- #1
abotiz
- 72
- 0
Hi,
There is this one thing that bugs me about photoelectric effect, everywhere I read it says [...] usually the K-shell electron is ejected [...], but there is not an explanation to why this is the case.
I know that the photon's energy needs to be a "little bit" larger than the binding energy and for the photon to be absorbed the nucleus/atom needs to receive a recoil (extra) energy. But this could apply for any shell electron, L,M etc.
The only way I can make sense of all this, is if the nucleus needs to be "close" to the electron (or where the interaction takes place) so that the nucleus "can" receive the recoil energy. But it sounds weird.
I think I heard somewhere that it is usually the K-shell due to the higher binding energy, "does not move around", so the photon has a "clear shot". But the elctron is not a " particle" right? Its like a cloud around the nucleus?
We need an explanation once and for all!
Thank you!
There is this one thing that bugs me about photoelectric effect, everywhere I read it says [...] usually the K-shell electron is ejected [...], but there is not an explanation to why this is the case.
I know that the photon's energy needs to be a "little bit" larger than the binding energy and for the photon to be absorbed the nucleus/atom needs to receive a recoil (extra) energy. But this could apply for any shell electron, L,M etc.
The only way I can make sense of all this, is if the nucleus needs to be "close" to the electron (or where the interaction takes place) so that the nucleus "can" receive the recoil energy. But it sounds weird.
I think I heard somewhere that it is usually the K-shell due to the higher binding energy, "does not move around", so the photon has a "clear shot". But the elctron is not a " particle" right? Its like a cloud around the nucleus?
We need an explanation once and for all!
Thank you!
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