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)
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.