There is a simple reason why a free election can't absorb a photon completely: you can't conserve both energy and momentum for the system of you start off with an electron and a photon and and up with only an electron. You need a final photon as well for conservation of energy and momentum to be satisfied.

For the photoelectric effect things are different. We don't have a free electron ; it is bound to a nucleus. Consequently it can only have certain precise energy values (and won't interact with photons that would move it to a non existent energy level, which is the basic argument for quantum behaviour, but not really comparable to Compton scattering).

In the situation analagous to Compton scattering, the photon has more energy than the binding energy of the electron, and so we wind up with a free electron. But in this case energy and momentum can be conserved without a final photon since the nucleus is also involved in the interaction. The initial state is {photon, bound electron, nucleus } and the final state {free electron, nucleus}.

It is possible to construct Feynman diagrams with a final photon present too, but since they have an extra vertex they happen less often by a factor of roughly the fine structure constant $\alpha\sim\frac{1}{137}.$

Or to put it another way, sometimes in the photoelectric effect where you end up with a free electron you *do* get a final photon, but it is in less than 1% of cases (unless I've overlooked some reason why it can't happen).

## Best Answer

The two processes are different.

In the Compton effect the energy of the incident photon (~ many keV or Mev) is very much larger than the binding energy of the electron in the atom (~ eV) and so a target electron bound in an atom can be considered as essentially free.

In the Compton effect the target electron is ejected from the atom and the process can be treated using "billiard ball" dynamics.

The equation for the change in wavelength derived by use of the conservation of momentum and energy shows that there is a finite maximum amount of energy that can be transferred from the photon to an electron in this process.

$\lambda' - \lambda = \dfrac {h}{m_{\rm e} c}(1 - \cos \theta)$ where $\lambda' - \lambda$ is the change in wavelength of the photon and $\theta$ is the scattering angle whose maximum value is $180^\circ$.

The scattered photon having lost some of its energy can then undergo further collisions with electrons in the material.

The photons responsible for the photoelectric effect have energies of order of a few electron-volts ($450\; \rm nm\approx 2.8 \;eV$) and for metals it is the conduction (free) electrons which are ejected. A free electron is given enough energy by the photon it has absorbed to overcome a potential barrier (work function energy) to escape with the rest of the energy of the photon from the surface of the metal.