1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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...

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?