1
$\begingroup$

I am reading on Compton scattering, more specifically on how Compton interpreted his results. He observed that the frequency of the scattered radiation was independent of the material used, so he assumed that the scattering was due to photons hitting electrons. The textbook from where I am reading this from states that Compton had some a priori justification from considering the fact that a x-ray photon has more energy than a ultraviolet photon. Furthermore, from the photoelectric effect the energy of an ultraviolet photon is comparable to the minimum energy with which an electron is bound in metal. How is this a good justification? Wouldn't it be enough to base it from Einstein's prediction of the photoelectric effect? That way only one photon produces an electron and is therefore surface independent.

$\endgroup$
2
$\begingroup$

Compton observed that the scatter angle of a photon was the same regardless of the material used to effect the scattering. He concluded (correctly) that whatever it was that was doing the scattering did not seem to be specific to the material in question.

So what are the candidates in the material?

  1. The nucleus. Definitely different for different materials
  2. Bound electrons. Like the electrons that give rise to photoelectric effect, their energy is a function of the material.
  3. Free electrons. Unlike the bound electrons, free electrons "don't care what material they are in" - they behave in the same way.

This makes the assumption that free electrons are responsible for Compton scatter very reasonable. And since Xray photons have much more energy than UV photons, we can further declare that "pretty much all electrons appear to be free for the purpose of Compton scatter" - because whatever little binding energy they have (relative to the energy of the gamma ray) it will not really make a difference to the outcome.

The above is not rigorous - I am just trying to address your concern on a level I hope you are comfortable with.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.