Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Compton Scattering of a photon by a moving electron

  1. Dec 22, 2011 #1
    1. The problem statement, all variables and given/known data
    If the total energy of the electron E>>m[itex]c^{2}[/itex]
    Prove that the formula of wavelength would be
    [itex]\lambda '[/itex]=[itex]\frac{hc}{E}[/itex](1+[itex]\frac{m^{2}c^{4}\lambda}{4hcE}[/itex])
    where [itex]\lambda '[/itex] is the wavelenght after scattering
    m is the mass of electron

    2. Relevant equations



    3. The attempt at a solution
    I have proved that [itex]\lambda '[/itex]=[itex]\frac{\lambda(E-Pc)+2hc}{E+Pc}[/itex]
    In my equation, P is the initial momentum of the electron.
    I haven't use the given assumption E>>m[itex]c^{2}[/itex]. So we must use it to eliminate P.

    But if E>>m[itex]c^{2}[/itex][itex]\Rightarrow[/itex] E[itex]\approx[/itex]Pc
    this cannot get the answer
     
  2. jcsd
  3. Dec 22, 2011 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    There's no angular dependence? The problem seems a bit ambiguous to me. What are the directions of the photon and electron initially?
     
  4. Dec 23, 2011 #3
    they are both along x axis, even after collision

    The origin question should be
     
    Last edited: Dec 23, 2011
  5. Dec 23, 2011 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    OK, you have
    $$\lambda' = \frac{\lambda(E-Pc)+2hc}{E+Pc} = \frac{2hc}{E+Pc} + \lambda\frac{E-Pc}{E+Pc}$$Now multiply the second term by (E+Pc)/(E+Pc) and then use the approximation ##E\approx Pc##.
     
  6. Dec 23, 2011 #5
    The second term will become zero after using E=Pc, since the second term has the factor E-Pc.
     
  7. Dec 23, 2011 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    That's not quite true. You need to multiply by (E+Pc)/(E+Pc) and simplify the numerator and then let E=pC.
     
    Last edited: Dec 23, 2011
  8. Dec 23, 2011 #7
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook