Physics 3401, Quiz 6

1. What fraction of 4 MeV alpha particles will scatter between 30° and 40° when directed at a 1 micrometer
thick gold foil? Zgold=79, Agold=197, density=1.93x10^4 kg/m3.

For 30 degrees, sigma = pi*(2*79*1.44 MeV fm/(2*4 MeV))^2*(1 + cos 30)/(1 - cos 30)
 35391.9223 fm^2 = 353.92 barns

For 40 degrees, sigma = pi*(2*79*1.44 MeV fm/(2*4 MeV))^2*(1 + cos 40)/(1 - cos 40)
 19181.26831 fm^2 = 191.8 barns
Net cross section for scattering between 30 and 40 degrees =    162.10 barns

The scattered fraction is then = N*L*density*sigma/(A*1E-3)
 Scattered fraction =  0.00095

2. a. If 6 MeV alpha particles are back scattered from an aluminum foil, what upper bound on the nuclear
radius of the aluminum nucleus does this yield?(Zalum=14, Aalum=27, density=2.7x103 kg/m3,
1 micrometer thick)

Set the kinetic energy equal to the potential energy
6 MeV = 2Zke^2/r , so r = 2*14*1.44 MeV fm/6 MeV=    6.72 fm

b. If a 6 MeV alpha scatters at 60° when passing through an aluminum foil, what is its closest approach
 to the nucleus?

 The impact parameter for an angle theta is
 b=(kq1q2/mv^2)*(1 + cos theta)/(1 - cos theta) and for 60 degrees
 b= ((2)(14)(1.44 MeV fermi)/(2*6 MeV))*sqrt((1 + cos 60)/(1 - cos 60)) =
 b= 5.82 fermi

 b. The closest approach is given by
 rm = Zke^2/KE + .5*sqrt((2Zke^2/KE)^2 + 4b^2) =
  =14*1.44/6 + sqrt((2*14*1.44/6)^2 +4b^2) =    10.08 fermi

3. If the kinetic energy of a scattered particle is large enough, then its scattering can be treated just like
the Compton scattering of a photon.

a. For scattering of a 4 MeV electron from a medical radiation therapy accelerator off a proton, what
percentage error is made by assuming that KE=pc?
pc=sqrt((4 MeV+.511MeV)^2-.511MeV^2) =    4.481963855 MeV
   12.025 % error

b. Using the Compton scattering approach, what kinetic energy will a scattered electron have if it scatters
 off a proton at 60°?
Using the assumption pc=KE before and after, even knowing that there is significant error, gives:
 hc/pcf - hc/pci=(h/mc)*(1-cos 60)
 hc/pcf= 310.661
 pcf= 3.9915 MeV
The only rationale for carrying out such a calculation is that it is much easier than carrying out
the full relativistic conservation of momentum and energy calculation for a particle for which v/c=.99356 .

It does in fact provide useful information: that the scattering is essentially elastic since there is very little
 momentum change for the electron.