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Compton Shift and Momentum

  1. Oct 18, 2015 #1
    1. The problem statement, all variables and given/known data

    X-rays with wavelength 55pm are scattered from a graphite target. Consider an x-ray which is scattered from a valence electron at an angle of 180 degrees (back towards the x-ray source).

    a. Is the wavelength of the X-ray greater or less than before the collision?
    b. What is the momentum of the recoiling electrons?


    2. Relevant equations

    ##\Delta\lambda=\frac {h} {mc}(1-cos(\theta))##
    ##p=\frac {h} {\lambda}##


    3. The attempt at a solution

    Part a. seems obvious. The X-ray collides with an electron, transfers some kinetic energy to it, losing some energy in the process. Less energy, longer wavelength, right?

    Part b however, is giving me a few issues.
    I start by finding the Compton shift for 180 degrees, which gives [itex]\Delta\lambda=4.85pm[/itex].
    This means that the scattered X-ray should have a wavelength of 55+4.85=59.85pm, correct?
    This is where things get a bit hairy. The solution for this problem however, states that the recoiling X-ray has a wavelength of 50.14 pm. This doesn't seem to make sense, as the energy of the X-ray has increased, despite transferring energy to the electron (these solutions have had issues in the past). Anyway,they then simply use [itex]p=\frac{h}{\lambda}[/itex] to calculate the momentum of the recoiling electron, yielding (with their wavelength), 1.32E-23 Ns.
    I however have tried the following:
    ##p_{i}=p_{f}##
    ##\frac{h}{\lambda_{i}}=p_{electron}-\frac{h}{\lambda_{f}}##
    ##p_{electron}=h(\frac{1}{\lambda_{i}}+\frac{1}{\lambda_{f}})##
    ##p_{electron}=2.31E-23Ns## (using my values for wavelength, 55pm and 59.85pm)
    OR:
    ##p_{electron}=2.53E-23Ns##(using their values)
    Which of course, doesn't agree with the solutions.
    So, I was wondering, am I missing something fundamental and conceptual in regards to energy and wavelength shift and, is my momentum approach valid?
     
  2. jcsd
  3. Oct 18, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Right.
    I didn't check the number but the approach is right.
    50.14 pm is clearly wrong.
     
  4. Oct 19, 2015 #3
    Alright, thanks, I was just worrying I'd managed to miss something fundamental!
     
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