This is a problem in my textbook and I've shown it this way:
$E_{initial}=\frac{hc}{\lambda} + mc^2$
$p_{initial}=h/\lambda$
After collision with photon having zero energy we get
$p_{final}=h/\lambda$
$E_{final}=\sqrt{(\frac{hc}{\lambda})^2+(mc^2)^2}$
Which is in contradiction with the conservation of energy.
Now, this result is I think contradictory to Einstein's explanation of the photoelectric effect.
In the photoelectric effect the photon is absorbed by the free electron and this is what makes it have kinetic energy.
What am I interpreting wrong? The problem comes from the context of the Compton Effect, by the way.
Best Answer
The original problem can be seen in terms of energy and momentum conservation.
Before scatter, there are two particles in the center of mass and the center of mass has an invariant mass larger than the mass of the electron. For total absorption of the photon there would be only the electron left. As the electron has a fixed mass and at the center of mass it should be at rest, the reaction cannot happen. It can only happen if a third particle is involved to conserve the overall energy and momentum , and this is what is happening with the photoelectric effect.
The incoming photon interacts with an electron that is tied to the atom by a virtual photon . The whole system takes up the energy and momentum conservation.
The inverse problem happens with a gamma generating an e+e- pair. The gamma has zero invariant mass, the pair will have at least two electon masses at the center of mass, so a gamma cannot turn into an electron positron pair , a third particle has to be involved. The simplest is a virtual photon from some nucleus too. The Feynman diagram is the same as the one above with a different interpretation ( incoming e- is read as outgoing e+)
a) diagram copied