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Photoeffect and compton scattering

  1. Nov 1, 2006 #1
    I was asked question: "Why in case of photoelectric effect electrons closer to the atom are ejected and in case of Compton effect electrons in the outer shells are emitted from nucleus by x-ray photon?".

    Well I know that this topic is very deep and one has to carefully select words to explain this but I think that in case of CE photon has far too much energy and momentum to eject outer shell electron which is not too tightly bound to the atom. And in case of PE the photon is absorbed completely and for that to happen it has to encounter electron which is closer to the atom and therefore has higher binding energy.

    Am I thinking wrong? If so then smarter opinions are welcome.
  2. jcsd
  3. Nov 1, 2006 #2
    I don't understand this. Please explain
  4. Nov 1, 2006 #3
    sorry, i meant in case of CE it is not possible for photon to dissapear (absorbed by electron) so it only gives some part of energy and momentum to electron which is sufficient to eject it from the atom.
  5. Nov 1, 2006 #4
    The more I read this forum, especially topics concerning CE and PE I'm getting more and more confused.
  6. Nov 1, 2006 #5
    Well, I tend to oversimplify things intentionally because I think there is a natural human tendency to obfuscate.

    That said, I would suggest that the photoelectric effect and compton scattering are the same thing - photons hitting matter.
  7. Nov 2, 2006 #6
    yes, but still I dont understand why Compton effect involves free or loosely bound electrons and not electrons from inner shells. Can u enlighten me on that one?
  8. Nov 2, 2006 #7
    When one does compton scattering the photon energy is far beyond the absorption edges of whatever elements you're shooting the photons at. The probability that the photon interacts with atom via the photoelectric effect is much lower then.
  9. Nov 2, 2006 #8
    Do I understand this? If the target is tightly bound, then and interaction is more likely with a low energy photon. If the target is loosely bound, then the interaction is more likely with a high energy photon.
  10. Nov 3, 2006 #9
    Not really. It's more about how far off resonance the photon energy is from the absorption edges. I've seen a nice schematic picture of the interaction probabilities vs energy in the x-ray regime, I'll try to find it for you. You could also look up the Kramers-Heiseberg formula which describes the physics involved well.

    edit: the pic seems to be very evasive. too evasive considering that I've seen it atleast a gazillion times.
    Last edited: Nov 3, 2006
  11. Nov 3, 2006 #10
    >>interaction probabilities vs energy in the x-ray regime <<

    How about this: High energy photons are more like little bullets and interact more rarely, but low energy photons are more spread out and therefore more likely to interact?
  12. Nov 22, 2006 #11
    in case of compton effect,elastic collision between x-ray photon and target electron takes place. The electron close to the nucleus of the atoms are tightly bound. When such bound electrons scatter x-rays,the collision is considered to be taking place between the photon and the whole atom. the whole atom recoils instead of the individual electron.The compton shift in this case is given by same expression except that the mass of electron is replaced with that of the atom. The mass of the atom is about ten thousand times greater than mass of electron. Therefore the compton shift is so small as to be unobservable. this process gives unmodified line and we say that compton effect involves outer electrons only.
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