Sometimes it is shown that in a Compton scattering it is not possible that the photon transfers all it's momentum and energy to the electron, see for example here:
If one assumes complete energy and momentum transfer the corresponding energy equation is:
$$
h\nu + m_0c^2 = 0 + \sqrt{m_0^2 c^4 + p^2c^2}
$$
where $\nu$ is the frequency of the incoming photon and $m_0$ the rest mass of the electron and $p$ the momentum of the electron after the scattering.
The momentum equation is:
$$
\frac{h\nu}{c} + 0 = 0 + p
$$
Substituting for this for $pc$ in the energy relation gives:
$$
h\nu + m_0 c^2 = \sqrt{m_0^2 c^4 + h^2\nu^2}
$$
Squaring both sides results by the binomial formula in $2h\nu m_0 c^2 = 0$ which is a contradiction since $\nu \neq 0$ and $m_0c^2 \neq 0$.
In the photoelectric effect however the photon seems to give all it's momentum and energy to the electron? I guess that the corresponding conservation laws are nevertheless valid because the atom or complete metal will get some energy any momentum, but how can one see this and the difference to the compton effect in formulas?
Best Answer
No, the photoelectrons are emitted with a range of energies. The well known expression:
$$ E = h\nu - \phi $$
gives the maximum energy, but photoelectrons are emitted with energies ranging from zero to this maximum value.
The emission of photoelectrons is a messy business. The incident photon creates a photoelectron with a high quantum yield, but this electron has its momentum pointing in the same direction as the incident photon i.e. down into the metal. Most of the time the electron just transfers energy to the lattice, but around $1$ time in $10^5$ to $10^6$ the electron will either ricochet out, or it will hit other electrons and eject them from the metal surface. That's when we observe a photoelectron, and given the random nature of the process it's no surprise that the ejected electrons have a spread of energies.