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How you could find E'/E at the min of 180 degrees

  1. Apr 4, 2005 #1
    From the Comptom Scattering formula, you get
    (E-E')/E E' = (1/mc^2)(1-cos@).

    Can someone tell me how you could find E'/E at the min of 180 degrees. I've tried using the conjugate, and other methods, but I cant get E'/E out of it. I must be doing something wrong. Thanks.
     
  2. jcsd
  3. Apr 4, 2005 #2

    dextercioby

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    [tex] \frac{1}{E'}-\frac{1}{E}=\frac{2}{mc^{2}} \Rightarrow \frac{E'}{E}=\frac{mc^{2}}{2E+mc^{2}} [/tex]

    Daniel.
     
  4. Apr 5, 2005 #3
    Wait

    .... is the answer just E'/E = mc^2/(2E + mc^2) or
    mc^2/(2E + mc^2) (1- cos@).
     
  5. Apr 5, 2005 #4

    dextercioby

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    You said about [itex] 180 \mbox{deg} [/itex],right...?I assumed you did,and used this fact.How would my formula change,if,instead of that particular value for the scattering angle,you'd use the general case?
    It's not difficult,it's simple algebra.

    Daniel.
     
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