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Compton scattering - energy of the scattered photon

  1. Nov 18, 2013 #1
    [Mentor's note: This thread does not use the template because it started in one of the non-homework forums. I moved it here instead of deleting it and asking the poster to repost here, because it had accumulated several useful replies.]

    Hi.
    I have the exact same problem that ZachWeiner had in here:
    A photon whose energy equals the rest energy of the electron undergoes a Compton collision with an electron. If the electron moves off at an angle of 40 degrees with the original photon direction, what is the energy of the scattered photon?

    Unfortunately the answers don't help me much. I would need more information how to get the energy of the scattered photon.

    Thanks.
     
    Last edited by a moderator: Nov 19, 2013
  2. jcsd
  3. Nov 18, 2013 #2
    Write down your conservation of momentum and energy equations.

    If an object is scattered with some y momentum. To conserve overall y momentum the object that was at rest must have equal and opposite momentum in y.

    This should be done using 4 momentum.
     
  4. Nov 18, 2013 #3

    mathman

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    Step 1. get the angle (x) of the photon after scatter.
    Step 2. increase in Compton wavelength is given by 1-cos(x).
    Step 3. get energy from Compton wavelength.

    Note: Your original photon energy has a Compton wavelength of 1 (corresponding to 511 kev). So the final energy of the photon will be 511/(2-cos(x)) kev.
     
  5. Nov 19, 2013 #4
    how

    How do I get x?
    I know this:
    y = 40°
    hc/lam = 0,511 MeV

    In x direction: h/lam = h/lam' cos(x) + p_e cos(y)
    In y direction: h/lam' sin(x) = p_e sin(y)

    We also know: lam' - lam = lam_c (1 - cos(x))
    Where:
    lam = wavelength before scattering
    lam' = wavelength after scattering
    x = angle of the scattered photon
    y = angle of the electron

    I don't know how to get x.
     
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