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Compton Scattering and maximum energy

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data
    If the maximum kinetic energy given to the electrons in a Compton scattering experiment is 10 keV, what is the wavelength of the incident X-Rays?


    2. Relevant equations
    [tex]\Delta[/tex][tex]\lambda[/tex] = (h/mc)*(1-cos[tex]\theta[/tex])
    E = hc/[tex]\lambda[/tex]
    [tex]\Delta[/tex][tex]\lambda[/tex] = [tex]\lambda[/tex]scattered - [tex]\lambda[/tex]incident

    [c]3. The attempt at a solution[/b]
    I think I made this question more complex than it is...
    So I know that [tex]\Delta[/tex][tex]\lambda[/tex] = .00243 nm and Einitial=Ephoton + Eelectron, and Eelectron =10 keV.
    I made [tex]\lambda[/tex]scattered = hc/Ephoton.
    This equals [tex]\Delta[/tex][tex]\lambda[/tex] = hc/(Einitial - 10 keV), and then I plugged hc/[tex]\lambda[/tex]incident for Einitial.

    My final equation is .00243 - [tex]\lambda[/tex]incident = 1240 eVnm/(hc/[tex]\lambda[/tex]incident - 10 kEv).

    The solution is .0239 nm, however the correct answer in the book is .022 nm.
    Although the solution is close, I feel that it's more due to luck than to actually doing the correct methodology. Could anyone help me solve this problem? Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 28, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    so if you're happy with the wavelength shift to find the energy shift knowing

    [tex] E = \frac{hc}{\lambda} [/tex]

    differentiating and assuming only small changes, which probably isn't too far form the truth..
    [tex] dE = -\frac{hc}{\lambda^2}d\lambda [/tex]

    if you want the exact answer you need to solve:
    [tex] \Delta E = \frac{hc}{\lambda} -\frac{hc}{\lambda + \Delta \lambda}[/tex]
    you should be able to solve for lambda, by multiplying through by the denominators & rearranging to give a quadratic in terms of lambda
     
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