1. SCATTERING OF PARTICLES
Khemendra
 Shukla
MSc (Applied Physics) sem. I
B.B.A. University, Lucknow
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2. 2

3. 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.”
Thomson’s atomic model
Fig. 1.1 3

4.  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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5.  Born: 30 August 1871, a New Zealand-British
chemist & physicist.
 Known as father of nuclear physics.
(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
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ions. Continued…

6.  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.”
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7.  Rutherford model was accepted because he proved
scattering of 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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8.  Experiment
 Theory
 Mathematical analysis
 Experimental verification
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9. Small aperture
Fig. 1.2 9

10.  Most of the – 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.
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11.  Assuming -particles and nucleus are point charges.
 Impact parameter b is the minimum distance to which
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.
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Continued…

12.  Scattering angle is the angle between the asymptotic
direction of approach of -particle and the asymptotic
direction in which it recedes.
P2 b
P
2
2
-particle
P1
b=impact parameter
Head on collision
Fig. 1.3
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13. At the instant of closest approach KE of -particle
2
1 2Ze
KEinitial PE
4 0 D
2Ze2
D
4 0 KEinitial
is the no. of particles N(
per unit area that reach the
screen at scattering angle of
N 180 o is this no. of backward
scattering.
N 180 o
60 0 800 1000 1200 1400 160o 180o
Fig. 1.4
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14. As a result of impulse F dt given it by the nucleus, the
momentum of the -particle changes by P from
initial value P1 to the final value P2.
P P2 P1 Fdt
Hence the magnitude of its momentum is also same
before and after, and
P1 P2 mv
We have magnitude for momentum change
P 2mv sin
2
Impulse Fdt is in same direction as P
Fdt F cos dt
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15. P1 P2 mv
P
P mv P2
sin sin P1
2 2
F
P2 b
P
-particle
P1 2
b=impact parameter

16. 2mv sin F cos dt
2
2 dt
2mv sin F cos d
2 2 d
2 d 2
m r mr mvb
dt
dt r2
d vb
2
2
2mv b sin Fr 2 cos d
2 2
4 0 mv 2b 2
2
sin cos d 2 cos
Ze 2 2 2
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17.  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
Continued…

18. The centripetal force holding the electron in
mv 2
FC
2
an orbit r from the nucleus is provided r
by the electric force F
1 e2 e F
Fe 2
4 0r
Electron
The condition for dynamical stable orbit Fig. 1.2
is F Fc e
mv 2 1 e2 e
v
2 0 r2 4 0 mr
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Continued…

19. Total energy of atom (hydrogen) E
E KE PE
2 2
mv e
E
2 4 0
e2 e2
E (Putting value of ‘v’ )
8 0 r 4 0 r
e2
E
8 0 r
The total energy is –ve this holds for every atomic electron.
This reflects the fact that it is bound to the nucleus.
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20.  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.
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21. Thank You

22. P = P = mv
1 2
Fig. 1.3b
vo
m
Impact
parameter
Fig. 1.3a
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