Pdt Unit16 Df024(student)

UNIT 16: ATOMIC PHYSICS

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16.1 Atomic models and energy levels in atom 16.1.1 Thomson’s model of the atom


In 1898, Joseph John Thomson suggested a model of an atom that consists of homogenous positively charged spheres with tiny negatively charged electrons embedded throughout the sphere as shown in figure 16a.





positively charged sphere

The electrons much likes currants in a plum pudding. This model of the atom is called ‘plum pudding’ model of the atom.

electron Fig. 16a

16.1.2 Rutherford’s model of the atom


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In 1911, Ernest Rutherford performed a critical experiment that showed the Thomson’s model is not correct and proposed his new atomic model known as Rutherford’s planetary model of the atom as shown in figure 16b. 2







nucleus




electron


Fig. 16b




-e




+Ze

The atom was pictured as electrons orbiting around a central nucleus which concentrated of positive charge. The electrons are accelerating because their directions are constantly changing as they circle the nucleus. Based on Maxwell’s electromagnetic theory, an accelerating charge emits energy. Hence the electrons must emit the e.m. radiation as they revolve around the nucleus. As a result of the continuous loss of energy, the radii of the electron orbits will be decreased steadily. This would lead the electrons spiral and falls into the nucleus, hence the atom would collapse as shown in figure 16c.

‘plop’

energy loss SF027

Fig. 16c

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16.1.3 Bohr’s Model of Hydrogen Atom



In 1913, Neils Bohr proposed a new atomic model based on hydrogen atom. According to Bohr’s Model, he assumes that each electron moves in a circular orbit which is centered on the nucleus, the necessary centripetal force being provided by the electrostatic force of attraction between the positively charged nucleus and the negatively charged electron as shown in figure 16d.


-e


r FE

+e

r

r v

On this basis he was able to show that the energy of an orbiting electron depends on the radius of its orbit. This model has several features which are described by the postulates (assumptions) stated below : 1.

The electrons move only in certain circular orbits, called STATIONARY STATES or ENERGY LEVELS. LEVELS When it is in one of these orbits, it does not radiate energy. energy

Fig. 16d

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2.

The only permissible orbits are those in the discrete set for which the angular momentum of the electron L equals an integer times h/2π .

nh L= 2π = mvrn ; n = 1,2,3,.. 3.

Emission or absorption of radiation occurs only when an electron makes a transition from one orbit to another. The frequency f of the emitted (absorbed) radiation is given by where

E f : final energy state

Ei : initial energy state

1eV = 1.6 × 10 −19 J Note :   SF027

If ∆E ⇒ negative value If ∆E ⇒ positive value 5

16.2 Electron transitions in hydrogen atom





The energies of the electron in an atom can have only certain values. This values are called the energy level of the atom. The energy level of hydrogen atom can be calculated by using eq. (16.2a) which is

When n = 1, the ground state (the state of the lowest energy level) level

n = 2, the first excited state, n = 3, the second excited state, n = 4, the third excited state, n = ∞, SF027

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Ionization energy

Figure 12.3a shows diagrammatically the various energy levels in the hydrogen atom.

n ∞

En (eV ) 0.0 Free electron

5 4 3

− 0.54 − 0.85 − 1.51

4th excited state 3rd excited state 2nd excited state

2

− 3.39

1st excited state

excited state

Excitation energy - the energy required by an electron that raises it to an excited state from its ground state.

− 13.6

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Fig. 16e

Ground state 7




Example 1 : The electron in the hydrogen atom makes a transition from E2= -3.40 eV energy state to the ground state with E1= -13.6 eV. Calculate a. the energy in Joule, and b. the wavelength of the emitted photon. (Given h = 6.63 x 10-34 J s and c = 3.00 x 108 m s-1) Solution: a. The energy of the emitted photon is

∆E = E f − Ei ∆E = E1 − E2

b. The wavelength of the photon is

hc ∆E = λ

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Example 2 : The lowest energy state for hydrogen atom is E1= -13.6 eV. Calculate the frequency of the photon required to ionize the atom. (Given h = 6.63 x 10-34 J s and c = 3.00 x 108 m s-1) Solution: An atom where its electron at ground state is raised to the zero energy level said to be ionized. The ionization energy is

∆E = E f − Ei

The frequency of the photon required to ionize the atom is

∆E = hf

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Example 3 : A hydrogen atom emits radiation of wavelengths 121.5 nm and 102.4 nm when the electrons make transitions from the 1st excited state and 2nd excited state respectively to the ground state. Calculate: a. the energy of a photon for each of the wavelengths above; b. the wavelength emitted by the photon when the electron makes a transition from the 2nd excited state to the 1st excited state. (Given h = 6.63 x 10-34 J s and c = 3.00 x 108 m s-1) Solution: λ1=121.5x10-9 m, λ2=102.4x10-9 m a. The energy of the photon due to transition from 1st excited state to the ground state is hc

∆E1 =

λ1

The energy of the photon due to transition from 2nd excited state to the ground state is

hc ∆E2 = λ2 SF027

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b. ∆E3

2nd excited state 1st excited state

∆E1

∆E2 Ground state

Therefore the wavelength of the emitted photon due to the transition from 2nd excited state to the 1st excited state is

hc ∆E3 = λ3

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Example 4 : The electron of an excited hydrogen atom make a transition from the ground state to the 4th excited state. Determine the energy absorbs by the atom. (Given h = 6.63 x 10-34 J s and c = 3.00 x 108 m s-1) Solution: a. By applying the equation of energy level in hydrogen atom,

Ground state, n = 1

E1 = −

13 . 6 eV 12

4th excited state, n = 5

E5 = −

13 . 6 eV 52

13 . 6 eV En = − n2

Hence the energy absorbs by the atom is

∆E = E f − Ei

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Emission Line Spectrum



The emission lines correspond to photons of discrete energies that are emitted when excited atomic states in the gas make transitions back to lower energy levels. Figure 12.4a shows line spectra produced by emission in the visible range for hydrogen (H), mercury (Hg) and neon (Ne).

Fig. 12.4a 12.4.1 Hydrogen Emission Line Spectrum
Emission processes in hydrogen give rise to series, which are sequences of lines corresponding to atomic transitions as shown in figure 12.4b.

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Additional notes on Atomic Physics As a knowledge

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Bohr’s Radius in Hydrogen atom





Consider one electron of charge –e and mass m moves in a circular orbit of radius r around a positively charged nucleus with a velocity v (Figure 12.2a). The electrostatic force between electron and nucleus contributes the centripetal force as write in relation below :

electrostatic force




FE = Fc kq1q2 mv 2 = 2 r r

centripetal force

q1 = e ; q2 = e nh mvr = 2π

where

From the Bohr’s second postulate :

1 and k = 4πε0

By taking square of both side2of the equation, we get

e mv = 4πε0 r

(1)

2 2 n h 2 2 2 mv r = 4π 2

(2)

2

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By dividing eq. (12.2d) and (12.2c), thus

n 2 h 2ε 0 mr = 2 eπ n 2 h 2ε 0 r= ; n = 1,2,3,... 2 me π


(3)

which r are radii of the permissible orbits for the Bohr atom. Eq. (12.2e) can be written as 2

r = a0 n ; n = 1,2,3,...with 2

h ε0 a0 = me 2 π

Where a0 is called the Bohr’s radius of hydrogen atom. It is defined as the radius of the lowest orbit or ground state (n=1) and is given by 2

( 6.63 × 10 ) (8.85 × 10 ) r = a (1) = (9.11 × 10 )(1.60 × 10 ) π − 34

−12

2

0

− 31

−19 2

a0 = 5.31 × 10 −11 m
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The radii of the orbits associated with allowed orbits or states n

2,3,… are 4a0,9a0,…, hence the atomic radius is quantized.

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Discrete Energy in Hydrogen atom


The total energy E of the system is given by Kinetic energy of the electron

E = K +U

(12.2f)

Potential energy of the electron The potential energy, U of the electron is

1 kq1q2 q = − e q = e ; where and k= U= 2 1 4πε0 r 2 e (12.2g) U =− 4 πε0 r and the kinetic energy, K of the electron is

1 2 K = mv 2

(12.2h)

By substituting eq. (12.2c) into eq. (12.2h), thus

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1  e2   K =  2  4πε0 r  e2 K= 8 πε0 r

(12.2i)

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By substituting eq. (12.2g) and (12.2i) into eq. (12.2f), therefore the total energy E of the system is

 e2   E= +  − 8πε 0 r  4πε 0 r  e2 (12.2j) E=− 8 πε0 r 2 Ze In general, En = − 8 πε0 r where Z : atomic number e2







(12.2k)

From de Broglie’s relation, the electron is to be regarded as a wave, then its stable orbits in an atom are those satisfy the conditions of a standing (stationary) wave as shown in figures 12.2b, 12.2c, 12.2d , 12.2e and 12.2f. λ λ λ orbital

n=1

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Equation of discrete energy in hydrogen atom only

λ = 2πr1 Fig. 12.2b

n=2

2λ = 2πr2 Fig. 12.2c

n=3

3λ = 2πr3 Fig. 12.2d

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λ







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λ n=4

n=5

4λ = 2πr4

5λ = 2πr5

Fig. 12.2e

Fig. 12.2f

If there are n waves in the orbital and λ is wavelength of wave properties of electron therefore

nλ = 2πrn (12.2l) th where rn : radius of the n orbit n = 1,2 ,3,... h Since λ = then eq. (12.2l) can be written as mv  h  n  = 2πrn  mv  nh mvrn = Bohr’s second postulates 2π

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The energy level of hydrogen atom is given by

n 2 h 2 ε0 Ze 2 where Z = 1 and r = En = − me 2 π 8 πε0 r (hydrogen atom) ( 1)e 2 En = −  n 2 h 2 ε0   8πε0  2  me π  me 4 me 4 En = − 2 2 2 and E1 = 2 2 = 13.6 eV 8 ε0 h 8 ε0 h n 13.6 eV E1 (12.2m) En = − 2 or En = − 2 n n th where En : energy level of the n orbit(state) E1 : energy level of ground state n = 1,2 ,3,... The negative sign means that work has to be done to remove the electron to infinity, where it is considered to have zero energy, i.e. the electron is bound to the atom. SF027

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