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.