clc9/15/02
RUTHERFORD
SCATTERING
The equipment includes:
-Scattering
chamber with strong non-monoenergetic 241Am source and
appropriate collimators.
-Suitable
electronics for surface barrier detector , PC based mutichannel analyzer.
One of the
most profound experiments ever performed in nuclear science was the scattering
of alpha particles from a thin hammered gold foil. This experiment was performed in 1911 by Lord Ernest Rutherford
and Marsden in the Cavendish Laboratory in England. At the time of this experiment, the approximate size of the atom
was known from the properties of gases and kinetic theory. Sir J.J. Thompson had proposed a model of
the atom in which the electrons and other nucleons were mixed up in an ordered
fashion in the atom. In this model the
electrons maintained an equilibrium position in this matrix and variations in
this average position through excitation would yield emission of light in the
rearrangement process. When Rutherford
tried to use this model to understand the scattering of alpha particles from
gold, he found that it was not possible to explain the large back angle
scattering cross-section that was observed in the experiments. In the Thompson model the positive charges
were distributed throughout the atom and hence the electric field was not large
enough in any location within the atom to produce the necessary force on the
alpha particle for observed backscattering.
Rutherford knew from his experiments that the positive charges had to be
clustered more closely to explain his results.
Through his tremendous insight into physical phenomena, Rutherford
finally developed a model in which all of the positive charge of the atom was
concentrated in a hard core called the nucleus and the electrons rotated about
the nucleus in a planetary fashion.
This hard core nucleus had a radius of approximately 10-13 cm
which was 1/100,000 the known diameter of atoms. Rutherford's calculations from the Coulomb field for this atomic
model gave the necessary forces to explain the relatively large back angle
scattering for the experiment. Later
experiments would also prove that this new model would give the correct
frequencies of light from the Bohr theory.
This simple alpha particle scattering experiment gave us a correct
picture of the atom and certainly formed a foundation upon which modern nuclear
science is based.
A Rutherford scattering experiment is
written up in detail in Melissinos pp
226-252. The details of your experiment differ from that, especially the
detector and data taking, but the outlines of the experiment are the same. You
should read through this section and use it as reference material as you do
this experiment.
The physical
process of Rutherford scattering is the
deflection of of a charged particle, originally incident at a certain impact
parameter b, by an angle q
by the Coulomb interaction with a second charged particle. This process
underlies not only nuclear scattering but also the energy loss of charged
particles interacting with matter (here the collision partner is an electron)
and is at the heart of every atomic collisions process. The relationship
between b and q can be shown to be
cot
(q/2) = 2b/ r0 (1)
where r0 is the collision diameter= Z1
Z2/(4peoE) (MKS),
Z1 and Z2 are the charges of the interacting
particles and E is the collision energy
. The collision diameter is the distance of closest approach in a head-on
collision. If the target is much (infinitely) heavier than the projectile, then
the energy and angle entering in equation (1) are laboratory values. If the
target and projectile masses are M1 and M2, non infinite, then E must be replaced by the
center of mass energy and q becomes the center of mass angle.
For small values of q , equation (1) becomes
q
= r0 /b (2)
Interestingly enough, this small angle result is correct
in either laboratory or center of mass coordinates. This small angle result is
at the heart of the derivation of the expression for energy loss of charged
particles moving in matter which you studied in the Particle Detectors
laboratory, and can be readily derived by an impulse approximation treatment of
Coulomb scattering.
The differential cross section for Rutherford scattering
is obtained by asking into what solid angle particles will be scattered if they
are incident at impact parameters between b and b+db. That is, particles
entering a circular ring of area 2pbdb
will be scattered into a solid angle of size 2psin(q)dq, so that the differential cross section
for differential scattering , area per solid angle, is given by
ds/dw =
2pbdb/2psin(q)dq=[ b/ sin(q) ] db/dq (3)
Evaluating the derivative from (1) yields the
differential cross section
ds/dw = r02 /
16sin4(q/2) (4)
This is the non-relativistic result for Coulomb or
Rutherford scattering. Further details of this derivation are given in
Melissinos, pp 231 ff. Remember that the energies and angles are still in the
center of mass. For two body scattering, the center of mass energy Ecm
is related to the laboratory energy E by
Ecm=[
M2/( M1+ M2) ] E (5)
sin
(qcm-q) = (M1/M2 ) sin (q) (6)
which reduces to qcm = [M1+
M2/ M2 ] q for
small angles. (7)
For small angles, expression (4) is remarkably good when
laboratory angles and energies are used. Note two important results:
1) Rutherford scattering is very forward peaked, going
as 1/ sin4(q/2)
at forward angles.
2) Rutherford scattering rises with decreasing energy as
1/E2 .
Further detail can be found in most classical mechanics
textbooks and in “The Atomic Nucleus”, Evans, McGraw Hill, 1955, Appendix B.
The
relationship between yield (Y, counts per second) of scattered particles and
cross section is
Y
= Nb nt ds/dw Dw (8)
Here Nb is the number of alpha particles per
second in the beam, nt is the number of target particles per unit
area in the target, Dw is the solid angle
subtended by the detector. From a measured Y you can deduce ds/dw and compare it to that of
expression (4).
You
will repeat Rutherford's experiment using a strong 241Am alpha particle source and a surface barrier detector
similar to that used in the particle detector experiment. You will measure the
absolute differential cross section for scattering of these particles in a gold
foil and compare your result to the expected absolute differential cross
section given in equation (4). The apparatus is similar to that used in the
particle detector experiment, so you are referred to that writeup for
discussion of the chamber, the detector,
the electronics , etc. You will evaluate the thickness of the gold foil
(from which you calculate nt ) by measuring the energy loss of the
5.4 MeV alphas passing directly through this foil, and using known stopping
powers. The solid angle of the detector is calculated from the geometrical
collimation. Use a vertical collimating about 5mm wide and 1 cm high. You will
be working at angles up to about 30 degrees (or more, depending on your
patience).
You should address the following points in your report:
1) How does the measured cross section vary with alpha
particle energy?
2) How does the measured cross section vary with
scattering angle?
3) How does the observed absolute value of the
scattering cross section compare to that predicted from equations (8) and (4)?
Some questions and problems you will have to
answer/solve include the following:
Q: The energy loss in the foil is not negligible. What
energy should you use in evaluating the collision diameter?
Complication:
The alpha particle source used has a broad energy spectrum due to the
fact that it is very thick and some particles come from deep layers. This means
the energy spectrum is wide, with a maximum at 5.4 MeV with with distribution
which extends much lower. This complicates the experiment because the energy is
not unique, but makes it more interesting because you can investigate equation
(1) as a function of alpha energy. That is, for each scattering angle you can
analyze several different bands of observed scattering energy (each of which
corresponds to some average incident energy), and thus obtain angular
distributions for several different energies. Do this.
Q: How is the observed energy related to the energy of
the incident alpha particle? Do you have to consider lab to cm angular
conversions?
Q: Derive the steps which yield (4) from (1) and be sure
you understand what it all means.