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X-Ray Spectra and Photon Energy

  1. Oct 26, 2012 #1
    1. The problem statement, all variables and given/known data
    In a particluar x-ray tube, an electron approaches the target moving at 2.35 108 m/s. It slows down on being deflected by a nucleus of the target, emitting a photon of energy 40 keV. Ignoring the nuclear recoil, but not relativity, compute the final speed of the electron.

    2. Relevant equations
    Equation for photon Energy: Ephoton = Ei - Ef


    3. The attempt at a solution
    Ephoton = Ei - Ef

    Using the emitted photon energy as the change in energy (40 KeV = 6.409x10-15 Joules)
    and the relativistic equations for energy: E = pc = mevc.
    We have: 6.409x10-15 = mec(vi - vf)
    Where mec = (9.109-31)(3x108)
    Then we have: 2.345x107 = vi - vf
    We can use 2.35x108 as the electron's initial velocity:
    so vf = 2.35 108 - 2.345x107 = 2.1155x108.
    However, the answer is 2.27x108. So I am off by a small amount.

    What am I doing wrong?
     
  2. jcsd
  3. Oct 26, 2012 #2
    Try the relativistic formulation of momentum

    p = [itex]\gamma[/itex]mov
    where
    [itex]\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/itex]
     
    Last edited: Oct 26, 2012
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