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I would like to know, please, regarding the photoelectric effect and the Compton effect, why is the following process dynamically impossible. I would like to know how to justify by words and also if there is some kind of mathematical demonstration. The teacher can ask me this question on a university test, so I need a complete answer, please.

$$\gamma + e^- \rightarrow e^-$$

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  • $\begingroup$ The reaction you describe is exactly what happens during a photoelectric absorption. Can you specify whether you mean "for the case of an unbound electron"? The basic problem is one of conservation of momentum and energy. In the above reaction, you can only conserve momentum by having a second mass (the atom in which the electron is bound) to absorb the rest of the change in momentum. Do you think you can work it out from there? $\endgroup$
    – Floris
    Apr 19 '17 at 20:36
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You cannot simultaneously conserve momentum and energy.

The "reaction" must take place in a straight line. Consider the reaction in the initial rest frame of the electron. Conservation of momentum says that $$ p_{\gamma} = p_e,$$ where $p_{e}$ is the momentum of the electron afterwards.

Conservation of energy says $$ p_{\gamma}c + m c^2 = (p_e^{2}c^2 +m^2c^4)^{1/2}$$

But using the first equation this means $$p_e^{2} c^2 + m^2c^4 +2p_e m c^3 =p_e^{2}c^2 +m^2c^4$$ This can only be true if the momentum of the electron after the interaction $p_e =0$ and thus that the initial photon had zero momentum.

Any finite momentum for the initial photon and energy and momentum cannot be conserved.

In the photoelectric effect this problem is solved by transferring some momentum to the remaining ion. In the Compton effect, a lower energy photon appears after the interaction.

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  • $\begingroup$ Why in the start you consider the momentum of photon equals to momentum of electron? $\endgroup$ Apr 19 '17 at 21:00
  • $\begingroup$ @CarmenGonzález Because I consider the interaction in the initial rest frame of the electron. This is not a necessary step, but simplifies thealgebra. $\endgroup$
    – ProfRob
    Apr 19 '17 at 21:12
  • $\begingroup$ What is the electron frame? Sorry, English is not my mother tongue. Did you equalize the linear moments because you considered that the electron was at first stopped? (v=0 m/s) $\endgroup$ Apr 19 '17 at 21:18
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    $\begingroup$ @CarmenGonzález Yes. I can choose any frame of reference, and that one is the simplest. $\endgroup$
    – ProfRob
    Apr 19 '17 at 23:40
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Just use conservation of 4-momentum. Initially we have $p_I=(p+mc,p)$. After the absorption we have $p_F=(\gamma mc,\gamma mv)$. That means $$ \begin{align} p+mc = \gamma mc\\ p = \gamma mv \end{align} $$ Try solving these two equations simultaneously. You'll see that the only solution is when $v=0$, implying that the momentum of the photon is $0$, or, put another way, it doesn't exist.

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