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Electron scattering angles?

  1. Oct 21, 2009 #1
    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 [tex]\theta[/tex] 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,

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

    my attempted solution:

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

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

    then I looked at Fermi's Golden Rule:

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

    where the matrix element |Mfi| = [tex]\frac{4g^2\pi\hbar}{q^2 + (mc)^2}[/tex]
    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?

    thanks in advance.
     
  2. jcsd
  3. Oct 21, 2009 #2
    actually, ignore this question, i figured it out. the mass m = 0. i'm an idiot. can't even find a way to delete the question.
     
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