changes ‘flavour’. What change is needed? What charge must be carried away if this happens?

Making mesons Other, lighter ‘middle-weight’ particles called mesons can be made from pairs of quarks. But they have to be made from a special combination: a quark and an antiquark. There are now four particles to play with:

10.

•

Up quark u: charge +2/3 e

•

Down quark d: charge –1/3 e.

•

Antiup quark u : charge –2/3 e.

•

Antidown quark u : charge + 1/3 e. What is the charge on the combination u u ?

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11.

What is the charge on the combination d d ?

12.

What is the charge on the combination u d ?

13.

What is the charge on the combination d u ?

14.

Which combination could be the π+ meson?

15.

Which combination could be the π– meson?

16.

Which could be the neutral π0 meson?

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Practical advice These practice questions run through the various quark combinations, and the quark–antiquark combinations. They are intended to help students get familiar with how the combinations work, so that only integer multiples of the electronic charge appear.

Alternative approaches Perhaps a game with cards could be devised?

Social and human context It was extremely difficult for physicists to accept that there could be fractional charges. The origin of the number 3 in particle physics (number of colours, number of generations) is still far from clear.

Answers and worked solutions 1.

uuu has charge

+ 32 e + 32 e + 32 e = +2e. 2.

uud has charge =

3.

e + 32 e − 31 e = +1e.

udd has charge

= 4.

2 3

2 3

e − 31 e − 31 e = 0.

ddd has charge

− 31 e − 31 e − 31 e = −1e. 5.

Proton has charge +1e so could be uud.

6.

Neutron has charge 0 so could be udd.

7.

Δ– has charge –1e so could be ddd.

8.

Δ++ has charge +2e so could be uuu.

9.

A down quark, charge –1/3 e changes to an up quark, charge +2/3 e, so the charge carried away must be –1 e if charge is to be conserved.

10.

u u has charge +2/3 e –2/3 e = 0.

11.

d d has charge –1/3 e + 1/3 e = 0.

12.

u d has charge +2/3 e + 1/3 e = 1e.

13.

d u has charge –1/3 e – 2/3 e = –1e.

14.

The π+ with charge + 1e could be u d .

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15.

The π– with charge – 1e could be d u .

16.

The neutral π0 could be either or both of u u and d d . In fact it is an equal mixture of the two.

External reference This activity is taken from Advancing Physics chapter 17, 110S

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TAP 540- 4: Strange quarks An early classification of strange baryons by Murray Gell-Mann and Yuval Ne’emen gave this arrangement, called the baryon decuplet. The diagonal rows show baryons of the same charge. The Δ0 and Δ+ particles are more massive versions of the neutron and proton respectively

Δ-

Δ0 Σ-

Δ+ Σ0

Ξ-

Δ++ Σ+

Ξ0

strangeness 0 strangeness -1 strangeness -2

Ω-

strangeness -3

The Ω- was, in fact, predicted by Gell-mann from a gap in this pattern in much the same way as Mendeleyev predicted missing elements from gaps in his table. The subsequent discovery of the Ωconfirmed that particle physicists were on the right track with this classification, which led Gell-mann and Zweig to the quark theory. The strange quark s is a more massive version of the down quark d, and has the same charge (-e/3). The presence of a strange quark gives a baryon or meson a strangeness of -1. 1.

The Δ0 and Δ+ particles are more massive versions of the neutron and proton respectively. Underneath each particle in the first row, write the three quarks that make up that baryon.

2.

The strange quark s is a more massive version of the down quark d, and has the same charge (-e/3). How many strange quarks must there be in

(a)

the Σ particles?

(b)

the Ξ particles?

(c)

the Ω− particle?

3.

Write under all the strange baryons the three quarks they must contain.

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The strange mesons fit a similar pattern: K0

π-

K+

π0 K-

strangeness +1

π+ K0

strangeness 0 strangeness -1

4.

If the strange quark s has a strangeness of -1, what is the strangeness of its anti-quark, s-bar?

5.

Each of these seven mesons consists of a u, d or s quark and a u-bar, d-bar or s-bar antiquark. Write down under each meson the quark and antiquark it contains.

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Answers and worked solutions An early classification of strange baryons by Murray Gell-Mann and Yuval Ne’emen gave this arrangement, called the baryon decuplet. The diagonal rows show baryons of the same charge. The Δ0 and Δ+ particles are more massive versions of the neutron and proton respectively. 1. and 3.

Δ-

Δ0

Δ+

Δ++

ddd

udd

uud

uuu

Σ-

Σ0

Σ+

dds

uds

uus

Ξ-

Ξ0

dss

uss

strangeness 0

strangeness -1

strangeness -2

Ωstrangeness -3

sss

2. (a)

1

(b)

2

(c)

3

4.

+1

5. K0

K+

d s-bar

u s-bar

strangeness +1

π−

π0

π+

d u-bar

d d-bar

u d-bar

strangeness 0

u u-bar s-sbar K

–

s u-bar

K0 s d-bar

strangeness -1

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TAP 540-5: Feynman diagrams with quarks Feynman diagrams can be used for strong interactions as well as weak and electromagnetic interactions. Quarks interact by the exchange of bosons called gluons. On Feynman diagrams, these are shown by curly lines. When a quark emits a gluon, it changes colour, and the quark absorbing it changes colour in the opposite way. Thus a red quark can change into a blue quark by emitting a red - anti-blue gluon. When a blue quark absorbs the red - anti-blue gluon, it turns into a red quark. Use the quark and gluon cards to construct Feynman diagrams for the interactions binding quarks together in baryons.

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Answers and worked solutions Feynman quark solutions The key point is that, taking this first solution as an example, a red-antiblue gluon carries the colours red and antiblue, so on leaving a red quark it carries away the red colour and leaves the absence of antiblue, i.e. blue, so the red quark it has left become blue. When it reaches a blue quark, the antiblueness of the gluon cancels the blueness and the redness of the gluon makes the target quark red.

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一些历史反应堆照片：Below are several schematic diagrams of a Boiling Water Reactor Pressure Vessel. This unit is about four stories all:

The above image is from a Nuclear Mafia web site.

-- The above image is from: The Silent Bomb: A Guide to the Nuclear Energy Controversy, Edited by Peter Faulkner, page 85,Vintage Books, NY, July, 1977 (colorization by this author).

Below is a schematic drawing of a GE BWR:

-- The above image is from: The Silent Bomb: A Guide to the Nuclear Energy Controversy, Edited by Peter Faulkner, page 283,Vintage Books, NY, July, 1977 (colorization by this author). The reactor cores remain radioactive for about a million years after their brief use in the reactor (two to six years). The waste must be carefully isolated from humans and other living things during this entire time. NO EXCEPTIONS. ----------------------------------------------------------

Below are schematic diagrams for PWR and BWR reactors, the two types used in the United States:

-- The above image is from: The Bridgeport (Connecticut) Telegram, April 30th, 1980, page 38 (colorization by this author). Many ideas for how to safely store the waste have been suggested, but none have been successfully implemented, and none solve the problem of transportation accidents along the way to the final repository, whatever it might be.

-- The above image is from: Northeast Utilities (Connecticut Yankee) Information Brochure, June, 1973, pages 10-11 (adjusted for the Internet and additional colorization by this author). ----------------------------------------------------------

-- The above image and subtitle is from a magazine article about nuclear power, probably 1956. (Title colorized by this author; juxtaposition was apparently unintentional.) ----------------------------------------------------------

Here are the reactor designs shown in the article from the 1950's:

-- The above images are from a magazine article about nuclear power, probably 1956.