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I made a scenario where a photon with energy and momentum $(E,p)$ hits an electron with mass $m$, and the electron absorbs the photon and got a contradiction. This implies that this scenario is impossible. I'm writing the contradiction down:

Using energy and momentum conservation:

$$E+mc^2 = E'$$ $$p=p'$$

where $E'$ and $p'$ are the energy and momentum of the moving electron.

Playing around with the formula, and using $E=pc$ for a photon, we get:

$$E+mc^2 = E' \implies pc + mc^2 = E' \implies p'c + mc^2 = E' \neq \sqrt{(p'c)^2+(mc^2)^2}$$

Using the contradiction above, we deduce that a free electron cannot absorb a photon. But how does this happen in the photo electric effect (or the bohr model but I think it's an incorrect model)? What changes in the photoelectric effect that makes the electron able to absorb the photon?

tryst with freedom
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Habouz
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    You are correct. It is not one electron that absorbs the energy and momentum of the photon. It is the entire system of the metal's lattice plus one of its electrons. If you add an additional (very heavy) lattice to your calculation, then you will see that the energy/momentum balance can be satisfied. A single electron and a single photon can only do a scattering process. – FlatterMann Oct 09 '22 at 12:31
  • So basically, the existence of potential energy in the lattice allows such interactions? – Habouz Oct 09 '22 at 12:37
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    The difference between the three body vs. the two body scattering process does. There are more degrees of freedom to satisfy both momentum and energy conservation at once. It is probably correct that there has to be a minimal work function to allow for the necessary coupling. – FlatterMann Oct 09 '22 at 12:45
  • @FlatterMann We very strongly discourage answering in comments. It leaves questions unanswered (from the system's point of view) and comments can be deleted by the system or moderators at any time. Also note that answer's can get you more reputation points than a comment. – StephenG - Help Ukraine Oct 09 '22 at 13:00
  • @StephenG-HelpUkraine I didn't feel like it's a complete answer. For that someone will have to actually show that my handwaving argument is correct. I am happy to delete my comment myself if it is deemed inappropriate by the standards. All I had in mind was to give a pointer to what was missing in the original Ansatz, which wasn't wrong, after all, just incomplete. – FlatterMann Oct 09 '22 at 13:22
  • @FlatterMann Your comment received two upvotes so I think it's a reasonable sign people think it's the core of an answer. I think in general that posting the guts of an answer is preferred to a question maybe not getting any answers (which happens). Your choice anyhow. Cheers. – StephenG - Help Ukraine Oct 09 '22 at 13:28

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A simple calculation gives: $$p'c+mc^{2}=E'=\gamma\, m'c^{2}=\sqrt{(p'c)^{2}+(m'c^{2})^{2}}$$

Which give:$$(p'c)^{2}+(mc^{2})^{2}+2p'c\;mc^{2}=(p'c)^{2}+(m'c^{2})^{2}$$ $$(m'c^{2})^{2}=(mc^{2})^{2}+2h\nu mc^{2}$$ $$m'=m\sqrt{1+\frac{2h\nu}{mc^{2}}}$$

The Tiler
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