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.

 

Introduction

      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.

 

Basic process

 

   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).

 

Experiment

 

            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.