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About Compton scattering

  1. Dec 23, 2008 #1
    Does the Compton scattering equation


    \lambda ' - \lambda = \frac{h}{m_{e} c} (1-cos \theta )


    work even when [tex] \theta = 180^{\circ}[/tex]?
  2. jcsd
  3. Dec 23, 2008 #2


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    yeah sure, why shouldn't it?
  4. Dec 23, 2008 #3
    I just did a textbook problem which asked for the percentage of a photon's energy that would be passed to an electron via Compton scattering given that it was the maximum amount possible, so I assumed the angle was 180 degrees and my answer did not exactly match the book's answer. Then I wondered if perhaps "scattering" implied a kind of deflection and not a complete rebound.

    Maybe the mismatch is just due to rounding..
  5. Dec 27, 2008 #4


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    Maximum transfer to electron is when E_prime is minimum
  6. Dec 27, 2008 #5
    At 180°, (E-E')/E = a/(1+a)
    E = photon's energy before scattering
    E' = photon's energy after scattering
    a = 2hv/mc^2 (v = photon's frequency before scattering).

    The maximun value of (E-E')/E is 100% for: a = +oo, that is for an infinite energy of the photon before scattering.

    What did you and your book get, instead?
    Last edited: Dec 27, 2008
  7. Dec 27, 2008 #6
    The book asks, "for the Compton effect, determine the percentage of energy transferred to the recoil electrons that acquire the maximum kinetic energy when [tex] \lambda = 1.21 x 10^{-10}m[/tex]."

    I computed [tex]1-\frac {\lambda}{\lambda '}[/tex]

    where [tex]\lambda ' = \lambda + \frac{h}{m_0 c}(1-cos \theta )[/tex]

    and [tex]\theta = 180^{\circ}[/tex] since that maximizes [tex]1-cos \theta[/tex].

    My answer: 3.85% (rounded from 3.85294..)
    Book's answer: 3.93%.
    Last edited: Dec 27, 2008
  8. Dec 27, 2008 #7


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    For what it's worth, I get 3.86% (rounded from 3.8558%). I used five significant figures for all constants, and did not round off any intermediate steps. I suspect that whoever calculated the book's answer either used less-precise values for the constants, or rounded off one or more intermediate steps.
  9. Dec 27, 2008 #8
    I guess that happens. Thanks for stepping in. I just looked at the Compton scattering section of another textbook and saw, "the scattered wavelength is angle-dependent and is greatest for scattering in the backward direction ([tex] \theta = 180^{\circ}[/tex])." So there you go.
  10. Dec 27, 2008 #9
    I get 3.86% using three significant figures for all constants:

    lambda = 1.21*10^(-10) m
    m =9.11*10^(-31) kg
    c = 3.00*10^8 m/s
    h = 6.63+10^(-34) J*s

    I get 3.90% if I take m = 9.00*10^(-31) kg
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