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I have encountered quite recently the "Compton edge", which made me review the Compton effect again. A photon with wave length $\lambda$ "bumps" into a charged particle (usually an electron) and passes some of its energy to the electron, while the remaining energy goes to another photon with wavelength $\lambda ' $. One can see by using conservation of energy and momentum that $$ \lambda' - \lambda = \frac{h}{mc}(1-\cos \theta)$$ Where $m$ is the particle's mass and $\theta$ is the angle between the incident photon's direction and the "new photon's" direction.

From this we can see that $$ \lambda \leq \lambda' \leq \lambda + \frac{2h}{mc} $$ The first inequality makes sense, because some of the energy is transferred to the electron so the wavelength can't shorten. It is equal when $\theta=0$ meaning the photon just "passed through" the electron with no interaction. The second is the what is called the "Compton edge" and it gives us a limit on how much energy can be given to the electron.

At this point I thought, how could this be consistent with the photoelectric effect? From what I know, in the photoelectric effect the whole energy is transferred to the electron and no "secondary photons" are created, which is like taking $\lambda \to \infty $, which isn't possible. So maybe this assumption that no energy going back the the electric and magnetic fields is like a photon with $\lambda \to \infty $ isn't valid, so lets attack the subject from the beginning with conservation laws:

The photon's momentum is transferred to the electron so $p=\frac{h}{\lambda}$ , but the energy is

$$ \frac{hc}{\lambda} = \sqrt{(pc)^2+(mc^2)^2}$$ And this equation can't be true because $pc = \frac{hc}{\lambda} $.

What is going on? I came to the conclusion that the photoelectric effect as I understand it can't be true. What am I missing? Thank you for your help!

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The photoelectric effect isn't a simple single photon-single electron interaction. The other electrons and the crystal lattice all participate so you can't apply the momentum conservation rules used in the Compton calculation (though energy conservation still applies). The end result is that the incoming photon can transfer all its energy to the photoelectron and we end up with:

$$ \tfrac{1}{2}m_ev_e{}^2 \le h\nu $$

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  • $\begingroup$ But if all of the energy in transferred, how come something else can get some of the momentum? A change in momentum comes with a change in energy. $\endgroup$ – Ofek Gillon Mar 26 '17 at 8:20

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