I'm having trouble understanding what my professor is getting at asking in this question. I just visited her office and her explanation minutely helped. I'm hoping to get a bit more clarity on what is being asked. Also, as a note, I got special placement in this class, and as a result, many of the equations being talked about (i.e., Compton Scattering) are new to me. I shouldn't have an issue learning them relatively quickly. Anyway:

Consider Compton scattering for x-rays, original wavelength 0.073 nm scattered to a wavelength 0.00243 nm longer. The scattered photon has a longer wavelength, therefore less energy. Assuming that all of that “lost” energy goes into the kinetic energy of the electron partner in this collision, at what fraction of the speed of light will the electron be moving? Be sure you think carefully about the difference in energy between the original photon and the scattered photon – that difference gets assigned to the KE of the electron.

I'm not looking to get the answer to the question, but maybe to have some help in working it out, or if I am just not recognizing something obvious it could be pointed out to me.

What I assume the process for figuring out the fractional velocity relative to $c$ will be is going to depend on my using the Compton scattering equation, then finding a relative difference in the $E$ of each wavelength and computing the velocity from that E relation. However, I don't know how Compton scattering plays a role in $E$.


1 Answer 1

  1. what is the energy of a photon?
  2. how much has that energy changed after interacting with the electron?
  3. what is the formula for the kinetic ennergy of an electron?
  4. if the electron was not moving, and all of the change of the photon energy changed th eelectron's kinetic energy, then what is the speed of the electron?
  5. divide the answer from 4. by c.
  • $\begingroup$ If you turn those ) into . it'll make a list :) $\endgroup$
    – Kyle Kanos
    Commented Sep 18, 2014 at 19:34
  • $\begingroup$ @KyleKanos I fixed it already. I was hoping the thing would autoformat. $\endgroup$ Commented Sep 18, 2014 at 19:35
  • $\begingroup$ OH, you're saying that it WOULD, if I had just formatted differently. $\endgroup$ Commented Sep 18, 2014 at 19:35
  • $\begingroup$ Right, it autoformats with specific conditions. I think only ` 1. word` and ` - word` work (spaces necessary, I think). $\endgroup$
    – Kyle Kanos
    Commented Sep 18, 2014 at 19:39

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