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Rutherford Scattering II
Rutherford Experiment: Extended version for 4th year.
Here are some definitions that may prove to be useful:
The number of scattering centres in the target is related to the thickness of the foil and the density of the material via N0 = (a x d) x ρ x A_{0}/ A , where
 A_{0} = Avogadro’s number = 6.0222 x 10^{23} atoms /mole.
 A = atomic weight of the target in g/mole.
 ρ = density of the material (19.3 g/cm^{3} for Au, 2.7 g/cm^{3} for Al).
 d = thickness of the target (2 micrometers = 2 microns for Au, 7 microns for Al).
 a = area of the target intercepted by the beam.
(N is the same as in Melissinos but note our N0 is not the same as Melissinos N0, which is the Avogadro’s number(A_{0} here)).
The activity of the AM241 alpha source is 330 kBq.
Key Concepts


Procedure
If you have not done the 3rd year experiment, please do so.
 This time around we would like you to improve on some of the measurements you have done in the previous year, as well as to use the 1mm slit. As you must have realized, it is really the large angle readings that are important. We would like to collect more data in the large angle regime. Begin by finding the zero of your apparatus again.
 Use the 5mm slit and the Gold foils for the following angles – for both positive and negative angles: 30º, 35º, 40º, 45º, 50º, 55º, 60º (be certain to record around 10 counts for each angle). How do you expect the rates to compare? (Answer: rate_{gold}/rate_{alum} = (Z_{gold}/Z_{alum})^{2} so it takes much longer to collect enough data for the Aluminium foil).
 Using the 5mm slit, collect background readings for the following angles. Note, at the higher angles your background readings will be zero  think about the geometry of the source and detector!
 Now mount the Gold foil with 1mm slit and again find the zero. Perform counts measurements for the following angles, 20º, 25º, 30º, 35º, (and if you feel enthusiastic 40º).
 Use the 1mm slit to take background measurements for the above angles. As you can see there is much less noise with the 1mm slit. However, this also means that it will take much longer to obtain the data. Why have we not assigned the 1mm slit with the Aluminium foil? (Hint: We do want you to graduate one day).
From the above measurements, verify that the cross section does indeed vary as 1/sin^{4}θ/2, and find the value of Z for gold.
References
 Melissinos, Experiments in Modern Physics, Academic Press.
 Preston and Deitz, The Art of Experimental Physics, Wiley and Sons.
 H. Frauenfelder and E. Henley, Subatomic Physics, Prentice Hall.
 A. Das and T. Ferbel, Introduction to Nuclear and Particle Physics, J. Wiley.