Physics 3401, Homework 6
Due Monday, November 8
Rohlf Chapter 6, #11,13,14,15,17,18,24

#11 Compare alpha scattering to Davisson-Germer
Compare the alpha wavelength to nuclear size and separation between nuclei.
Compare electron wavelength in Davisson-Germer to atom size and separation between atoms.

The alpha wavelength is comparable to nuclear size, but the separation of atoms is four orders of
magnitude larger, so it can be considered to interact with one nucleus at a time.
In Davisson-Germer, the wavelength of the electron was on the order of the spacing
between atomic layers, so it interacted with multiple atoms.

13.Beam of protons with pc=200 MeV scattering off of aluminum nuclei.
a. Cross-section for scattering angles greater than 10 degrees.
The electron kinetic energy is needed. For pc=200 MeV
 KE=sqrt((200 MeV)^2+938.27^2) - 938.27 =    21.08 MeV
 For aluminum with Z=13, the cross section is given by
 sigma= pi b^2 = pi(Zke^2/2 KE)^2(1 + cos theta)/(1 - cos theta)
  = pi*(13*1.44 MeV fm)/(2*21 MeV))^2*(1 + cos 10)/(1-cos 10) =
    = 81.00 fm^2 or 0.810 barns

b. For foil 1E-6 m thick, what fraction scattered at greater than 10 degrees.
 The fraction scattered is
 Rs/Ri = sigma* Avogadro's number*pathlength*density/(atomic mass*1E-3 kg) p176 Rohlf
  = (81E-30*6E23*1E-6*2.7E3)/(27*1E-3) =    4.86E-06

#14. What thickness of air would have the same cross-section as a 0.2E-6 meter thick gold foil?
 The scattering rate is proportional to the pathlength times density.
 The density of air is about 1.2 kg/m^3 while that of gold is 19000 kg/m^3.
 The pathlength of air to be equivalent to the foil is then
 Air length equivalent = (2E-7 m)(19E3 kg/m^3)/(1.2 kg/m^3)*(14/197)* =     0.000225042 m
 Using 14 is treating air as mostly nitrogen, and you would scatter off one N, even in N2     0.225 mm
 So about .2 mm of air is equivalent in alpha scattering to the gold foil,
 so you would definitely have to perform the experiment in vacuum.

#15. Impact parameter and closest approach for a 10 MeV alpha scattering at 90 degrees.
 a. The impact parameter for an angle theta is
 b=(kq1q2/mv^2)*(1 + cos theta)/(1 - cos theta) but for 90 degrees
 b= (2)(47)(1.44 MeV fermi/20 MeV =   6.768 fermi

 b. The closest approach is given by
 rm = Zke^2/KE + .5*sqrt((2Zke^2/KE)^2 + 4b^2) =
  =47*1.44/10 + sqrt((2*47*1.44/10)^2 +4b^2) =    16.34 fermi

 an alternate approach to the closest approach is given by Fowles and Cassady, Analytical Mechanics

 rm= b cos(theta/2)/(1 - sin(theta/2)) =   16.34 fermi

#17. Calculate alpha incident rate from the scattering between 60 and 90 degrees.
Alpha particle energy is given as 5 MeV and the scattering rate as one per minute.
The scattering is off silver, which has atomic mass A=47
 The cross section is given by
 sigma = pi b2^2 - pi b1^2= pi(kq1q2/2 Ke)^2(T1 - T2)
 where T2= (1+cos 90)/(1-cos 90)=1
 T1=(1+cos60)/(1-cos 60) = 3
 sigma = pi*(2*47*1.44 MeV fm/(2*5 MeV))^2*(3-1)
 sigma = Rs/Ri= 1151.225921 fm^2 11.51 barns
After calculating the cross section, the incident flux can be calculated.
 Ri = Rs A(10^-3 kg)/(sigma *L* Avogadros's number*density)
 Ri =(1/60)*107.9*1E-3/(11.51E-28*1E-6*6.02E23*1.05E4) =    247.18 1/s
#18. Cross section for 5 MeV alpha to be scattered from platinum between 5 and 10 degrees.
 The common term for the impact parameters is
 Zke^2/KE = 78*1.44MeV fm/5 MeV =    22.464 fm
 the angular term (1+cos theta)/(1-cos theta) =   130.65 for 10 degrees
    524.58 for 5 degrees
 The net cross section is then given by
 sigma = pi*(22.46)^2(524.6-130.6) =   624525.45 fm^2 = 6245.25 barns

#24. Scattered 10 GeV electrons from protons at 30 degrees. Find energy of scattered electrons.
 This is like Compton scattering, so that the scattered energy is given by
 KE2 =KE1 m0c^2/(m0c^2 + KE1(1-cos theta))

 KE2 = 10 GeV*0.94 GeV/(0.94 GeV + 10 GeV*(1 - cos 30)) =    4.13 GeV