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Kinetic energy with compton effect

  1. Jan 22, 2012 #1
    Show that the recoiling electron has kinetic energy:
    K = Ephoton^2*(1-cos phi)/ mc^2 + Ephoton(1-cos phi)

    Now I have gotten
    K = hc/λ - hc/λ' = hc(λ'-λ)/λ'*λ = hc(Δλ)/λ'*λ
    and since λ' = λ + Δλ
    K = hc *Δλ/(λ(λ+Δλ)) = hf*Δλ/(λ+Δλ) = (Δλ/λ)/(1+Δλ)*hf

    and I know that Δλ = h/mc*(1-cos phi)

    but how do I go from what I have gotten to K = Ephoton^2*(1-cos phi)/ mc^2 + Ephoton(1-cos phi)???

    Thanks.
    Stephen
     
  2. jcsd
  3. Jan 23, 2012 #2
    bump....Please help
     
  4. Jan 24, 2012 #3

    vela

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    I don't think that expression for the kinetic energy is correct. I can show that
    $$K = \frac{E_\gamma^2}{mc^2}(1-\cos\phi) - \left(\frac{K}{mc^2}\right)E_\gamma(1-\cos\phi)$$For the expression you were given to show holds, you'd have to have K/mc2 = -1, which can't be true.
     
    Last edited: Jan 24, 2012
  5. Jan 24, 2012 #4
    That is the kinetic energy equation given on the homework that my professor gave....
     
  6. Jan 24, 2012 #5

    vela

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    I'll ask the other helpers to see if they have any insight on this problem. Just to clarify, when you say Ephoton, you're referring to the energy of the incident photon, correct? And phi is the angle of the scattered photon? Are you looking at a specific regime or are you supposed to use some approximation?
     
    Last edited: Jan 24, 2012
  7. Jan 24, 2012 #6
    Ephoton is incident photon and we are finding kinetic energy of the recoiling electron... Please help
     
  8. Jan 24, 2012 #7

    I like Serena

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    Hi StephenD420! :smile:

    I believe your formula should be (I think you dropped a couple of parentheses):
    $$K = {E_{photon}^2 (1-\cos \phi) \over m_e c^2 + E_{photon}(1-\cos \phi)}$$

    For the calculation I'd suggest that you use ##\lambda = {hc \over E_{photon}}## and ##\lambda' = \lambda + {h \over m_e c} (1 - \cos \phi)##.
     
  9. Jan 25, 2012 #8
    I looked back at the problem and no I did not drop any parentheses, my professor hand rights his homework so any ideas???
     
  10. Jan 25, 2012 #9

    I like Serena

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    In that case I think your professor dropped the parentheses.
     
  11. Jan 25, 2012 #10
    K = hc *Δλ/(λ(λ+Δλ)) = hf*Δλ/(λ+Δλ) = (Δλ/λ)/(1+Δλ)*hf

    so how do I go from above to what you got??
     
  12. Jan 25, 2012 #11
    ok so I have gotten thus far

    1/E' = [mc^2 + E(1-cos phi)]/E*mc^2

    now how do I go from here to what you got???
     
  13. Jan 25, 2012 #12

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    Calculate E - E', that is, E minus the inverse of what you got there.
    From there it is algebraic manipulations to subtract 2 fractions.
     
  14. Jan 25, 2012 #13
    ok now if I wanted to show that for very high energy photons and phi = 180 degrees the kinetic energy approaches K = Ephoton - 255.5KeV??
     
  15. Jan 25, 2012 #14

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    Since K=Ephoton - Ephoton', you will want to show that Ephoton' = 255.5 keV.

    The dependencies on Ephoton can be approximated by the fact that 1/Ephoton is relatively close to zero.
     
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