- #1
vabamyyr
- 66
- 0
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.
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.