ATOMIC NUCLEUS

RUTHERFORDS -PARTICLE SCATTERING EXPERIMENT

1. Suggested by- Rutherford

to investigate the structure

of atom.

2. Performed by- H.Geiger

and E.Marsden in 1911.

Assumptions

1. - Particle will suffer not

more than one scattering, as

gold foil is thin.

2. Nucleus is much heavier

than - particle therefore it remains at rest throughout the scattering process.

Experiment

1. Beam of 5.5 MeV - particles from Bi is directed on a thin gold foil.

2. - Particles scattered in different directions detected by rotatable detector.

3. Detector consists of ZnS screen and microscope.

4. When scattered - particles strike the screen then scintillations are produced which can be viewed

by microscope.

Observations:

Graph between - particles scattered and

scattering angle () is shown in fig.

1. Most of the - particles pass straight through

the fold foil.

2. Only few - particles were scattered by 900 or

more.

Scattering angle. The angle of deviation of an -

particle from its original direction.

RUTHERFORDS MODEL OF ATOM

(i) Every atom consists of a tiny central core, called nucleus which contains all its positive charge and

most of its mass .

!!Therefore, - particles traveling close to the nucleus will have large scattering angle. Thus

Rutherfords model explains large angle scattering of - particles.

(ii) The radius of the nucleus is of the order of 10-15 m and that of the atom is of the order of 10-10 m.

Therefore, the nucleus occupies only a small portion of the available space. Thus the atom mostly

consists of empty space.

!! Since very few particles were scattered through large angles, the probability of head-on approach is

small and indicates that the nucleus occupies only a small portion of the available space.

(iii) The electrons occupy the space outside the nucleus. Since an atom is electrically neutral, the

positive charge on the nucleus is equal to the negative charge on the electrons.

(iv) Electrons are not stationary but revolve around the nucleus in circular paths. In this way

Rutherford provided stability to the atom. It is because now the force of attraction between the

electrons and the nucleus provided the necessary centripetal force to the revolving electrons

otherwise if they are stationary then they would fall into nucleus.

Drawbacks: Rutherfords atomic model presented the following two main difficulties :

(i) According to Maxwells theory of electromagnetism an accelerated charge radiates energy as

electromagnetic waves. Since the electron moving around the nucleus is under constant acceleration

therefore, it should continuously lose energy. Due to this continuous loss of energy, the electrons

must fall into nucleus

Rutherford model failed to account for the stability of the atom.

(ii) During inward spiraling, the electrons angular frequency continuously increases. As a result,

electrons will radiate electromagnetic waves of all frequencies i.e. the spectrum will be continuous .

But Experiments show that an atom emits line spectra and each line corresponds to a particular

frequency/wavelength.

Rutherford model failed to explain the emission of line spectra.

IMPACT PARAMETER (b):The perpendicular distance of the initial velocity vector ( ) of the -

particle from the central line of the nucleus when the - particle is far away from the nucleus is called

impact parameter.

20

2

2

14

)2/cot(

mv

Zeb

)2/cot(b

(i) If b is large, then will be small. The - particle suffer small deflection.

(ii) If b is small, then will be large. The - particle suffer more deflection.

(iii) If b=0 ( - particle directed towards the nucleus), then =1800. The - particle will retrace its

path.

DISTANCE OF CLOSEST APPROACH (SIZE OF NUCLEUS)

The smallest distance between the nucleus and -particle fired for head on collision towards the

nucleus is called the distance of closest approach (r0).

Principle. As the -particle move towards the nucleus then it experiences a repulsive force and it

starts to slow down. Its kinetic energy starts to convert into potential energy. At r0 Its whole initial

kinetic energy is converted into potential energy. It comes to rest and then retrace its path due to

repulsion.

Mathematical expression for r0

00

04

1

r

eeZUisratEnergyPotential

This potential energy is converted in kinetic energy

2

00 2

1

4

1vm

r

eeZEU k

Distance of nearest approach 2

2

0

0

2

14

1

vm

eZr

It is approximately equal to the radius of the nucleus.

BOHRS MODEL OF ATOM

Salient features: Bohr retained the essential features of Rutherfords model, namely; a massive

positively charged nucleus and the electrons revolving around it in circular paths. However, he

modified Rutherfords model by introducing three postulates.

Postulate 1: The electrons can revolve around the nucleus only in certain stable orbits (stationary

states) without the emission of radiant energy.

+

b

Nucleus

Postulate 2: Only those stationary orbits are permissible in which angular momentum of electron is

an integral multiple of h/2 (h = planck's constant)

Thus, the angular momentum does not have continuous range of values i.e. it is quantised.

Postulate 3: The orbiting electron emits energy in the form of electromagnetic waves as it jumps from

outer stationary orbit to the inner orbit of lower energy. The energy radiated is equal to the

difference between the initial and final state.

If is the frequency of emitted radiation than

h = Ei - Ef

Bohr's Orbits (for Hydrogen and H2-like Atoms): Consider an electron of charge e and mass m revolving around nucleus with speed v in a circular orbit of radius r. (1) Radius of orbit : From first postulates of Bohr atom model, for an electron around a stationary nucleus the electrostatics force of attraction provides the necessary centripetal force i.e.

r

mv

r

eZe 2

20

)(

4

1

. (i)

From second postulates of Bohr atom model 2

hnrvm

rm

hnv

2 .(ii)

From equation (i) and (ii) radius of nth orbit

Z

n

eZm

hn

emZk

hnrn

2

2

022

22

22

53.04

(iii)

Z

nrn

2

(2) Speed of electron : From the above relations (ii) & (iii), speed of electron in nth orbit can be calculated as

secmn

Z

n

Zc

hn

eZ

hn

eZkvn /102.2.

1372

2 6

0

22

where (c = speed of light 3 108 m/s)

(3) Total energy : From eq (i) 2

0

)(

4

1mv

r

eZe

so kinetic energy

r

eZeK

)(

4

1

2

1

0 .

Electrostatic potential energy, r

eZeU

))((

4

1

0

Total energy (E) is the sum of potential energy and kinetic energy i.e.

r

eZe

r

eZeUKE

)(

4

1

2

1))((

4

1

00

nr

ZeE

2

04

1

2

1

From eq (iii) 2

022

ezm

hnrn

.

Hence 2

2

220

4

.8 n

z

h

emE

(iv) OR eV

n

ZEn 2

2

6.13

v

L

r

m, e

E2

E1 E2 E1 = h

Note: Total energy 2

UKE &

320

4

8 hc

emR

= Rydberg's constant = 1.09 107 per m.

(4) Transition of Electron: When an electron makes transition from higher energy level having energy E2(n2) to a lower energy

level having energy E1 (n1) then a photon of frequency is emitted

(a) Energy of emitted radiation : From eq (iv)

12 EEE 21

2

220

4

22

2

220

4

.8

.8 n

z

h

em

n

z

h

em

22

21

220

42 11

8 nnh

eZmE

OR

22

21

2 116.13nn

ZE

(b) Frequency of emitted radiation; hE

2

2

2

122

0

42 11

8 nnh

eZmvh

The frequency of the emitted radiations can be found from the following relation

2

2

2

132

0

42 11

8 nnh

eZmv

(c) Wave number/wavelength: If c be the velocity of light and its wavelength, then v = c/

2

2

2

132

0

4 11

8 nnh

mecv

or

2

2

2

1

32

0

4 11

8

1

nnch

me

=

2

2

2

1

11

nnR

where R = ,8 3

2

0

4

ch

me

R is known as Rydbergs constant and its value is .10097.1 17 m

Wave number is the number of waves in unit length. It is reciprocal of wavelength is given by

2

2

2

1

111

nnR

This equation is the general expression for the wave number of radiation emitted by the electron when it jumps from higher orbit n2 to lower orbit n1.

ELECTRON ENERGY LEVELS IN HYDROGEN ATOM -The orbit energy of an electron revolving in nth orbit

is given by 222

0

4

8 nh

meEn

=

2234212

41931

)1062.6()10854.8(8

)106.1()1011.9(

n

=

2

19107.21

n

joule

= eVn 219

19 1

106.1

107.21

= eVn2

6.13 (1) ( 1 eV = 1.6 10-19 joule)

The significance of negative sign in eq. (1) is that electron is bound to the nucleus by attractive forces and to separate the electron from the nucleus energy must be supplied to it. Giving different values to n, we can calculate the orbital energy or binding energy of the electron in different orbitals.

eVE 6.131 when n = 1 (K-shell)

eVE 4.32 n = 2 (L-shell)

eVE 5.13 n = 3 (M-shell)

eVE 0 n = (Limiting case) The lowest energy level (n = 1) corresponds to normal unexcited state of hydrogen. This state is also called as ground state. In energy level diagram the lower energies (more negative) are at the bottom while higher energies (less negative) are at the top. By such a consideration the various electron jumps between allowed orbits will be vertical arrows between different energy levels. The energy of radiated photon is greater when the length of arrow is greater. SPECTRAL SERIES OF HYDROGEN ATOM -Wavelength of different members of the series can be found from the following relation.

v =

2

2

2

1

111

nnR

This relation explains the complete spectrum of hydrogen.

(a) Lyman series - The series consists of all wavelengths which are emitted when electron jumps from an outer orbit to the first orbit i.e., the electronic jumps to K orbit gives rise to Lyman series.

Here n1 = 1 and n2 = 2, 3, 4, , The wavelengths of different members of Lyman series are :

(i) First member - In this case n1 = 1 and n2 = 2, it is called line of Lyman series, hence

4

3

2

1

1

1122

RR

or

R3

4 or m10

7101216

10097.13

4

= 1216

(ii) Second member - In this case n1 = 1 and n2 = 3, it is called line of Lyman series, hence

9

8

3

1

1

1122

RR

or

R8

9 or m10

7101026

10097.18

9

= 1026

Similarly, the wavelengths of other members can be calculated.

(iii) Limiting member - In this case n1 = 1 and n2 = , hence

RR

1

1

112

or R

1 or

m10

710912

10097.1

1 912

This series lies in ultraviolet region. (b) Balmer series - When an electron jumps from an outer orbit to the second orbit i.e., the electron jumps to L orbits give rise to Balmer sereis.

Here n1 = 2 and n2 = 3, 4, 5, , . The wavelengths of different members of Balmer series are :

(i) First member - In this case n1 = 2 and n2 = 3, it is called line of Balmer series, hence

36

5

3

1

2

1122

RR

or

R5

36 or m10

7106563

10097.15

36

= 6563 .

(ii) Second member - In this case n1 = 2 and n2 = 4, it is called line of Balmer series hence

+ +

Photon of

wavelength Spectrum

+

16

3

4

1

2

1122

RR

or

R3

16 or m10

7104861

10097.13

16

= 4861 .

Similarly the wavelengths of other members can be calculated.

(iii) Limiting case - In this case n1 = 2 and n2 = , hence

4

1

2

112

RR

or

R

4 = 3646 .

This series lies in visible and near ultraviolet region. (c) Paschen series - electron jumps from an outer orbit to the third orbit i.e., the electronic jumps to M orbit give rise to Paschen series.

Here n1 = 3 and n2 = 4, 5, 6, , . ,

2

2

2

1

3

11

nR

where n2 = 4, 5, 6, , .

For the first member, the wavelength is 18750 . This series lies in infra-red region. (d) Brackett series - when an electron jumps from an outer orbit to the fourth orbit i.e., the electronic jumps to N orbit give rise to Brackett series.

Here n1 = 4 and n2 = 5, 6, 7, , .

2

2

2

1

4

11

nR

where n2 = 5, 6, 7, , .

This series lies in infra-red region. (e) Pfund series - The series consists of all wavelengths which are emitted when an electron jumps from an outer orbit to the fifth orbit i.e., the electronic jumps to O orbit give rise to Pfund series.

Here n1 = 5 and n2 = 6, 7, 8, , . The different wavelengths of this series can be obtained from the formula

2

2

2

1

5

11

nR

where n2 = 6, 7, 8, , . This series lies in the infra-red region of the spectrum.

S.No. Series observed Value of n1 Value of n2 Position in the spectrum

1. Lyman series 1 2, 3, 4, , Ultraviolet

2. Balmer series 2 3, 4, 5, , Visible

3. Paschen series 3 4, 5, 6, , Infra-red

4. Brackett series 4 5, 6, 7, , Infra-red

5. Pfund series 5 6, 7, 8, , Infra-red

SUCCESS OF BOHRS THEORY

(i) Introduced quantum mechanics: Bohr explained that sub-atomic particles (e.g. electrons) are

governed by the laws of quantum mechanics and not by the classical laws of electrodynamics as

assumed by Rutherford.

(ii) Made atom stable: According to this theory, an electron moving in the permitted (or quantized)

orbits cannot loose energy even though under constant acceleration.

(iii) Gave mathematical explanation of hydrogen series: The hydrogen series found by various

scientists were based on empirical relations but had no mathematical explanation. However, these

relations were easily derived by applying Bohrs theory. Further, the size of hydrogen atom as

calculated from this theory agreed very closely with the experimental value.

DE BROGLIES EXPLANATION OF BOHRS SECOND POSTULATE OF QUANTISATION From knowledge of standing wave, for an electron moving in n

th

circular orbit of radius rn, the total distance is the circumference

of the orbit, 2rn should be an integral multiple of wavelength

2rn = n, n = 1, 2, 3...

--------(1)

From De-Broglie hypothesis,

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

From(1) and (2),

LIMITATIONS OF BOHRS THEORY

(i) It could only partially explain hydrogen atom. For example, this theory does not explain the fine

structure of spectral lines in the hydrogen atom.

(ii) It could not explain the difference in the intensities of emitted radiations.

(iii) It is silent about the wave properties of electrons.

(iv) It could not explain experimentally observed phenomena such as Zeeman effect, Stark effect etc.

(v) Bohrs model is applicable to simplest atoms like hydrogen with Z = 1. It fails for other elements.

(vi) Bohrs model does not explain why the orbits are circular while elliptical orbits are also possible.

Questions for Practice:

1. Explain the Rutherford Experiment. How will you determine the distance of closet approach by alpha particle experiment. Prove it mathematically.

2. What is impact parameter Write its expression. How angle of scattering varies with impact parameter.

3. What is the shortest wavelength present in the Paschen series of spectral lines?

4. 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?

5. The ground state energy of hydrogen atom is 13.6 eV. What are the kinetic and potential energies of the electron in this state? (K=13.6eV,U=-27.2eV)

6. 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.

7. (a)Using the Bohrs 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.

8. The radius of the innermost electron orbit of a hydrogen atom is 5.31011 m. What are the radii of the n = 2 and n =3 orbits?

9. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

10. 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? 11. In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a

7.7 MeV -particle before it comes momentarily to rest and reverses its direction? 12. 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 (ii) Highest permitted energy level to the second permitted level.(3:1)

13. What is the longest wavelength photon that can ionize a hydrogen atom in its ground state? Specify the type of radiation.

14. (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? (b) Which transition corresponds to emission of radiation of maximum wavelength?

15. The radius of the innermost orbit of a hydrogen atom is 5.3 x 10-11m. What is the radius of orbit

in the second excited state?(21.2 x 10-11m) 16. The ground state energy of hydrogen atom is 13.6 eV.

(a)What is the kinetic energy,Potential energy of an electron in the 2nd excited state? (n=3)(K=1.51eV)(U=-3.02eV) (b) IF the electron jumps to the ground state from the 2nd excited state, calculate wavelength of the spectral line emitted. (1.023 x 10-7m)

17. In hydrogen atom, if the electron is replaced by a particle which is 200 times heavier but has the same charge, how would it radius change?(1/200)