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Compton Scattering formula for 180 degree scattering.

  1. Jan 20, 2010 #1
    I'm trying to derive a simple formula for 180° scattering.

    I've got to this stage and I really can't figure out how to simplify it further.

    [tex]\[ \frac{1}{\lambda}-\frac{1}{\lambda'} = \frac{2m_ec}{h} \][/tex]

    What I actually need is:
    [tex]\[ \lambda' - \lambda = \frac{2h}{m_ec} \][/tex]

    I'm pretty sure the first formula is right but I can't seem to simplify it into the second!

    Thanks in advance.
  2. jcsd
  3. Jan 20, 2010 #2


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    Hi durand! :smile:

    (have a lambda: λ :wink:)

    If you multiply them together, you get (λ' - λ)2 = 4λ'λ, or λ'/λ = 3 ± 2√2 :redface:

    How did you get your equation?
  4. Jan 20, 2010 #3
    Uhm, I derived the first using conservation of momentum and energy at non relativistic speeds, when the photon bounces back. The second comes from the standard compton scattering formula.
  5. Jan 20, 2010 #4
    CoM: h/λ = mv + h/λ'
    CoE: hc/λ = hc/λ' + 0.5mv²

    By substituting one into the other, I reach the formula I mentioned in my first post.
  6. Jan 20, 2010 #5
  7. Jan 21, 2010 #6
    Yeah, I did find that, however, it uses a relativistic derivation so I can't really see how to do the last step as it's totally different to mine :/ Thanks anyway.

    My exam's in an hour so it doesn't really matter now. Thanks everyone for your help :)
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