photoeletric effect and compton effect - why the reaction photon + electron --> electron is impossible 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.
$$\gamma + e^- \rightarrow e^-$$
 A: 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.
A: 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. 
