SCATTERING OF a-

Khemendra Shukla MSc (Applied Physics) sem.

IB.B.A. University, Lucknow1

RUTHERFORD SCATTERING

PARTICLES

2

If we are talking about Rutherford Scattering, we should know one thing i.e., what was the first step about the nuclear atom.

Sir J.J. Thomson discovered electron in

1897. After one year he suggested an

atomic model.

“Atoms are just positively charged lumps of

matter in which electrons are embedded in

them.”

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Origin of the Atomic Model

Thomson’s atomic modelFig. 1.1

This model could not explain all features of visible

spectrum of hydrogen atom and other elements.

Rutherford performed a number of experiment

with Geiger & Marsden on the scattering of alpha

particles by a very thin gold foil. Thomson’s model

couldn’t explain the experimental results.

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Why was it not correct?

Born: 30 August 1871, a New Zealand-

British chemist & physicist.

Known as father of nuclear physics.

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Ernest Rutherford: Glance at scientific life

(1871 – 1937)@ McGill University, discovered the concept of

radioactive half-life, proved that radioactivity

involved the transmutation of one chemical

element to another, and also differentiated and

named alpha and beta radiation, proving that

the former was essentially helium ions.Continued…

Awarded with Nobel Prize in Chemistry in

1908 "for his investigations into the

disintegration of the elements, and the

chemistry of radioactive substances.“

He remains the only Nobel Prize winner to

performed his most famous work after

receiving the prize.

“That was Rutherford Atomic Model and

obviously discovery of NUCLEUS.” 6

Rutherford model was accepted because he

proved scattering of a particles with

mathematical formula.

Atom consists of central massive nucleus in

which all the positive charge and most of the

mass are concentrated.

A cloud of negatively charged electrons

surrounds this nucleus. They are moving

around the nucleus.

Most of the space in atom is empty.

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Rutherford Nuclear Atom Model

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Rutherford‘s experiment on scattering of a-particles

Experiment

Theory

Mathematical analysis

Experimental verification

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Experimental arrangement

Small aperture

Fig. 1.2

Most of the a – particle were scattered by

small deviations while passing through gold

foil.

There were a few particles that were

scattered through large angle.

One of about 8000 particles suffered angles

of scattering >90

A few of them go head on collision. 10

Observations

Assuming a-particles and nucleus are point

charges.

Impact parameter b is the minimum distance

to which -a particle would approach the

nucleus if there is no forces between them.

b= 0 for head on collision

Distance of closest approach D is the

minimum distance to which particle

approaches nucleus head on. 11

Theory Of - a Particle Scattering

Continued…

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Scattering angle q is the angle between the

asymptotic direction of approach of a-particle

and the asymptotic direction in which it recedes.

Head on collision

q

b=impact parameter

b

a-particle 1P

2PP

Fig. 1.3

2

2

At the instant of closest approach KE of a-particle

q is the no. of -a particles

per unit area that reach the

screen at scattering angle of .q

is this no. of backward

scattering.

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D

ZePEKEinitial

2

0

2

4

1

initialKE

ZeD

0

2

4

2

060 080 0100 01200140 o160 o180

N( )q

Fig. 1.4

oN 180

oN 180

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As a result of impulse F dt given it by the nucleus, the momentum of the a-particle changes by from initial value to the final value .

Hence the magnitude of its momentum is also same before and after, and

We have magnitude for momentum change

Impulse is in same direction as

P

1P 2P

FdtPPP 12

mvPP 21

2sin2

mvP

PFdt

dtFFdt cos

q

b=impact parameter

b

a-particle 1P

2PPF

P

2

q

1P

2P

2sinsin

mvP

mvPP 21

2

16

2

2cos

2sin2

cos2

sin2

dd

dtFmv

dtFmv

2

22

20

2

2

22

2

22

2cos2cos

2sin

4

cos2

sin2

dZe

bmv

dFrbmv

vb

r

d

dt

mvbdt

dmrrm

The electron can not be stationary in this

model because there is nothing to balance the

force of nucleus.

If the electron in motion, however ,

dynamically stable orbits are possible like

those of the planets around the sun.

Ruther ford assumed circular orbit for

convenience, though it might be reasonably

be assumed to be elliptical in shape. 18

Electron orbits: The planetary model

Continued…

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2

2mvFC

2

2

04

1

r

eFe

ec FF

2

2

0

2 1

2 r

emv

The centripetal force holding the electron in

an orbit r from the nucleus is provided by the electric force

The condition for dynamical stable orbit is

eF

F

Electron

r

mr

ev

04

Continued…

Fig. 1.2

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Total energy of atom (hydrogen) E

0

22

42 emv

E

PEKEE

r

eE

r

e

r

eE

0

2

0

2

0

2

8

48

(Putting value of ‘v’ )

The total energy is –ve this holds for every atomic electron.This reflects the fact that it is bound to the nucleus.

0

22

42 emv

E

PEKEE

r

eE

r

e

r

eE

0

2

0

2

0

2

8

48

(Putting value of ‘v’ )

By the application of newton’s law of motions

and Coulomb’s law experimental observation

that atoms are stable. But,

“Accelerated Electric Charges Radiates Energy

In The Form Of EM Waves.”

An electron accelerating in curved path

should continuously lose energy, spiraling

into the nucleus in a fraction of a second.

But atoms do not collapse.21

Failure Of Rutherford’s Model

Thank You

12

vom

Impactparameter

Fig. 1.3a

Fig. 1.3b

P = P = mv1 2