Question about Photoelectric Effect versus Compton Scattering.

[uart]

In the context of interaction of photons (with energies from around visible light and upwards) and the electrons of solid matter. I've read that at the lower energy levels that the photoelectric effect is more likely to occur during such an interaction and that Compton scattering is more likely for higher photon energies like x-rays. (... and pair production at even higher energies).

I was just wondering why this is. Looking at it from a simplistic point of view I would have expected the opposite. That is, I would have expected that the higher energy photons would have been more likely to completely knock an electron out of the solid (PE effect) and that the lower energy photons would have been more likely to just make the electrons recoil as in Compton scattering.

Any simple explanations?

 

[mathman]

The photoelectric effect involves the electron absorbing the photon. Compton scattering involves the photon giving up part of its energy to the electron, since it has too much energy to get absorbed.

 

[uart]

Thanks mathman, I guess I was wondering why the electron can't just absorb the higher energy photon and and get ejected at with more KE.

Anyway it's making more sense to me now. So I'm guessing it's momentum considerations that makes it unlikely for a higher energy photon to give up all it's energy. If a high energy photon were to give all of it's energy to an electron then the electron would end up with more momentum than the photon could provide, so the lattice would have to make up the difference. Does this sound correct?

 

[ZapperZ]

Thanks mathman, I guess I was wondering why the electron can't just absorb the higher energy photon and and get ejected at with more KE.

Actually, it can, but there's a limit to this where, after a certain level, the increase will not be noticeable.

As you increase the photon energy, you not only can get electrons close to the Fermi energy, you also start to probe deeper into the band structure. So not only are you getting higher energy photoelectrons, but you're also getting more low-energy electrons coming from deeper in the band. However, at some point, you will no longer get any more significant photoelectrons because the penetration depth of the photon is now longer than the escape depth of the photoelectrons. This is especially true for metals where the electrons trying to escape the metal have a higher probability of losing its energy via collisions with other conduction electrons. So when the penetration depth is longer than the electrons' escape depth, you'll not get any more electrons.

Zz.

 

[lightarrow]

In the context of interaction of photons (with energies from around visible light and upwards) and the electrons of solid matter. I've read that at the lower energy levels that the photoelectric effect is more likely to occur during such an interaction and that Compton scattering is more likely for higher photon energies like x-rays. (... and pair production at even higher energies).

I was just wondering why this is. Looking at it from a simplistic point of view I would have expected the opposite. That is, I would have expected that the higher energy photons would have been more likely to completely knock an electron out of the solid (PE effect) and that the lower energy photons would have been more likely to just make the electrons recoil as in Compton scattering.

Any simple explanations?

About photoelectric effect ZapperZ answered you, about Compton scattering, think that only for X-rays the photon starts to have enough energy to do something to the electron; for a photon in the visible range, a collision with an electron is something like a baseball colliding with a truck.