# Homework Help: Electron scattering angles?

1. Oct 21, 2009

### jeebs

hi,
here's the problem:

imagine a heavy nucleus and an electron approaching it with some momentum pi.
it scatters elastically off the nucleus at some angle $$\theta$$ and momentum pf. the nucleus can be considered to remain at rest, since it is so massive.

I have to start with the definition of vector q = pi - pf, get an expression for q^2 (the square of the magnitude of the vector q), and show that the angular dependence of the scattering is given by the Rutherford formula,

$$\frac{d\sigma}{d\Omega}$$ $$\propto 1/(sin^4(\theta/2))$$

my attempted solution:

so far i've worked out that q2 = (pi - pf)2 = pi2 - pf2 - 2pi.pf
ie. q2 = 2p2 - 2p2cos($$\theta$$)
= 2p2(1-cos($$\theta$$))
since the magnitudes of pi and pf are equal.

then I got 1 - cos($$\theta$$) = 2sin2($$\theta/2$$)
by using cos(A+B) = cos(A)cos(B) - sin(A)sin(B) where A = B = $$\theta/2$$

then I looked at Fermi's Golden Rule:

$$\frac{d\sigma}{d\Omega}$$ = ($$2\pi$$/$$\hbar$$)|Mfi|2Df

where the matrix element |Mfi| = $$\frac{4g^2\pi\hbar}{q^2 + (mc)^2}$$
which is a result that is given to me in a previous part of the problem.

clearly when used in Fermi's Golden Rule, the square of the matrix element gives 3 terms, one of which is the 1/q4 term I am looking for, but also 2 other terms. Is there a way for me to get rid of the 2 extra terms or am I barking up the wrong tree here?