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Compton Scattering (Back-Scattering)

  1. Nov 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider the situation where high-energy gamma rays are striking a detector after
    scattering off of the electrons in the material surrounding a detector. Show that if
    the gamma rays are perfectly back-scattered by the material, such that the scattering
    angle is θ = 180 degrees, then the detected scattered photons will all have essentially
    the same energy of 0.26 MeV, independent of the precise energy of the incident gamma
    rays as long as the incident gamma ray energy is much larger than the rest-mass energy
    of an electron.


    2. Relevant equations

    [tex]\lambda[/tex]final - [tex]\lambda[/tex]initial = [tex]\frac{h}{(me)c}[/tex] (1 - cos[tex]\theta[/tex])


    3. The attempt at a solution

    Well for [tex]\theta[/tex] = 180 degrees, the compton scattering eqn becomes
    [tex]\lambda[/tex]final - [tex]\lambda[/tex]initial = [tex]\frac{2h}{(me)c}[/tex].
    I'm confused about where to go from here since [tex]\lambda[/tex]final - [tex]\lambda[/tex]initial needs to be proven it is 0, and also that [tex]\lambda[/tex]final = [tex]\lambda[/tex]initial = 0.26MeV.
    I have a feeling I need to to use the binomial expansion somewhere since it is saying that the incident gamma ray energy has to be much larger than the rest-mass energy
    of an electron. Can someone push me in the right direction?
     
  2. jcsd
  3. Nov 22, 2009 #2
    Hi person in the same PHYS242 class at Queen's I'm in.

    You're on the right track. Just convert the equation you have right now to an energy equation using ∆E and you will get ∆E = 1/2mec2. Solve this and you will get 0.26 MeV with the assumptions that your initial lambda is much much smaller than your final lambda (and therefore your ∆E can be considered a final E rather than delta).
     
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