Rutherford scattering show
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Transcript of Rutherford scattering show
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.”
3
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.
4
Why was it not correct?
Born: 30 August 1871, a New Zealand-
British chemist & physicist.
Known as father of nuclear physics.
5
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.
7
Rutherford Nuclear Atom Model
8
Rutherford‘s experiment on scattering of a-particles
Experiment
Theory
Mathematical analysis
Experimental verification
9
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…
12
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.
13
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
14
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…
19
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
20
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