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Photo-electric effect, Compton Scattering

  1. Apr 28, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the maximum kinetic energy of electrons knocked out of a thin copper foil by Compston scattering of an incident beam of 17.5 KeV rays? Assume the work function is negligible.


    2. Relevant equations
    Δλ = h/mc (1-cosθ)


    3. The attempt at a solution

    I reasoned that the greatest energy transfer to an electron will occur when the x-ray rebounds at 180 degrees, in which case the change in wavelength is 4.85 x 10^-12 m.

    I figured that the wavelength of the x-rays increases by this amount, thereby decreasing in energy. I thought I could therefore take this change in wavelength and calculate the energy associated with it using E = hc/Δλ, and I got E = 256 KeV.

    However, the answer key requires that you calculate the wavelength of the original x-ray, add Δλ, then calculate the energy and deduct the original energy. It yields a different answer, 1.1 KeV.

    I don't understand why you can't say that the energy loss associated with the increase in the wavelength of the x-ray is completely transferred to the electron and then be done with it. What am I missing? Thanks!!!!
     
  2. jcsd
  3. Apr 29, 2012 #2
    Does anybody have any insight into what I'm doing wrong here? I'd appreciate any and all help :).
     
  4. Apr 29, 2012 #3

    Redbelly98

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    Science Advisor
    Homework Helper

    So far, so good.
    Not quite. First, note that it's impossible for a 17.5 keV photon to lose 256 keV of energy.

    You have two energies, E1 = hc/λ1 and E2 = hc/λ2.

    The energy difference E1-E2 is the difference between the hc/λ terms, not hc/Δλ. What you did is equivalent to saying
    (1/5) - (1/3) = 1/(5-3) = 1/2,​
    which is not true.

    Can you take it from here?

    Yes, that is the basic idea.
     
  5. Apr 29, 2012 #4
    Thanks, that makes sense!
     
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