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Relative angle after Compton scattering?

  1. Nov 14, 2013 #1
    1. The problem statement, all variables and given/known data
    Photon with ##\lambda =10^{-12} m## hits an electron (Compton scattering). After the interaction the photon and electron move under relative angle of 90°. Calculate the kinetic energy of electron. Hint: First find the relation between ##\theta ## and ##\varphi##. (##\theta +\varphi = 90°##)



    2. Relevant equations



    3. The attempt at a solution

    I can't find the relation!!! How on earth can I find it?

    The only thing I get from components of momentum is: ##\frac{hc}{\lambda ^{'}}sin\theta =cp_esin\varphi ##

    But this doesn't help me much... or does it? o_O

    Please help.
     
  2. jcsd
  3. Nov 14, 2013 #2

    mfb

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    Staff: Mentor

    Use energy conservation together with momentum conservation. Alternatively, find a formula for the electron momentum and λ′ as function of the two angles.
     
  4. Nov 14, 2013 #3
    Ok...

    ##cp_esin\varphi =E_{\gamma }^{'}sin\theta ## and therefore ##\frac{cp_esin\varphi }{sin\theta }=E_{\gamma }^{'}##

    momentum conservation along x axis:

    ##E_{\gamma }=cp_ecos\varphi +E_{\gamma }^{'}cos\theta ## using the last expression above:

    ##cp_e=\frac{E_\gamma }{cos\varphi +sin\varphi ctg\theta }##

    so If I now insert this to ##cp_esin\varphi =E_{\gamma }^{'}sin\theta ## than:

    ##E^{'}=\frac{E_\gamma }{sin(\varphi +\theta )}##

    But I can't see how this helps me?
     
  5. Nov 15, 2013 #4

    mfb

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    sin(φ+θ) = sin(90°) = ?

    Hmm, that doesn't look right. Where does that come from?
     
  6. Nov 15, 2013 #5
    ##cp_e=\frac{E_\gamma }{cos\varphi +sin\varphi ctg\theta}## and ##cp_esin\varphi =E_{\gamma }^{'}sin\theta##

    so

    ##\frac{E_\gamma }{cos\varphi +sin\varphi ctg\theta}sin\varphi =E_{\gamma }^{'}sin\theta##

    ##\frac{E_\gamma }{sin\theta cos\varphi +sin\theta sin\varphi ctg\theta}sin\varphi =E_{\gamma }^{'}##

    ##sin\theta cos\varphi +sin\theta sin\varphi ctg\theta=sin\theta cos\varphi+sin\varphi cos\theta =sin(\varphi + \theta )##

    ##\frac{E_\gamma }{sin(\theta +\varphi )}sin\varphi =E_{\gamma }^{'}##

    I know it doesn't like right... :D That's why I am here asking.. :/
     
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