Worksheet-xii (atoms & Nuclie).docx

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Rutherford’s 𝛼- Particle scattering experiment

page-01

Rutherford’s 𝜶-Particle scattering experiment (Geiger – Marsden experiment) A schematic arrangement of Rutherford’s 𝛼-particle scattering experiment (Geiger-Marsden experiment) is shown below.

214 A source of 𝛼 –Particles 222 86𝑅𝑛 or 83𝐵𝑖 is enclosed in a thin lead block, provided with a narrow opening. 𝛼 –Particles from this source are collimated as a beam and then allowed to fall on a very thin gold foil (of thickness 2.1 X 10-7 m)as shown. Scattered 𝛼- particles in different directions are observed by a rotating detector consisting of ZnS screen and a microscope. The entire apparatus is enclosed in a vacuum chamber to avoid scattering of 𝛼 –Particles by air molecules.

Observations and conclusions (i) Most of the 𝛼-particles passed through gold foil undeviated, it indicates that most of the space in an atom is empty. (ii) 𝛼-particles are scattered through all angles. Some particles (nearly 1 in 2000) suffers scattering through angles more than 900, while a still smaller number (nearly 1 in 8000) scattered at 1800. This implies that when fast moving positevely charged 𝛼 − particles come near gold atom, then a few of them experience such a strong repulsive force that they turn back. On this basis Rutherford concluded that, whole of the positive charge of atom is cocentrated in a small centre core, known as nucleus. (iii) Scattering of 𝛼-particles by heavy nuclei is in accordance with coulomb’s law. Rutherford observed that number of 𝛼-particles scattered is given by

N∝

1 𝑠𝑖𝑛4 𝜃⁄ 2

Distance of closest approach : Estimation of size of nucleus K.E. of electron = P.E.

½ m v2 =

𝑟0 =

1

𝑍𝑒 𝑋 2𝑒

4𝜋𝜀0

𝑟0

1 𝑍𝑒 𝑋 2𝑒 4𝜋𝜀0 1 m 𝑣 2 2

N

𝜽

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Rutherford’s 𝛼- Particle scattering experiment

page-02

Impact Parameter (b) The impact parameter is defined as the perpendicular distance of the velocity vector of the 𝛼- particle from the centre of the nucleus, when it is far away from the atom, and is given by

b

𝑍𝑒 2 cot𝜃 ⁄2

=

(i) For large value of b, cot

1 2

4𝜋𝜀0 ( 𝑚 𝑢2 )

𝜃 2

is large and the scattering angle

(θ) is small. i,e, 𝛼- particles travelling far away from the nucleus suffer small deflections. (ii) For small value of b, cot 𝜃⁄2 is small and the scattering angle (θ) is large. i,e, 𝛼- particles travelling close to the nucleus suffer large deflections. (iii) for the 𝛼- particles directed towards the centre of the nucleus, b = 0, then cot 𝜃⁄2 = 0 or 𝜃⁄2 = 900 or θ = 1800 The 𝛼- particle retrace its path.

Rutherford’s atomic model (i) An atom is a hollow sphere consists of a small massive central core in which entire positive charge and almost the whole mass of the atom are concentrated. This core is called the nucleus. (ii) The size of the nucleus (≈ 10−15 𝑚) is very small as compared to the size of the atom(≈ 10−10 𝑚). (iii) Electrons revolve around the nucleus in various circular orbits, for which the necessary centripetal force is provided by the electrostatic force of attraction between electron and nucleus. (iv) Number of electrons in an atom is such that it neutralises the positive charge so that the atom remains neutral.

Drawbacks or limitations of Rutherford’s atomic model (i) According to electromagnetic theory, an accelerated charged particle always radiates energy.The negatively charged electron revolving around the nucleus posseses centripetal acceleration and would lose its energy continously. The radius of its orbit, therefore, would go on decreasing and it would finally spiral in to the nucleus, resulting in an ultimate collapse of the atom. (ii) An electron can revolve in any orbit, according to electromagnetic theory, it must emit continous radiations of all Wavelengths (frequencies). But an atom like hydrogen, always emits a descrete line spectrum.

Bohr’s atomic model (i) An atom is a hollow sphere (≈ 10−10 𝑚) consists of a small massive central core in which entire positive charge and almost the whole mass of the atom are concentrated. This core is called the nucleus (≈ 10−15 𝑚). (ii) Electrons revolve around the nucleus in various circular orbits, for which the necessary centripetal force is provided by the electrostatic force of attraction between electron and nucleus. (iii) Electrons can revolve only in those circular orbits in which the angular momentum of an electron is an integral multiple of

ℎ 2𝜋

;

h being the Planck’s constant. i,e,

mvr = n

ℎ 2𝜋

where n = 1,2,3,4,------------- & is principal quantum number.

This is called Bohr’s quantisation condition of angular momentum. While revolving in these permissible orbits , an electron does not radiate energy. These non-radiating orbits are are called stationary orbits. (v) An atom can emit or absorb radiation in the form of descrete energy photons only when an electron jumps from a higher to a lower orbit or from a lower to a higher orbit respectively. i,e, ℎ𝜈 = 𝐸2 − 𝐸1 where h is called Planck’s constant.

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Bohr’s Theory of hydrogen atom

Bohr’s Theory of Hydrogen Atom Let an electron of mass ‘m’ is revolving around the nucleus with velocity ‘v’ as shown in figure. The electrostatic force of attraction between electron and nucleus provides the necessary centripetal force 𝑚𝑣 2

i,e,

=

𝑟

1

(𝑍𝑒)(𝑒)

4𝜋𝜀0

𝑟2

1

𝑍 𝑒2

4𝜋𝜀0

𝑟

𝑚𝑣 2 =

⇨

-----------------------(1)

Now according to Bohr’s quantum condition mvr = n v =n

⇨

ℎ

where n = 1,2,3,4, --------------------------

2𝜋 ℎ

----------------------(2)

2𝜋𝑚𝑟

On putting this value of v in equation (1) we get m (n

2 1 𝑍 𝑒2 ℎ ) = 2𝜋𝑚𝑟 4𝜋𝜀0 𝑟

𝑚𝑛2 ℎ2

⇨

=

4𝜋 2 𝑚2 𝑟 2

𝑟 ∝ 𝑛

4𝜋𝜀0 𝑟 2

(4𝜋 𝜀0 )𝑛2 ℎ2

𝑟=

⇨ ⇨

𝑍 𝑒2

1

4𝜋 2 𝑚 𝑍𝑒 2

------------------(3)

2

For hydrogen atom Z = 1, and for innermost orbit, n=1

⇨ from eqn (3) we get, 𝑟0

=

(4𝜋 𝜀0 )ℎ2 4𝜋 2 𝑚 𝑒 2

= 0.53 A0

This is called Bohr’s orbit. On putting the value of r from equation (3) in eqn (2) we get, 𝑛ℎ V= (4𝜋 𝜀0 ) 𝑛2 ℎ2 2𝜋𝑚 2 2 4𝜋

𝑚 𝑍𝑒

2

⇨

v

=

2𝜋 𝑍𝑒 2 (4𝜋 𝜀0 )𝑛 ℎ

⇨ ⇨

v ∝ v = Where 𝛼 =

For innermost orbit n =1, ⇨ 𝑣0

-----------------------(4)

1

𝑛 2𝜋 𝑍𝑒 2 (4𝜋 𝜀0 )𝑐 ℎ

𝑐

2𝜋 𝑍𝑒 2 1 = 137 (4𝜋 𝜀0 )𝑐 ℎ 𝑐

=

𝑐

X𝑛= 𝛼𝑛 is called fine structure constant.

137

Thus the speed of electron in the innermost orbit of hydrogen atom is

1 137

of the speed of light in vacuum

page-03

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Bohr’s theory of hydrogen atom

page-04

Energy of electron Kinetic energy of electron 1

𝐸𝐾 =

𝑚𝑣 2

2

On putting the value of mv2 from eqn (1) we get,

⇨

𝐸𝐾 =

1 2

(

1

𝑍 𝑒2

4𝜋𝜀0

𝑟

)

and potential energy of electron U=

1

(𝑍𝑒)(−𝑒)

4𝜋𝜀0

𝑟

=−

1

(𝑍 𝑒 2 )

4𝜋𝜀0

𝑟

1

(𝑍 𝑒 2 )

4𝜋𝜀0

𝑟

⇨ total energy of electron in nth orbit is En = Ek + U 1

⇨

En =

⇨

En =

⇨

En = −

2

(

1

𝑍 𝑒2

4𝜋𝜀0

𝑟

) −

(𝑍 𝑒 2 ) 1

1 4𝜋𝜀0

(2 - 1)

𝑟

1

1

(𝑍 𝑒 2 )

2

4𝜋𝜀0

𝑟

On putting the value of r from eqn (3) in above eqn we get,

En = −

1

(𝑍 𝑒 2 )

1

2 4𝜋𝜀0 (4𝜋 𝜀0 )𝑛2 ℎ2 4𝜋2 𝑚 𝑍𝑒2

⇨

⇨

En = −

1

1

(𝑍 𝑒 2 ) 4𝜋2 𝑚 𝑍𝑒 2

2

4𝜋𝜀0

(4𝜋 𝜀0 )𝑛2 ℎ2

En = − En = − En = −

⇨

En = −

𝒎 𝒁 𝟐 𝒆𝟒 𝟏

( )

𝟖𝜺𝟐𝟎 𝒉𝟐 𝒏𝟐 𝒎 𝒁 𝟐 𝒆𝟒 𝟏

𝑐ℎ

𝟖𝜺𝟐𝟎 𝒉𝟐 𝒏𝟐

𝑐ℎ

( )x

𝒎 𝒁 𝟐 𝒆𝟒

En = −

Where

R=

𝒁𝟐 𝑹𝒄𝒉 𝒏𝟐

𝒏𝟐

8𝜀02 𝑐ℎ3

En = −

--------------------------(6)

𝑹𝒄𝒉

𝑚 𝑒4

⇨

𝒄𝒉

( )

𝟖𝜺𝟐𝟎 𝒄𝒉𝟑 𝒏𝟐

For hydrogen atom Z =1

⇨

-----------------------(5)

𝟏𝟑.𝟔 𝒏𝟐

= 1.097 X 107 m-1 and is called Rydberg’s constant. eV

----------------------------(7)

The negative sign of total energy shows that electron is bound with the nucleus by electrostatic force and some work is to be done to pull it away from the nucleus.

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Spectral series of hydrogen atom

page-05

Spectral Series of Hydrogen Atom

We know that from Bohr’s theory of hydrogen atom, energy of an electron in nth orbit is given by 𝑚 𝑍2𝑒4 1

En = −

( )

-------------------------(1)

8𝜀02 ℎ2 𝑛2

According to Bohr’s quantum condition when an electron makes a transition from higher orbit n2 lower orbit n1, then the frequency of emitted photon is given by

h𝜈 = 𝐸𝑛2 −

𝐸𝑛1

Using eqn (1) we get, 𝑚 𝑍 2𝑒 4

h𝜈 = −

⇨ ⇨ ⇨

Where

h

8𝜀02 ℎ2 𝑚 𝑍 2𝑒 4

𝑐 𝜆

=

1 𝜆

=

1 𝜆

=

R=

8𝜀02 ℎ2 𝑚 𝑍 2𝑒 4 8𝜀02

𝑐ℎ3

1

𝑚 𝑍 2𝑒 4

2

8𝜀02 ℎ2

(𝑛 2) − [− 1

1

1

𝑛2 2

1

1

1

𝑛2 2

[𝑛 2 −

[𝑛 2 − 1

1

1

𝑛2 2

𝑍2 𝑅 [𝑛 2 −

𝑚 𝑒4 8𝜀02 𝑐ℎ3

or where 𝜈̅ is called wave number.

1 𝜆

=

𝜈̅ =

1

] ]

-----------------------------(2)

= 1.097 X 107 m-1 is called Rydberg’s constant.

For hydrogen atom Z = 1,

⇨

]

1

(𝑛 2 )]

1

1

1

𝑛2 2

𝑅 [𝑛 2 −

]

1

1

1

𝑛2 2

𝑅 [𝑛 2 −

]

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Spectral series of hydrogen atom

page-06

(i) Lyman Series When an electron jumps from any higher energy level n2 =2,3,4,--- to a lower energy level n1=1, we get a set of spectral lines, called Lyman series.This series is given by 1 𝜆

1

1

= 𝑅 [12 −

𝑛2

where n = 2,3,4,---------

]

This series belongs to ultraviolet region (UV region) and so it is not visible. Obviously, 1 𝜆𝑚𝑎𝑥

⇨

=

1

1

𝑅 [12 −

3

22

] = 1.097 X 107 [1 − 14] = 1.097 X 107 X 4

𝜆𝑚𝑎𝑥 = 1215 A0

&

⇨

1 𝜆𝑚𝑖𝑛

=

1

𝑅 [12 −

1 ∞2

] = 1.097 X 107 [1 − 0] = 1.097 X 107

𝜆𝑚𝑖𝑛 = 912 A0

(ii) Balmer Series When an electron jumps from any higher energy level n2 =3,4,5--- to a lower energy level n1=2, then the spectral series obtained is called Balmer series.This series is given by 1 𝜆

1

1

= 𝑅 [22 −

𝑛2

where n = 3,4,5---------

]

This series lies in visible region and so it is visible. Obviously, 1 𝜆𝑚𝑎𝑥

⇨ &

⇨

=

1

𝑅 [22 −

1 32

5

] = 1.097 X 107 [14 − 19] = 1.097 X 107 X 36

𝜆𝑚𝑎𝑥 = 6563 A0 1 𝜆𝑚𝑖𝑛

=

1

𝑅 [22 −

1 ∞2

] = 1.097 X 107 [14 − 0] = 1.097 X 107 X 14

𝜆𝑚𝑖𝑛 = 3646 A0

(iii) Paschen Series When an electron jumps from any higher energy level n2 =4,5,6--- to a lower energy level n1=3, then the spectral series obtained is called Paschen series.This series is given by 1 𝜆

=

1

𝑅 [32 −

1 𝑛2

where n = 4,5,6---------

]

This series lies in Infrared region and so it is not visible. Obviously, 1 𝜆𝑚𝑎𝑥

⇨ &

⇨

=

1

𝑅 [32 −

1 42

] = 1.097 X 107 [19 −

1 16

] = 1.097 X 107 X

7 116

𝜆𝑚𝑎𝑥 = 18752 A0 1 𝜆𝑚𝑖𝑛

=

1

𝑅 [32 −

1 ∞2

] = 1.097 X 107 [19 − 0] = 1.097 X 107 X 19

𝜆𝑚𝑖𝑛 = 8204 A0

(iv) Brackett Series When an electron jumps from any higher energy level n2 =5,6,7--- to a lower energy level n1=4, then the spectral series obtained is called Brackett series.This series is given by 1 𝜆

=

1

𝑅 [4 2 −

1 𝑛2

]

where n = 5,6,7---------

This series lies in Infrared region and so it is not visible

(v) Pfund Series When an electron jumps from any higher energy level n2 =6,7,8--- to a lower energy level n1=5, then the spectral series obtained is called Pfund series. It is given by 1 𝜆

=

1

𝑅 [52 −

1 𝑛2

]

This series lies in far infrared region and not visible.

where n = 6,7,8---------

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Spectral series of hydrogen atom

page-07

Energy level diagram for hydrogen atom It is a diagram in which the energies of the different stationary states of an atom are represented by parallel horizontal lines, drawing acording to some suitable energy scale. We know that for hydrogen atom, energy of an electron in nth orbit is given by

En = − ⇨

E1 = − E2 = − E3 = − E4 = − E5 = −

13.6 eV 𝑛2 13.6 = − 13.6 eV 12 13.6 = − 3.4 eV 22 13.6 = − 1.51 eV 32 13.6 = − 0.85 eV 42 13.6 = − 0.54 eV 52

-------------------------------E∞ = −

13.6 ∞2

= 0 eV

Note- If electron gets excited to nth state, then the number of spectral lines in the emission spectrum = For example – if electron gets excited to n=4 state,then total number of spectral lines in emission spectrum=

𝒏(𝒏−𝟏) 𝟐

=

𝟒 (𝟒−𝟏) 𝟐

=

𝟒𝑿𝟑 𝟐

=6

𝒏(𝒏−𝟏) 𝟐

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Atoms

page-08

Excitation,Excitation nergy & Excitation potential The Process of absorption of energy due to which an electron in an atom, raises from lower energy level to higher energy level is called excitation. The potential difference through which an electron in an atom has to be accelerated,so that it raises from a lower energy state to higher energy state(or the atom just raises from ground state to an excited state), is called excitation potential. The minimum energy required so as an electron in an atom just raises from a lower energy level to higher energy level (or the atom just raises from ground state to an excited state), is called excitation energy.

Ionisation, Ionisation energy & Ionisation potential The Process of knocking out of an electron from an atom is called ionisation. The potential difference through which an electron in an atom has to be accelerated,so that it is knocked out from the atom is called ionisation potential. The minimum energy required so that an electron in an atom just knocked out from it , is called ionisation energy.

Limitations of Bohr’s atomic model (i) It explains spectra of atoms having one electron only (H-atom) and it fails in explaining the spectra of atoms having more than one electron. (ii) It fails to explain fine structure of spectral lines. (iii) It can not explain Stark & Zeeman effect in spectral lines. (iv) It does not explain the relative intensities of spectral lines.

NOTE-

Bohr’s quantisation condition of angular momentum

Let us consider the motion of an electron in a circular orbit of radius r around the nucleus of the atom. According to de-Broglie hypothesis, this electron is also associated with wave character. Hence a circular orbit can be taken to be a stationary energy state only if it contains an integral number of de-Broglie wavelengths i,e, we must have

2𝜋𝑟 = nλ

------------------(1)

ℎ

But

λ =

⇨

2𝜋𝑟 = n

⇨

𝑚𝑣𝑟 = n 2𝜋,

𝑚𝑣 ℎ 𝑚𝑣 ℎ

This is famous Bohr’s quantisation condition for angular momentum.

n= 1,2,3 -----------------

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Atoms

page-09

Q.No. 01- Draw a schematic arrangement of the Geiger – Marsden experiment for studying 𝛼 −particle scattering by a thin foil of gold. Describe briefly, by drawing trajectories of the scattered 𝛼 −particles, how this study can be used to estimate the size of the nucleus ? (F-2010) Q.No. 02- Draw a schematic arrangement of the Geiger – Marsden experiment .How did the scattering of 𝛼 −particle by a thin foil of gold provide an imporatant way to determine an upper limit of the size of the nucleus. Explain briefly. (AI-2009) Q.No. 03- Draw a schematic arrangement of the Geiger – Marsden experiment .Describe briefly how the information on the size of the target nucleus was obtained by studying the scattering of 𝛼 −particles on a thin foil of gold ? (F-2008) Q.No. 04- Draw a labelled diagram of experimental set-up of Rutherford’s 𝛼 −particle scattering experiment. Write two important inferences drawn from this experiment. (AIC-2005, D-2005) Q.No. 05-Draw a labelled diagram for 𝛼 −particle scattering experiment .Give Rutherford’s observations & discuss the significance of this experiment. Obtain the expression which helps us to get an idea of the size of a nucleus, using these observations. (F-2003) Q.No. 06-In the Rutherford’s scattering experiment the distance of closest approach for an 𝛼 −particle is d0. If 𝛼 −particle is replaced by a proton, how much kinetic energy in comparison to 𝛼 −particle will it require to have the same distance of closest approach d0 ? [Ans. ½] (F-2009) Q.No. 07-What is the ratio of radii of the orbits corresponding to first excited state and ground state in a hydrogen atom ? [Ans. 4:1] (D-2010) Q.No. 08-The radius of innermost electron orbit of a hydrogen atom is 5.3 X 10-11 m. What is the radius of orbit in the second excited state ? (D-2010) Q.No. 09-Write the expression for Bohr’s radius in hydrogen atom. (D-2010) Q.No. 10-In hydrogen atom, if the electron is replaced by a particle which is 200 times heavier but have the same charge, how would its radius change ? (F-2008) Q.No. 11-Define ionisation energy. What is its value for a hydrogen atom ? [Ans. 13.6 eV] (AI-2010) Q.No. 12-Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its – (i) second permitted energy level to the first level, and

[Ans.

51 68

] (AI-2010)

(ii) the highest permitted energy level to the first permitted level. Q.No. 13-Calculate the ratio of energies of photons produced due to transition of electron of hydrogen atom from its(i) second permitted energy level to the first level and [Ans. 3] (SP-2010) (ii) highest permitted energy level to the second permitted level. Q.No. 14-The spectrum of a star in the visible and the ultraviolet region was observed and the wavelength of some of the lines that could be identified were found to be (SP-2010) 824 A0, 970 A0, 1120 A0, 2504 A0, 5173 A0,6100 A0 Which of these lines can not belong to hydrogen spectrum ? Given Rydberg constant R = 1.03 x 107 m-1 &

1 𝑅

= 970 A0

Support your answer with suitable calculations. [Ans. 970 A0,1120 A0 falls in Lyman series, 5173 A0, 6100 A0 falls in Balmer series and the remaining 824 A0 & 2504 A0 are not the part of hydrogen spectrum as they do not fall in any spectral series.]

Q.No. 15-State Bohr’s postulate for the permitted orbits for the electron in a hydrogen atom. Use this postulate to prove that the circumference of the nth permitted orbit for the electron can contain exactly ‘n’ wavelengths of the de-Broglie wavelength associated with the electron in that orbit. (SP-2010) Q.No. 16-The value of ground state energy of hydrogen atom is −13.6 eV. (F-2009,AI-2000) (i) what does the negative sign signify ? [Ans (ii) 10.2 eV] (ii) How much energy is required to take an electron in this atom from the ground state to the first excited state ? Q.No. 17-The ground state energy of hydrogen atom is −13.6 eV. (AI-2008) (i) what is the kinetic energy of an electron in the 2nd excited state ? [Ans. (i) -1.51 eV (ii) λ = 1.02 X 10-7 m] (ii) If the electron jumps to the ground state from 2nd excited state, calculate the wavelength of the spectral line emitted.

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Atoms

page-10

Q.No. 18-The ground state energy of hydrogen atom is −13.6 eV. (AI-2008) (i) what is the potential energy of an electron in the 3rd excited state ? [Ans. (i) -1.75 eV (ii) λ = 970 A0] (ii) If the electron jumps to the ground state from 3rd excited state, calculate the wavelength of the photon emitted. Q.No. 19-The ground state energy of hydrogen atom is −13.6 eV. What are the kinetic and potential energies of electron in this state ? [Ans. K.E. = 13.6 eV, P.E. = −27.2 eV] (AI-2000) Q.No. 20-The total energy of an electron in the first excited state of hydrogen atom is - 3.4 eV. What is its kinetic energy ? [Ans. 3.4 eV] (D-1995,90) Q.No. 21-The total energy of an electron in the first excited state of hydrogen atom is - 3.4 eV. What is potential energy of the electron in this state ? [Ans. -6.8 eV] (D-1995,DC-1993) Q.No. 22-(a) The energy levels of an atom are as shown below. Which of them will result in the transition of a photon of wavelength 275 nm ? [Ans. B] (D-2009,D-1997,AI-1999,DC-1993) (b) Which transition corresponds to emission of radiation of maximum wavelength ? [Ans. A] A B 0 eV -2.0 eV C

-4.5 eV

D

-10 eV

Q.No. 23- The energy levels of an element are given below : Identify, by doing necessary calculations, The transition, which corresponds to the emission of a spectral line of wavelength 482 nm ? [Ans. C] (D-2008) -0.85 eV A -1.5 eV C -3.4 eV B

D

-13.6 eV

Q.No. 24- The energy level diagram of an element is given below. Identify, by doing necessary calculations, which transition corresponds to the emission of a spectral line of wavelength 102.7 nm ? [Ans. D] (D-2008) A B -0.85 eV -1.5 eV -3.4 eV D

-13.6 eV Q.No. 25-The energy levels of an atom of an element are as shown in the following diagram. Which of thelevel transitions will result in the emission of photons of wavelength 620 nm ? Support your answer with mathematical calculations. A B C [Ans. D] (D-1997) 0 eV -1 eV D -3 eV

E

-10 eV

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Atoms

page-11

Q.No. 26-The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy ? [Ans. III] (AIEEE-2005) n =4 n =3 n =2

n =1 I

II

III

IV

Q.No. 27-Photons, with a continous range of frequencies, are made to pass through a sample of rarefied hydrogen. The transitions, shown in figure below indicate three of the spectral absorption lines in the continous spectrum. [Ans. (ii) III] (DC-2009)

(i) Identify the spectral series, of the hydrogen emission spectrum,to which each of these three lines correspond. (ii) Which of these lines corresponds to the absorption of radiation of maximum wavelength ? [Ans. (i) line I-Lyman series, line II- Balmer series, Line III- Paschen series (ii) III]

Q.No. 28-The transition from the state n =4 to n= 3 in a hydrogen like atom results in ultraviolet radiation . Infrared radiation will be obtained in the transition , [Ans (d) ] (IIT JEE-2001) (a) 2 1 (b) 3 2 (C) 4 2 (d) 5 4

NCERT PROBLEMS Q.No. 01-A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level. [ Ans. 5.6 X 1014 Hz.] Q.No. 02-The ground state energy of hydrogen atom is – 13.6 eV. What are the kinetic and potential energies of the electron in this state ? [Ans. K.E. = 13.6 eV, P.E.= - 27.2 eV] Q.No. 03-A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n=4 level. Determine the wavelength and frequency of photon. [Ans. λ = 974.4 A0 , 𝜈 = 3.078 X 1015 Hz] Q.No. 04-(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n=1,2 and 3 levels. (b) Calculate the orbital period in each of these levels. [Ans.(a) v1 = 2.186 X 106 m/s , v2 = 1.093 X 106 m/s, v3 = 0.729 X 106 m/s (b) T1 = 1.52 X 10-16 s, As Tn = n3 T ⇨ T2 = 1.22 X 10-15 s, T3 = 4.10 X 10-15 s] Q.No. 05-The radius of the innermost electron orbit of a hydrogen atom is 5.3 X 10-11m. What are the radii of the n=2 and n=3 orbits ? [Ans. as rn = n2 r, r2 = 2.12 X 10-10 m, r3 = 4.77 X 10-10m ] Q.No. 06-A 12.75 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelength will be emitted ? [Ans. 970.6 A0, 4852.9 A0, 6547.6 A0, 1023.6 A0, 1213.2 A0, 28409 A0] Q.No. 07-The total energy of an electron in the first excited state of the hydrogen atom is about −3.4 eV. (a) What is the kinetic energy of the electron in this state ? (b) What is the potential energy of the electron in this state ? (c) Which of the answers above would change if the choice of the zero of potential energy is changed ? [Ans. (a) 3.4 eV (b) -6.8 eV (c) K.E. does not change but P.E. and total energy will change.] ------------------------------------------------------

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Atomc Nuclei

page-01

Atomic Mass Unit (u) One atomic mass unit is defined as i,e,

1u=

1 12

1 12

th of the actual mass of c-12 atom.

X mass of C-12 atom =

1 12

X 1.992678 X 10-26 kg = 1.66 X 10-27 kg.

Electron Volt (eV) It is the energy acquired by an electron when it is accelerated through a potential difference of 1 volt. 1 eV = 1.6 X 10-19 J & 1 MeV = 1.6 X 10-13 J

Relation Between amu & MeV We know, 1 u = 1.66 X 10-27 kg According to Einstien’s mass energy relation E = m c2 = 1.66 x 10-27 x (3 x 108)2 = 1.66 X 9 X 10-11 J =

1.66 𝑋 10−11 1.6 𝑋 10−19

≈ 931 MeV

Nuclear Size

It has been found that, the radius of nucleus is given by R ∝ A1/3 R = R0 A1/3 Where R0 = 1.1 X 10-15 m Nuclear size is measured in Femi, and 1 Fermi (f) = 10-15 m

⇨

Nuclear Density (𝝆) The ratio of mass of the nucleus to its volume is called nuclear density. i,e,

𝜌= = = =

𝑀 𝑉 𝐴(𝑎𝑚𝑢) 4 3

𝜋 𝑟3

𝐴 𝑋 1.66 𝑋 10−27 4 3

𝜋 ( 𝑅0 𝐴1/3 )3

𝐴 𝑋 1.66 𝑋 10−27 4 3

𝜋 𝑅03 𝐴

=

3 𝑋 1.66 𝑋 10−27 4 𝜋 𝑅03

=

3 𝑋 1.66 𝑋 10−27 4 𝑋 3.14 𝑋 (1.1 𝑋 10−15 )3

= 2.3 X 1017 Kg/m3 obviously, nuclear density is independent of mass number A. NoteDensity of ordinary matter is 103 Kg/m3

⇨

𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑛𝑢𝑐𝑙𝑒𝑎𝑟 𝑚𝑎𝑡𝑡𝑒𝑟 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑜𝑟𝑑𝑖𝑛𝑎𝑟𝑦 𝑚𝑎𝑡𝑡𝑒𝑟

=

2.3 𝑋 1017 103

= 1014 : 1

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Atomc Nuclei

page-02

Isotopes The atoms of an element, which have the same atomic number but different mass numbers are called isotopes. For examples Hydrogen has three isotopes 11𝐻 , 21𝐻 & 31𝐻 Lithium has two isotopes 63𝐿𝑖 & 73𝐿𝑖 37 Chlorine has two isotopes 35 17𝐶𝑙 & 17𝐶𝑙

Isobars

The atoms having the same mass number but different atomic numbers are called isobars. 40 40 3 3 37 37 For examples 1𝐻 & 2𝐻𝑒 , 17𝐶𝑙 & 16𝑆 , 20𝐶𝑎 & 18𝐴𝑟

Isotones The nuclids having the same number of neutrons are called isotones. 39 198 197 37 For examples 17𝐶𝑙 & 19𝐾 , 80𝐻𝑔 & 79𝑃𝑢

Isomers These are the nuclei having the same atomic number & same mass number but existing in different energy states. For example A nucleus in its ground state and the identical nucleus in metastable excited state, are isomers.

Nuclear Forces The strong attractive forces,which firmly holds the nucleons together in a tiny nucleus of an atom, are called nuclear forces. Properties of nuclear Forces (i) Nuclear forces are very short range attractive forces. (ii) Nuclear forces are charge independent. (iii) Nuclear forces are non-central forces. (iv) Nuclear forces do not obey inverse square law. (v) Nuclear forces are dependent on spin or angular momentum. Difference between Nuclear force & Coulomb’s Force S.No. 1 2 3 4

It It It It

Nuclear Force is a force between two nucleons. is the strongest force in nature. is a very short range (2-3 fermi) force. does not obey inverse square law.

It It It It

Coulomb’s Force is a force between two charged particles. is 10-2 times weaker than the nuclear force. is a very long range (∞) force. obeys inverse square law.

Nuclear force as a separation between two nucleons

(i) Nuclear forces are very short range forces (nearly 2-3 fermi) (ii) Nuclear forces are negligible if r > 10 fermi.

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Atomc Nuclei

page-03

Potential energy of a pair of nucleons as a separation between two nucleons

Important conclusions drawn from above potential energy curve(i) Nuclear forces are repulsive when r < r0 and attractive when r > r0. (ii) Nuclear force between two nucleons falls rapidly to zero as their distance is more that a few fermis. This explains constancy of the binding energy per nucleon for large size nucleus. (iii) Negative potential energy shows that binding force between the nucleons is strong. Nuclear Forces are exchange forces : Nature of nuclear forces According to Yukava’s theory nuclear forces arise due to continous exchange of particles (called mesons) between nucleons. All nucleons consists of identical cores surrounded by a cloud of one or more than one type of mesons. Mesons have the charges as 𝜋 + ,𝜋 − or 𝜋 0 . The difference between proton and neutron is due to difference in composition of their respective meson clouds. Mesons are continously emitted of absorbed by every nucleon. The transfer of momentum associated with the shift of mesons is equivalent to the action of a force, called nuclear force. A neutron becomes a proton inside the nucleus by emitting 𝜋 − meson. Similarly a proton becomes a neutron by absorbing a 𝜋 − meson. n p + 𝜋− and p + 𝜋− n

Also,

p

n + 𝜋+

and

n + 𝜋+

n

n’ + 𝜋 0

and

p

p p’ + 𝜋 0

Where n’ and p’ are neutrons and protons of different states respectively. ---------------------------------------------------------------------------------------------------------------------------------------------Q.No. 01- Define mass number ‘A’ of an atomic nucleus. Assuming the nucleus to be spherical, give the relation between mass number (A) and radius (R) of the nucleus. (SP-2005) Q.No. 02- How does the size of a nucleus depend upon its mass number ? Hence explain why the density of a nuclear matter should be independent of size of the nucleus? (AIC-2004) Q.No. 03- Calculate the density of nuclear matter. Radius of nucleus of 11𝐻 = 1.1 X 10-5 A0, what is the ratio of order of magnitude of density of nuclear matter and density of ordinary matter ? (SP-2005)

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Atomc Nuclei

page-04

Q.No. 04- Prove that the density of matter of a nucleus is independent of mass number A.

(AIC-1997)

Q.No. 05- Group the following nucleides in to three paires of (i) isotopes (ii) isotones & (iii) isobars.

(AIC-2004)

198 197 12 4 3 14 6𝐶 , 2𝐻𝑒 , 80𝐻𝑔 , 1𝐻 , 29𝐴𝑢 , 6𝐶

Q.No. 06- Compare the radii of two nuclei with mass numbers 1 and 27 respectively.

(AI-2000)

Q.No. 07- Two nuclei have mass numbers in the ratio 1:8. What is the ratio of their nuclear densities ?

(F-2001)

Q.No. 08- Two nuclei have mass numbers in the ratio 1 : 8. What is the ratio of their nuclear radii ?

(AI-2009)

Q.No. 09- Two nuclei have mass numbers in the ratio 8:125. What is the ratio of their nuclear radii ?

(AI-2009)

Q.No. 10- Two nuclei have mass numbers in the ratio 27:125. What is the ratio of their nuclear radii ?

(AI-2009)

Q.No. 11- Two nuclei have mass numbers in the ratio 1:2. What is the ratio of their nuclear densities ?

(D-2009)

Q.No. 12- Two nuclei have mass numbers in the ratio 1:3. What is the ratio of their nuclear densities ?

(D-2009)

Q.No. 13- Two nuclei have mass numbers in the ratio 2:5. What is the ratio of their nuclear densities ?

(D-2009)

Q.No. 14- Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions which you can draw regarding the nature of nuclear forces. (AI-2010) Q.No. 15- Plot a graph showing the variation of potential energy of a pair of nucleons as a function of their separation. (AI-2009) Q.No. 16- Draw a graph showing the variation of potential energy of a pair of nucleons as a function of their separation. Indicate the regions in which nuclear force is- (i) attractive , and (ii) repulsive. (AI-2007,D-2005) Q.No. 17- Draw a plt of potential energy of a pair of nucleons as a function of their separation. What is the significance of negative potential energy in the graph drawn ? (F-2006,D-2005) Q.No. 18- (a) State two properties of nuclear forces. (b) State four characteristics of nuclear forces.

(DC-2007,2003) (AI-2008,SP-2010)

(c) Write two characteristic features of nuclear force which distinguish it from Coulomb’s force. (D-2007,AIC-2003) ---------------------------------------------------------------------------------------------------------------------------

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Atomc Nuclei

page-05

Mass Defect (∆𝒎) It is defined as the difference between the sum of masses of protons and neutrons forming a nucleus and actual mass of the nucleus. i,e, mass difference = [Mass of protons + Mass of neutrons] – [Mass of nucleus] ∆𝒎 = [Z mp + (A – Z) Mn ] − MN

Packing fraction (P.F.) It is defined as the mass defect per nucleon. i,e,

P.F. =

∆𝑚 𝐴

Nucleus is stable if P.F.>1 & unstable if P.F.< 1

Binding Energy (B.E.) The binding energy of a nucleus may be defined as the energy required to break up a nucleus in to its constituent protons and neutrons and to separate them to such a large distance that they may not interact with each other.It is equivalent energy of mass defect. i,e,

B.E. = ∆𝑚 X c2

⇨

B.E. = [{Z mp + (A – Z) Mn} − MN ] x c2

Binding Energy per nucleon It is the average energy required to extract one nucleon from the nucleus to infinite distance and is given by the total binding energy divided by the mass number of the nucleus. i,e,

B.E. per nucleon =

= =

𝐵.𝐸. 𝐴 ∆𝑚 𝑐 2 𝐴

[{Z mp + (A – Z) Mn} − MN ] x c2 A

Note(i) A nucleus is more stable if, it has (a) high value of B.E./A (b) greater n/p ratio, and (c) even-even nucleus. For example- Out of two nuclei 73𝑋 & 43𝑌, nucleus 73𝑋 is more stable because n/p ratio for 73𝑋 is more than that for 43𝑌. (ii) Why is the mass of a nucleus is always less than the sum of the masses of its constituent,neutrons & protons ? [Ans. When nucleons approach each other to form a nucleus, they strongly attract each other. As a consequence their potential energy decreases and becomes negative. It is this potential energy which holds the nucleons together in the nucleus. The decrease in potential energy results in the decrease in the mass of the nucleons inside the nucleus.]

(iii) If the total number of neutrons & protons in a nuclear reaction is conserved, how is then the energy is absorved or evolved in the reaction ? [Ans. Since certain mass disappears in the formation of a nucleus (mass defect), it appears in the form of energy according to the relation E = ∆𝑚 c2. Thus the difference of B.E. of the two sides appear as energy released or absorbed in a nuclear reaction.]

(iv) If the nucleons of a nucleus are separated far apart from each other, the sum of the masses of all these nucleons is larger than the mass of the nucleus. Why ? [Ans. For the separation of nucleons to a distance far apart from each other, an energy equal to B.E. of the nucleus is given to these nucleons. From E = ∆𝑚 c2, this mass difference comes.] Q.No.01- The binding energies of deutron ( 21𝐻 ) & 𝛼-particle ( 42𝐻𝑒 ) are 1.25 & 7.25 MeV/ nucleon respectively. Which nucleus is

more stable & why ? [Ans. ( 42𝐻𝑒) is more stable as it has more B.E./nucleon.

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Atomc Nuclei

page-06

Binding Energy Curve The value of binding energy per nucleon of a nucleus gives a measure of the stability of that nucleus. greater is the binding energy per nucleon of a nucleus, more stable is that nucleus.

Salient features of binding energy curve (i) Light nuclei like 11𝐻 , 21𝐻 & 31𝐻 have small values of binding energy per nucleon. (ii) For mass numbers from 2 to 20, there are sharp peaks corresponding to nuclei having mass numbers in multiple of 4 (i,e, containing equal numbers of protons & neutrons) e,g, 42𝐻𝑒 , 84𝐵𝑒 & 126𝐶 , 168𝑂, 20 10𝑁𝑒 . Peaks shows more B.E. per nucleon, it means these nuclei are more stable. (iii) For mass numbers 30 to 120, B.E. curve has a broad maximum. Average binding energy per nucleon corresponds to this range is 8.5 MeV. Peak value is 8.5 MeV for 𝐹𝑒 56 . Elements belonging to this range are highly stable & nonradioactive. (iv) After the mass number 120, binding energy per nucleon decreases gradually to 7.6 MeV per nucleon(𝑈 238 ). Coulomb repulsion between the protons starts dominating the nuclear force as a result heavy nuclei are always less stable. Imortance of binding energy curve This curve can be used to explain the phenomenon of nuclear fission & fusion. (i) Nuclear Fission There is overall gain in the binding energy per nucleon, when we move from heavy to medium range nuclei, hence release of energy. This indicates that energy can be released when a heavy nucleus is broken in to small fragments. This is called nuclear fission. (ii) Nuclear Fusion Similarly there is overall gain in the binding energy per nucleon, when we move from lighter to medium range nuclei, hence release of energy. This indicates that energy can be released when two or more lighter nuclei fuse together to form a heavy nucleus. This is called nuclear fusion.

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Atomc Nuclei

page-07

Q.No. 01- Draw a plot showing the variation of binding energy per nucleon as a function of mass number for a large number of nuclei, 2 ≤ A ≤ 240. How do you explain the constancy of binding energy per nucleon in the range 30
Q.No. 02- Draw a plot showing the variation of binding energy per nucleon versus the mass number A. Explain with the help of this Plot the release of energy in the processes of nuclear fission & fusion. (AI-2009) Q.No. 03- Sketch the graph showing the variation of binding energy per nucleon as a function of mass number A, for large number of nuclei. State briefly,from which reason of the graph,can release of energy in the process of nuclear fusion can be explained ? (F-2008) Q.No. 04- Draw a graph showing the variation of binding energy per nucleon with mass number for different nuclei. Explain with the help of this graph release of energy in the process of nuclear fusion. (D-2006) Q.No. 05- Draw a curve showing the variation of binding energy / nucleon with mass number for different nuclei. Briefly state how nuclear fusion & nuclear fission can be explained on the basis of this graph ? (AIC-2006,2004) Q.No. 06- Draw a plot showing the variation of binding energy per nucleon with mass number for different nuclei. Write two important conclusions , which you can draw from this plot.Explain with the help of this plot, the release of energy in the process of nuclear fission & fusion. (F-2005) Q.No. 07- Draw a diagram to show the variation of binding energy per nucleon with mass number for different nuclei. State why heavy nuclei undergo fission & light nuclei usually undergo fusion. (AI-2001,D-2004) Q.No. 08- Draw the graph showing the variation of binding energy per nucleon with mass number for different nuclei. Give reason for the decrease of binding energy per nucleon for nuclei with high mass number. (DC-2006, D-2004) Q.No. 09- Draw a diagram to show the variation of binding energy per nucleon with mass number for different nuclei. State two inferences drawn from this graph. (DC-2006, D-2004) Q.No. 10- Binding energy per nucleon Vs. mass number curve for nuclei is shown in the figure. W,X,Y & Z are four nuclei indicated on the curve. [Ans. (c) ] (IIT JEE-1999)

The process that would release energy is – (a) Y 2Z (b) W

X+Z

(c) W

2Y

(d)

X

Y+Z

Q.No. 11- Given below is a plot of binding energy per nucleon Eb, against the nuclear mass M; A,B,C,D,E,F correspond to different nuclei. Consider four reactions : [Ans. (a) ] (AIEEE-2009) (i) A + B C+𝜖 (ii) C A+B+𝜖 (iii) D + E F+𝜖 (iv) F D+E+𝜖

Where 𝜖 is the energy released. In which reactions is 𝜖 positive ? (a) (i) and (iv) (b) (ii) and (iv) (c) (i) and (iii)

(d) (ii) and (iii)

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Atomc Nuclei

page-08

Q.No. 12- Binding energy per nucleon plot against the mass number for stable nuclei is shown in the figure. Which curve is correct ? (a) A (b) B (c) C (d) D [Ans. (c) ] (DCE-2003,IPUEE-2004)

Q.No. 13- The dependent of binding energy per nucleon (BN) on the mass number A is represented by [Ans. (a) ]

(AIIMS-2004)

Q.No. 14- Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in figure.

Use this plot to choose the correct choice(s) given below. [Ans. (b), (d) ] (a) Fusion of two nuclei with mass numbers lying in the range of 1 < A < 50 will release energy.

(IIT JEE-2008)

(b) Fusion of two nuclei with mass numbers lying in the range of 51 < A < 100 will release energy. (c) Fision of a nucleus lying in the mass range of 100 < A < 200 will release energy when broken in to two equal fragments. (d) Fision of a nucleus lying in the mass range of 200 < A < 260 will release energy when broken in to two equal fragments. ----------------------------------------------------------------------------------------------------------------------------------------

Numericals Q.No. 01- Calculate the binding energy per Given, Mass of Mass of Mass of Mass of

nucleon (in MeV) for 42𝐻𝑒 & 1 1𝐻 = 1.00783 u 1 0𝑛 = 1.00867 u 3 2𝐻𝑒 = 3.01664 u 4 2𝐻𝑒 = 4.00387 u

3 2𝐻𝑒 .

(DC-2005) [Ans. 6.78 & 2.39 MeV]

Q.No. 02- Calculate the binding energy per nucleon of 40 20𝐶𝑎 nucleus. Given, Mass of 40 𝐶𝑎 = 39.962589 u 20 Mass of proton = 1.007825 u Mass of neutron = 1.008665 u & 1 u = 931 MeV/c2

(AI-2007,2004,2002,D-2002) [Ans. 8.547 MeV]

Q.No. 03- Calculate the binding energy per nucleon of. 56 26𝐹𝑒 nucleus. Given, Mass of 11𝐻 = 1.00783 u Mass of 10𝑛 = 1.00867 u 2 Mass of 56 26𝐹𝑒 = 55.934939 u , & 1 u = 931 MeV/c

(AI-2004,AIC-2000,DC-2005) [Ans. 8.79 MeV]

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Atomc Nuclei

Q.No. 04- Calculate the binding energy per nucleon of. 35 17𝐶𝑙 nucleus. Given, m ( 35 𝐶𝑙 ) = 34.980000 u 17 Mass of proton = 1.007825 u Mass of neutron = 1.008665 u & 1 u = 931 MeV/c2 Q.No. 05- Calculate the binding energy / nucleon of. 209 83𝐵𝑖 nucleus. Given, m ( 209 𝐵𝑖 ) = 208.980388 u 83 Mass of proton = 1.007825 u Mass of neutron = 1.008665 u & 1 u = 931 MeV/c2

(AI-1995) [Ans. 7.84 MeV]

Q.No. 06- Calculate the energy released if, 𝑈 238 , emits an 𝛼-particle. Given,

atomic mass of 𝑈

238

page-09 (AI-2002) [Ans. 8.22 MeV]

(AI-2007)

= 238.0508 u

[Ans. 4.24 MeV]

atomic mass of 𝑇ℎ234 = 234.04363 u atomic mass of 𝛼-particle = 4.00260 u & 1 atomic mass unit (1u) = 931 MeV/c2 Q.No. 07- Calculate the energy released in MeV in the following nuclear reaction 238 92U

Given,

234 90Th

+ 2He4 + Q

238 92U

= 238.05079 u

mass of

234 90Th

= 234.043630 u

mass of

2He

mass of

4

= 4.002600 u = 931 MeV/c2

& 1u

Q.No. 08- Calculate the amount of energy released in the 𝛼- decay of 238 92U

Given,

90Th

atomic mass of

(AI-2008) [Ans. 4.24 MeV]

238 92U

234

(D-2007)

+ 2He4

[Ans. 4.25 MeV]

= 238.05079 u

234 = 234.04363 u 90Th 4 atomic mass of 2He = 4.00260 u

atomic mass of

& 1u

= 931.5 MeV/c2

Q.No. 09- Calculate the energy released in the following nuclear reaction 2 1H

+

Given,

3 1H

mass of mass of mass of mass of

2He 0n

1

2 1H 3 1H

2He

4

4

+ 0n1 + Q

= 1.008665 = 2.014102 = 3.016049 = 4.002603

u u u u

Q.No. 10- Calculate the energy released in the following nuclear reaction 6 3Li

Given,

+

0n

1

2He

4

+ 1H3 + Q

mass of

0n

1

= 1.008665 u

mass of mass of mass of

6 3Li

= 6.015126 u = 3.016049 u = 4.002603 u

1H

3

2He

4

(DC-2003) [Ans.17.59 MeV]

(AI-2002,DC-2003) [Ans.10.41 MeV]

Q.No. 11- A neutron is absorbed by a 3Li6 nucleus with the subsequent emission of an alfa particle. Write the corresponding nuclear reaction. Calculate the energy released in this nuclear reaction. (AI-2006,D-2005,DC-2000) 6 Given, m(3Li ) = 6.015126 u [Ans.10.415 MeV] m(0n1) = 1.008665 u

m(2He4) = 4.0026044 u m(1H3) = 3.016049 u 1u = 931

MeV/c2

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Atomc Nuclei

page-10

Q.No. 12- A nucleus 10Ne23, 𝛽-decays to give the nucleus of 11Na23. Write down the 𝛽-decay equation. Calculate the kinetic energy of electron emitted. (Rest mass of electron may be ignored.) (AI-2004) Given, m(10Ne23) = 22.994466 u [Ans.4.37 MeV] m(11Na23) = 22.9897704 u Q.No. 13- When a deutron of mass 2.0141 u & negligible kinetic energy is absorbed by a Lithium (3Li6) nucleus of mass 6.0155 u, the compound nucleus disingrates spontaneously in to two alpha particles, each of mass 4.0026 u. Calculate the energy in Joules carried by each alpha particle. [Ans. 0.17568 X 10-11 J] (AI-2004) Q.No. 14- A nucleus of an atom of

92Y

235

(SP-2010) [Ans. 1.55 X 107 m/s]

, initially at rest, decays by emitting an 𝛼-particle as per the equation

92Y

235

231 90X

+ 2He4 + energy

It is given that the binding energies per nucleon of the parent & daughter nuclei are 7.8 MeV & 7.835 MeV respectively and that of 𝛼-particle is 7.07 MeV/nucleon. Assuming the daughter nucleus to be formed in the unexcited state & neglecting its share in the energy of the reaction. Calculate the speed of the emitted 𝛼-particle. ( take the mass of 𝛼-particle to be 6.68 X 10-27 Kg). [Sol. Energy released = [ m (90X231) + m (2He4) − m (92Y235) ] X c2 = [ (231 X 7.835) + (4 X 7.07) – (235 X 7.8) ] MeV = 8.072 X 10-13 J

⇨

1 2

m v2 = 8.072 X 10-13

v = 1.55 X 107 m/s]

⇨

NCERT PROBLEMS Q.No. 01- Obtain the binding energy (in Mev) of a Nitrogen nucleus (7N14).

[Ans 104.67 MeV]

14

m (7N ) = 14.00307 u

Given,

= 1.00783 u mn = 1.00867 u 209 Q.No. 02- Obtain the binding energy of the nuclei 56 26𝐹𝑒 and 83𝐵𝑖 in nits of MeV from the following datamH

m ( 56 26𝐹𝑒) = 55.934939 u m ( 209 83𝐵𝑖 ) = 208.980388 u mH

[Ans. B.E. of B.E. of

56 26𝐹𝑒 = 492.26 MeV] 209 83𝐵𝑖 = 1640.3 MeV

= 1.00783 u

mn = 1.00867 u Q.No. 03- A given coin has a mass of 3.0g. Calculate the nuclear energy that would be required to separate all the neutrons & protons from each other. For simplicity assume that the coin is entirely made of 63 29𝐶𝑢 atoms(of mass 62.92960 u). The mass of proton & neutron are 1.00783 & 1.00867 u respectively. [Ans. 1.58 X 1025 MeV] [Sol. Number of atoms in 3g coin

= (6.023 X 1023 X 3)/63 = 2.688 X 1022

Mass defect of each atom ∆m = [{Z mp + (A – Z) mn } − MN] = 0.59225 u Total mass defect of all the atoms = 0.59225 X 2.688 X 1022 ⇨ required nuclear energy = 0.59225 X 2.688 X 1022 X 931 = 1.58 X 1025 MeV] Q.No. 04- A star converts all its hydrogen to helium achieving 100% helium composition. It then converts helium to carbon via the 12 reaction 3 2He4 + Q (7.27 MeV) 6C

The mass of the star is 5.0 X 1032 Kg & it generates energy at the rate of 5 X 10 3 Watt. How long will it take to convert all the helium in to carbon at this rate ? [Ans. 1.85 X 108 years] [Sol. Mass of each helium atom

= 4 amu = 4 X 1.66 X 10-27 Kg

Number of helium atoms in given mass Number of reactions ⇨

= 5.0 X 1032 / 4 X 1.66 X 10-27 = =

1 3

X

5 6.64

5 6.64

X 1059

X 1059

Now given that energy released in each reaction = 7.27 MeV = 7.27 X 1.6 X 10-13 J Total energy released (E) = No. of reactions X energy relesed in one reaction = Required time t =

1 3 𝐸 𝑃

X

5 6.64

X 1059 X 7.27 X 1.6 X 10-13 = 2.96 X 1046 J

= 2.96 X 1046/ 5 X 103 = 1.85 X 108 years]

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Radioactivity

page-01

Radioactivity The phenomenon of spontaneous and continuous distintegration of the nucleus of an atom of a heavy element on its own with the emission of certain type of radiations, is called radioactivity.

Main features of phenomenon of radioactivity (i) It is a spontenuous nuclear phenomenon.i,e, it does depend on temperature, pressure or chemical reactions. (ii) It is a statistical process ; we can not precisely predict the timing of a particular radioactivity of a particular nucleus. The nucleus can disintegrate immediately or it may take infinite time. We can predict the probability of the number of nuclei being disintegrate at an instant. (iii) The energy liberated during radioactive decay comes from within individual nuclei. (iv) when a nuclei undergoes 𝛼 or 𝛽 decay its atomic number changes and it transforms in to a new element. (v) It obeys exponential decay law.

Why heavy nucleus show the phenomenon of radioactivity ? A nucleus becomes unstable when it has too many protons in relation to the number of neutrons. Under such situation, the electrostatic repulsive force between the protons exceeds the nuclear force holding the nucleons together. The nucleus corrects this excess proton instability by emitting suitable radiations. This is radioactivity.

Why radioactivity is considered to be a nuclear phenomenon ? As we know the radioactive elements decay spontaneously to attain stability. This disintegration of radioactive elements is not affected by external conditions such as temperature, pressure, chemical reactions etc, because the energy in any physical or chemical process changes by 1-2 eV, while the energy librated in nuclear disintegration is of order of 10 6 eV. This automatically implies that the radioactivity is a nuclear phenomenon.

𝜶 −particles,𝜷-particles & 𝜸 −rays When radiations emitted by a radioactive element is placed under electric field or magnetic field, radiations split in to three rays, classified as 𝛼 −particles, 𝛽-particles & 𝛾 −rays.

𝜶 −particles An 𝛼 −particle is the nucleus of helium or it is a doubly ionised He- atom. It is denoted by 42𝐻𝑒. Charge on 𝛼 −particle = + 2e =3.2 X 10-19 C. Mass of 𝛼 −particle = 6.645 X 10-27 Kg.

𝜷 −particles 𝛽 −particles are fast moving electrons. It is denoted by Charge on 𝛽 −particle = - 1.6 X 10-19 C. Mass of 𝛽 −particle = 9.1 X 10-31 Kg.

0 −1𝑒

or

0 −1𝛽 .

𝜸 −rays 𝛾 −rays are electromagnetic waves of wavelength 0.01 A0. Obviously these are not having any charge.

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Radioactivity

page-02

Properties of 𝜶 −particles (i) 𝛼 −particles are positively charged Helium nuclei or doubly ionised He-atom. (ii) 𝛼 −particles have velocity nearly

1 10

of velocity of light.

(iii) 𝛼 −particles are deflected by electric & magnetic fields. (iv) 𝛼 −particles have highest ionising power & lowest penetrating power. (v) 𝛼 −particles affect photographic plate. (vi) 𝛼 −particles produce flourence on ZnS. (vii) 𝛼 −particles cause burn on human body.

Properties of 𝜷 −particles (i) 𝛽 −particles are fast moving electrons of nuclear origin. (ii) 𝛽 −particles have velocity nearly

3 10

to

9 10

of velocity of light.

(iii) 𝛽 −particles are deflected by electric & magnetic fields. (iv) 𝛽 −particles have ionising power less than that of 𝛼 −particles but more than that of 𝛾 −rays. (v) 𝛽 −particles have penetrating power more than that of 𝛼 −particles but less than that of 𝛾 −rays. (vi) 𝛽 −particles affect photographic plate. (vii) 𝛽 −particles produce flourence on ZnS.

Properties of 𝜸 −rays (i) 𝛾 −rays are electromagnetic radiations of very high frequency. (ii) 𝛾 −rays have velocity equal to velocity of light. (iii) 𝛾 −rays are not deflected by electric & magnetic fields. (iv) 𝛾 −rays have lowest ionising power. (v) 𝛾 −rays have highest penetrating power. (vi) 𝛾 −rays affect photographic plate. (vii) 𝛾 −rays produce flourence on ZnS.

Difference between 𝜸 −rays & X-rays 𝛾 −rays & X-rays, both are electromagnetic radiations but they have the following differences(i) origin of 𝛾 −rays is nuclear but origin of X-rays is atomic de-excitation. (ii) frequency of 𝛾 −rays is much more than that of X-rays & hence penetrating power of 𝛾 −rays is much higher than that of X-rays.

Can a single nucleus emit 𝜶 −particle, 𝜷 −particle & 𝜸 −rays together ? No, a nucleus either emit an 𝛼 −particle or 𝛽 −particles and if left excited it may emit 𝛾 −rays.

A radioactive substance emits either 𝜶 −particle or 𝜷 −particle only, then why is radioactive radiation divided in to three parts on applying an electric or magnetic field ? The radioactive decay occurs in series where daughter product give rise to grand daughter product and so on. Some of them are emitters of 𝛼 −particles while others are emitters of 𝛽 −particles. If after 𝛼- emission or 𝛽- emission, nucleus is left in the excited state it may emit 𝛾 −rays. Therefore radioactive sample give out 𝛼 −particle, 𝛽 −particles and 𝛾 −rays together.

Why does 𝜶-decay occur ? A nucleus becomes unstable when it has too many protons in relation to the number of neutrons. Under such situation, the electrostatic repulsive force between the protons exceeds the nuclear force holding the nucleons together. The nucleus corrects this excess proton instability by emitting an 𝛼 −particle i,e, two protons & two neutrons bound together.

Why does 𝜷-decay occur ? A nucleus becomes unstable when it has too many neutrons in relation to the number of protons. Such a nucleus attains a more stable state when one of its neutrons changes in to a proton and an electron. When this happens, an electron is emitted as a 𝛽 −particle followed by emission of an antineutrino.

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Radioactivity

page-03

How is 𝜸 −ray emitted ? In both 𝛼 and 𝛽 emission, the parent nucleus undergoes a change in atomic number and therefore, becomes a nucleus of another element. The new nucleus is called daughter nucleus. In many cases this daughter nucleus is in excited state when it is formed. In such a case this nucleus reaches in ground state by emitting a 𝛾 −ray. If daughter nucleus is in ground state when it is formed then there is no emission of 𝛾 −rays.

Why do 𝜶 −particles have high ionising power ? Because of their large mass & large nuclear cross section 𝛼 −particles have highest ionising power.

The ionising power of 𝜷 −particles is less than that of 𝜶 −particles but their penetrating power is more than that of 𝜶 −particles. Why ? Since 𝛽 −particles are emitted at very high speed as compared to that of 𝛼 −particles, they spend very little time in the vicinity of a single atom and, therefore, produce much less ionisation than 𝛼 −particles. Because 𝛽 −particles move quickly through the medium, the loss of energy is very slow and they can penetrate a cosiderable distance through the medium.

How do 𝜷 −particles differ from electrons obtained in photoelectric emission ? (i) 𝛽 −particles is an electron that is emitted from nucleus when a neutron changes in to proton. In photoelectric effect, the emitted photoelectrons are the outer electrons of the atoms. (ii) 𝛽 −particles emitted from the nucleus have very high speed and therefore very high kinetic energy. However, electrons obtained in photoelectric effect have very small kinetic energy.

Explanation of process of 𝜶-decay 𝛼 -decay is a process in which an unstable nucleus transforms itself in to a new nucleus by emitting an 𝛼 −particle. For example238 92U

234 90Th

+ 2He4 + Q

In general, A ZX

+ 2He4 + Q Where Q is the energy released in 𝛼-decay and is shared by daughter nucleus and 𝛼 −particle. During the 𝛼 −decay ratio

Z-2Y

A-4

𝑛 increases. 𝑝

Theory A characteristic feature of 𝛼 –decay is that the K.E. of emitted 𝛼 −particles varies from 4-9 MeV. However the surface of the nucleus of 𝛼 −emitters presents a barrier of 25 MeV. Classically, this can not happen because the energy of emitted 𝛼 −particles is not enough to overcome the potential barrier at the nuclear surface. If it happens, it would violet the law of conservation of energy. This is explained by quantum mechanics. According to Gamow, Congdon & Gurney, the nuclei decay spontaneously by actually passing through the barrier in a process known as tunneling. Uncertainity principle tells us that energy conservation can be violated by an amount ∆E for a length of time ∆t given by (∆E) (∆t) ≤

ℎ 2𝜋

Thus quantum mechanically, there is a small but finite probability that the 𝛼 −particle may pass through the barrier even its energy is insufficient to cross the barrier height.

Explanation of process of 𝜷-decay The process of spontaneous emission of an electron (e-) or a positron (e+) from a nucleus is called 𝛽 −decay. For example15P

32

16S

32

+

-1e

+

-1e

0

+ 𝜈

In general, or

A ZX

Z+1Y

ZX

Z-1

A

𝑛 During the 𝛽 −decay ratio decreases. 𝑝

A

YA +

+ 𝜈

0

+1e

0

+ 𝜈

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Radioactivity

page-04

Theory In 𝛽-decay, a neutron inside the nucleus converted in to a proton & an electron. This electron comes out of the nucleus immediately as 𝛽-particle. n

p+

-1e

0

+ 𝜈

similarly during 𝛽 + decay a proton converted in to a neutron as given below. p

n+

+1e

0

+ 𝜈

Clearly a 𝛽- decay process involves the conversion of a neutron in to a proton or vice-versa. These nucleons have nearly equal masses. That is why, mass number of a nuclide undergoing 𝛽-decay does not change. It is found experimentally that energy of 𝛽 −particle emitted by the nucleus varies continously from zero to a maximum value, while energy of daughter nucleus remains constant. Hence result does not show the validity of both conservation of energy and conservation of momentum. This problem was solved by Pauli by postulating the existance of a new particle having zero rest mass, zero charge and spin equal to ½. Fermi called this particle as neutrino.It is found that both neutrino & antineutrino are involved in 𝛽 −decay. Thus energy shared by 𝛽 −particle and antineutrino randomly. In 𝛽- decay the disintegrating energy is shared in all proportionals between all the three particles (i,e, daughter nucleus, e0 & 𝜈 ). As a result kinetic energy of e0 is not fixed, it varies from 0 to a max value. Hence 𝛽 emission has a continous energy spectrum as shown.

Explanation of process of 𝜸-decay The process of spontaneous emission of a 𝛾-ray photon during the radioactive disintegration of a nucleus is called 𝛾-decay.

In general, A ZX

Z

XA + 𝛾

Theory After an 𝛼 or 𝛽 decay the daughter nucleus is usually left in excited state. It attains the ground state by emitting one or more photons. As the nuclear states have energies of order of several MeV, therefore photons emitted by the nuclei also have the energies of several MeV.The wavelength of such high energy photons is very small. These short wavelength electromagnetic radiations emitted by nuclei are called 𝛾-rays. An example of a 𝛾-decay is shown through an energy level diagram. In it an unstable 60 27𝐶𝑜 is transformed by a 𝛽-decay in to an 60 excited 28𝑁𝑖 nucleus, which in turn reaches the stable ground state by emitting photons of energies 1.17 MeV and 1.33 MeV, in two successive 𝛾-decay processes.

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Radioactivity

page-05

Radioactive decay The spontaneous emission of radiation from a radioactive element is called radioactive decay.

Decay Law The number of nuclei disintegrating per second of a radioactive sample at any instant is directly proportional to the number of undecayed nuclei present in the sample at that instant. i,e,

⇨

𝑑𝑁 𝑑𝑡

∝ N

𝑑𝑁 𝑑𝑡

=−λN

---------------------(1)

Where λ is constant of proportionality & is called decay constant. From equation (1) we have 𝑑𝑁 𝑑𝑡

=−λN

𝑑𝑁 𝑁

= − λ dt

⇨ Integrating on both sides we get,

∫

⇨

𝑑𝑁

= − λ ∫ 𝑑𝑡

𝑁

log 𝑒 𝑁 = − λ t + C

But, when

-------------------------(2)

t= 0, N= N0, therefore from equation (2) we get

N0

log 𝑒 𝑁0 = − λ X 0 + C

⇨

log 𝑒 𝑁0 = C

On putting this value of C in equation (2) we get

log 𝑒 𝑁 = − λ t + log 𝑒 𝑁0

⇨

N

log 𝑒 𝑁 − log 𝑒 𝑁0 = − λ t

⇨

log 𝑒

⇨

𝑁 𝑁0

⇨

𝑁 𝑁0

= −λt

= 𝑒 −𝜆𝑡

N = N0 𝑒 −𝜆𝑡

t

-----------------------(3)

This equation is known as decay equation.

N = N0 𝑒 −𝜆𝑡

From eqn (3) we have Substituting

t=

1 𝜆

in the above equation we get

N = N0 𝑒 −1 1

⇨

N = N0 ( 𝑒 )

Thus decay constant of a radioactive element may be defined as the reciprocal of the time in which number of undecayed nuclei of that radioactive elemnt falls to

1 times of its initial value. 𝑒

S.I. unit of decay constant is sec-1.

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Radioactivity

page-06

Half Life Period (T) The time interval in which one –half of the radioactive nuclei originally present in a radioactive sample disintegrate, is called half-life of that radioactive substance. S.I. unit of half life is second.

Relation between decay constant (𝝀)& Half Life Period (T) We know that,

N = N0 𝑒 −𝜆𝑡 When

t =T,

-------------------------(1)

𝑁0 N= , therefore, from equation (1) we get 2 𝑁0 = N0 𝑒 −𝜆𝑇 2

⇨

1 2

= 𝑒 −𝜆𝑇

⇨

2

= 𝑒 𝜆𝑇

⇨

𝜆𝑇 =

⇨

𝜆 =

----------------------------(2) N0

log 𝑒 2 log𝑒 2 𝑇

N ⇨

𝜆 =

2.303 𝑋 log10 2 𝑇

⇨

𝜆 =

2.303 𝑋 0.3010 𝑇

⇨

𝜆 =

0.6931 𝑇

Note-

𝑁0 2 𝑁0 4

t

We know that,

N = N0 𝑒 −𝜆𝑡 When

t =T,

-------------------------(1)

𝑁0 N= , therefore from equation (1) we get 2 𝑁0 = N0 𝑒 −𝜆𝑇 2

⇨

1 2

= 𝑒 −𝜆𝑇

----------------------------(2)

Let total time be’ t’ then the number of half lives in this time n

=

𝑡 𝑇

therefore from eqn (1) we can write,

N = N0 𝑒 (−𝜆𝑇)𝑋 ⇨

N = N0 𝑒 ( –𝜆𝑇)

⇨

N = N0 ( 2)𝑛

1

𝑡/𝑇

𝑛

[∵

1 2

= 𝑒 −𝜆𝑇 ]

Significance of Half Life Period (T) Significance of half life is that , it gives an idea of the relative stability of that isotope. The determination of half life of a radioactive substance is very useful for geologists in estimating the ages of mineral deposits, rocks, earth etc.

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Radioactivity

page-07

Mean Life(𝝉) The average time for which the nuclei of a radioactive sample exist is called mean life or average life of that sample. It is equal to the ratio of the combined age of all the nuclei to the total number of nuclei present in the given sample. i,e,

Mean life (𝜏)

𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑣𝑒𝑠 𝑜𝑓 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑛𝑢𝑐𝑙𝑒𝑖

=

𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑛𝑢𝑐𝑙𝑒𝑖

Now let, at t=0, N = N0 . After time t, this number reduces to N. Also let dN nuclei disintegrates in time from t to t + dt.As dt is very small so the life of each of the dN nuclei can be approximately taken equal to t. ∴

Total life of dN nuclei

= t dN

⇨

Total life of all the N0 nuclei

= ∫0 0 𝑡 𝑑𝑁

mean life

or

(𝜏)

𝑁

𝜏 = As

⇨

𝑇𝑜𝑡𝑎𝑙 𝑙𝑖𝑓𝑒 𝑜𝑓 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑁0 𝑛𝑢𝑐𝑙𝑒𝑖

=

𝑁0 1 𝑁0 ∫ 𝑡 𝑑𝑁 𝑁0 0

------------------------------(1)

N = N0 𝑒 −𝜆𝑡 dN = − 𝜆 N0

Also when t= 0 , N = N0 and when N = 0, t = ∞

𝜏 =

𝑒 −𝜆𝑡 dt changing limits of integration in terms of time we get,

1 ∞ ∫ 𝑡 𝜆 N0 𝑁0 0

𝑒 −𝜆𝑡 dt

𝜏 = 𝜆 ∫0∞ 𝑡 𝑒 −𝜆𝑡 dt −𝜆𝑡

∞

𝜏 = 𝜆 [{ 𝑡 𝑒−𝜆 }0 𝜏 = 𝜆 [0 +

−𝜆𝑡

− ∫0∞ ( 𝑒−𝜆 ) dt ]

1 ∞ −𝜆𝑡 dt ] ∫ 𝑒 𝜆 0

𝜏 = ∫0∞ 𝑒 −𝜆𝑡 dt 𝑒 −𝜆𝑡

𝜏 = [ ] −𝜆

⇨ ⇨

∞ 0

𝜏 =−

1 𝜆

[𝑒 −𝜆∞ − 𝑒 −0 ]

𝜏 =−

1 𝜆

[0 − 1]

𝜏 =

1 𝜆

𝜏 =

1 0.6931/𝑇

𝜏 =

𝑇 0.6931

𝜏 = 1.44 T

Mean life is also given as the reciprocal of its decay comstant. S.I. unit of mean life is second.

[∵ 𝜆 =

0.6931 𝑇

]

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Radioactivity

page-08

Activity or Decay rate (R) The activity or the rate of decay of a radioactive sample may be defined as the number of radioactive disintegrations taking place per second in the sample. 𝑑𝑁 𝑑𝑡

R=−

i,e,

R=−

𝑑 𝑑𝑡

(N0 𝑒 −𝜆𝑡 )

R = − N0

𝑑

[∵ N = N0 𝑒 −𝜆𝑡 ]

( 𝑒 −𝜆𝑡 )

𝑑𝑡

R = − N0 (−λ 𝑒 −𝜆𝑡 )

R0

R = λ N0 𝑒 −𝜆𝑡

R

R=𝜆N

⇨ ⇨

R = 𝜆 N0 𝑒 −𝜆𝑡

⇨

R = (𝜆 N0) 𝑒 −𝜆𝑡

t

⇨ R = R0 𝑒 −𝜆𝑡

[∵ 𝑅0 = 𝜆 N0 ]

Note R= 𝜆N

(i)

R= [

⇨ (ii)

We know,

⇨

log𝑒 2 ] 𝑇 𝑑𝑁 𝑑𝑡

R= 𝑑𝑅 𝑑𝑡

= =

𝑑 𝑑𝑡 𝑑 𝑑𝑡

=𝜆

(

N = 𝜆N

𝑑𝑁 𝑑𝑡

[∵ 𝜆 =

log𝑒 2 ] 𝑇

---------------------(1)

)

(𝜆 N) 𝑑 𝑑𝑡

(N)

= 𝜆 (𝜆 N) = 𝜆2 N log𝑒 2 2 ] 𝑇

=[

⇨

𝑑𝑅 𝑑𝑡

∝

N

1 𝑇2

Units of activity (1) Becquerel (Bq) One Becquerel is defined as the decay rate of one disintegration per second. i,e, 1 Bq = 1 decay/sec S.I. unit of activity is Becquerel. (2) Curie (Ci) One curie is the decay rate of 3.7 X 1010 disintegration per second. i,e, 1 Curie = 3.7 X 1010 decays/sec (3) Rutherford (Rd) One Rutherford is the decay rate of 106 disintegration per second. i,e, 1 Rutherford = 106 decays/sec

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Radioactivity

page-09

Q.No.01- Explain with the help of a nuclear reaction in each of the following cases, how the neutron to proton ratio changes during(i) 𝛼-decay (ii) 𝛽-decay. (D-2006,2004,AI-2003,F-2002) Q.No.02- Will the neutron to proton ratio increase or decrease in nucleus when (i) an electron (ii) a proton emitted ? Q.No.03- (i) what is 𝛽-decay ? (ii) Write the nuclear decay process of 𝛽-decay

(D-2003)

(F-2002) (AI-2010,2004,D-2004)

32 15𝑃

Q.No.04- Write the nuclear reactions for the following(i) 𝛼-decay of

204 84𝑃𝑜

(ii) 𝛽-decay of

32 15𝑃

(DC-2005)

(iii) 𝛽+ -decay of

11 6𝐶 .

Q.No.05- What is meant by the term radioactive decay ? 235 92𝑈

decays successively to form

(AIC-2002)

230 226 234 231 92𝑈, 91𝑃𝑎, 90𝑇ℎ, 88𝑅𝑎 .

Find the radiations emitted in each case.

Q.No.06- In a series of radioactive disintegration of 𝐴𝑍𝑋 first an 𝛼-particle and then a 𝛽- particle is emitted. What is the atomic number and mass number of new nucleus formed by the successive disintegrations. (D-2003) Q.No.07- Which of the following radiations 𝛼-rays, 𝛽-rays, 𝛾-rays(i) are similar to X-rays (ii) are easily absorbed by the matter (iii) travel with greatest speed (iv) are similar in nature to cathode rays.

(AI-2001)

Q.No.08- Explain how radioactive nuclei can emit 𝛽- particles even though atomic nuclei do not contain these particles ? Hence explain why the mass number of a radioactive nuclei does not change during 𝛽-decay ? (DC-2004) Q.No.09- (a) Draw the energy level diagram, showing the emission of followed by 𝛾-rays by a

60 27𝐶𝑜

nucleus.

(AI-2005)

(b) Plot the distribution of kinetic energy of 𝛽-particles and state why the energy of spectrum is continous ? Q.No.10- A radioactive nucleus A decays as given below0 -1e A

(AIC-2003) 𝛼 A1

A2

If the mass number and atomic number of A are 180 & 72 respectively, find the mass number & atomic number of A 2. Q.No.11- A radioactive nucleus decays as follows -

(AI-1999) +1e

176 71𝐴

0

𝛼 A1

A2

If the mass number and atomic number of A are 176 & 71 respectively, What are the mass number & atomic no. of A 1 & A2 ? Which of these elements are isobars ? Q.No.12- A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme𝛼 𝛽 𝛼 𝛾 A A1 A2 A3 A4

(D-2009,AIC-2002)

If the mass number and atomic number of A are 180 & 72 respectively, What are these numbers for A 4 ? Q.No.13- The sequence of stepwise decays of a radioctive nucleus is 𝛼 𝛽 𝛼 D D1 D2

(AIC-2005) D3

If the nucleon number and atomic number for D2 are 176 & 71 respectively, What are the corresponding values of D & D3 ? Justify your answer in each case. Q.No.14- The sequence of stepwise decays of a radioctive nucleus is (AI-1998) 𝛼 𝛽 𝛼 𝛼 D D1 D2 D3 D4 If the nucleon number and atomic number for D2 are 176 & 71 respectively, What are the corresponding values of D & D4 ? Justify your answer in each case.

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Radioactivity

Q.No.15- A radioactive isotope D decays according to the sequence 1 𝛼 0𝑛 D D1 D2

page-10 (AI-1998)

If the mass number and atomic number for D2 are 176 & 71 respectively, find the mass number and atomic number of D. Amongest D, D1, & D2, do we have any isotope or isobar ? Q.No.16- The sequence of stepwise decays of a radioctive nucleus is 𝛼 𝛽− D D1 D2

(D-2010)

If the atomic number and mass number of D2 are 71 & 176 respectively, What are their corresponding values for D ? Q.No.17- A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme𝛼 𝛽 𝛼 𝛾 A A1 A2 A3 A4

(D-2009)

If the mass number and atomic number of A are 190 & 75 respectively, What are these numbers for A 4 ? Q.No.18- Define the terms half life period & decay constant of a radioactive substance. Write their S.I. units. Establish the relation between them. (AI-2006,04, D-2005,01) Q.No.19- State the law of radioactive decay. Establish a mathematical relation between half life period & disintegration constant of a radioactive nucleus. (F-2007) Q.No.20- Use the basic law of radioactive decay to show that , radioactive nuclei follow an exponential decay law. Hence obtain a formula for the half life of a radioactive nuclei in terms of its disintegration constant. (DC-2004) Q.No.21- Derive an expression for the average life of a radionuclide. Give its relationship with the half life.

(AI-2010)

Q.No.22- Define the activity of a radioactive nuclide. Write its S.I. unit. Give a plot of the activity of a radioactive species versus time. How long will a radioactive isotope, whose half life is T years, take for its activity to reduce to 1/8 th of its initial value. (AI-2009) Q.No.23- Define the term activity of a radioactive substance. State its S.I. unit. Two different radioactive elements with half lives T1 and T2 have N1 and N2 (undecayed) atoms respectively present at a given instant. Determine the ratio of their activities at this instant. (SP-2010) Q.No.24- Prove that the instanteneous rate of change of the activity of a radioactive substance is inversely proportional to the square of its half life. (SP-2010) Q.No.25- (a) Define activity of a radioactive material and write its S.I. unit. (b) Plot a graph showing variation of activity of a given radioactive sample with time.

(D-2010)

Q.No.26- Complete the following decay process for 𝛽-decay of phosphorous -32.

(F-2004)

32 15𝑃

S + ------------------

The graph shows how the activity of a radioactive nucleus changes with time. Using the graph determine(i) half life of the nucleus , and (ii) its decay constant. [Ans. (i) T = 40 sec. (ii) λ = 0.0173 S-1]

Worksheet – XII (Physics) Unit- VIII (Atoms & Nuclei )

Topic- Radioactivity

page-11

Q.No.27- The graph shows how the activity of a sample of radon-220 changes with time. Use the graph to determine its half life. Also calculate the value of decay constant of radon -220. [Ans. (i) T = 16 days. (ii) λ = 0.043 day-1] (D-2004)

Q.No.28- Show that the decay rate ‘R’ of a sample of a radioactive nuclide is related to the number of radioactive nuclei ‘N’ at the same instant by the expression R = λ N. (D-2005) Q.No.29- (a) Define the activity of a radioactive nucleus and state its S.I. unit. (F-2005) (b) Two radioactive nuclei X and Y initially contain equal number of atoms. The half life is 1 hr & 2hr respectively. Calculate the ratio of their rates of disintegration after two hours. Q.No.30- Draw a plot representing the law of radioactive decay. Define the activity of a sample of a radioactive nucleus. Write its S.I.unit. (F-2008) Q.No.31- The activity of a radioactive element drops to

1 16

th of its initial value in 32 Years. Find the mean life of the sample.

[Ans. 𝜏 = 11.2 yrs] (AIC-1997) 20 against 𝛼-decay is 1.5 X 1017 sec. What is the activity of a sample of 238 92𝑈 having 25 X 10 atoms ? 4 [Ans. R = 1.15 X 10 Bq ] (D-2005) 238 9 Q.No.33- The half life of 238 92𝑈 against 𝛼-decay is 4.5 X 10 years. What is the activity of 1g sample of 92𝑈 ? 4 [Ans. R = 1.23 X 10 Bq ] (AI-2005,F-2006,NCERT) Q.No.32- The half life of

238 92𝑈

Q.No.34- A radioactive sample contains 2.2 mg of pure 116𝐶 which has half life period of 1224 sec. Calculate(AI-2005) 20 (i) the number of atoms present initially [Ans. (i) 𝑁0 = 1.2 X 10 (ii) R = 1.55 X 1014 Bq ] (ii) the activity when 5 𝜇𝑔 of the sample will be left. Q.No.35- (a) what is meant by half life of a radioactive element ? (b) the half life of a radioactive substance is 30 s. Calculate (i) the decay constant, and (ii) time taken for the sample to decay by 3/4th of its initial value. Q.No.36 A radioactive material is reduced to 4X

10-3

1 16

(F-2009,D-1999) [Ans. (i) λ = 0.00231 s-1 (ii) 60 s]

th of its original amount in 4 days. How much material should one begin with so that

Kg of the material is left after 6 days ?

[Ans. 𝑚0 = 0.256 Kg ] (SP-2010)

Q.No.37- The decay constant for a given radioactive nuclide has a value of 1.386 day-1. After how much time will a given sample of this radionuclide get reduced to only 6.25 % of its present number ? [Ans. t = 2 days] (AIC-2004) Q.No.38- The half life of a given radioactive nuclide is 138.6 days. What is the mean life of this nuclide ? After how much time will a given sample of this radioactive nuclide will get reduced to only 12.5 %, of its initial value ? [Ans. 𝜏 = 199.58 days, t= 415.8 days] (AIC-2004) ---------------------------------------------------x------------------------------------------------------------

Q.No. 01- The mass of a nucleus in its ground state is always less than the total mass of its constituents – neutrons and protons. Explain. (AI-2009) [Ans. When the neutrons and protons group to form a nucleus, energy is needed. This energy comes from mass defect created.]

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