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Homework Help: Photon Scattering

  1. Apr 27, 2008 #1
    1. The problem statement, all variables and given/known data
    A photon scatters in the backward direction ([tex]\theta[/tex]= 180) from a free proton that is initially at rest.

    What must the wavelength of the incident photon be if it is to undergo a 10.0% change in wavelength as a result of the scattering?

    2. Relevant equations

    [tex]\lambda[/tex]'-[tex]\lambda[/tex] = (h/mc)(1-cos([tex]\theta[/tex]))

    where the left side is the difference between scattered and incidence wavelengths.

    3. The attempt at a solution

    This seemed like a pretty straightforward problem. Since the photon undergoes a 10% change in wavelength, 1.1[tex]\lambda[/tex] = [tex]\lambda[/tex]'. Therefore .1[tex]\lambda[/tex] = (h/mc)(1-cos([tex]\theta[/tex])). Multiply by 10 and evaluate the cosine, and you get [tex]\lambda[/tex] = 20h/mc. However, when I substitute values into this and evaluate it I get the wrong answer. I have absolutely no clue what I am doing wrong here. This shouldn't be a difficult problem, but for some reason I am not getting the correct answer. Help would be appreciated. Thanks.
  2. jcsd
  3. Apr 27, 2008 #2

    Could you post a bit more of the working, as in the actual values you're putting in to evaluate it? & the answer if you have it.
  4. Apr 27, 2008 #3
    Sure. Here's my work:

    [tex]\lambda[/tex]'-[tex]\lambda[/tex] = (h/mc)(1-cos[tex]\theta[/tex])

    1.1[tex]\lambda[/tex]-[tex]\lambda[/tex] = (1 - cos180)(h/mc)

    .1[tex]\lambda[/tex] = 2h/mc

    [tex]\lambda[/tex] = 20h/mc = 20 * (6.626 * 10^-34)/(9.109 * 10^-31)(3.00 * 10^8) = .04852 nm
  5. Apr 27, 2008 #4
    Does anyone know what I'm doing wrong here?
  6. Apr 27, 2008 #5


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    Homework Helper

    Hi Fizzicist,

    You used the mass of an electron, but in this problem the scattering is from a proton.
  7. Apr 28, 2008 #6
    d'oh! haha....thanks...

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