what is the velocity distribution of photoelectrons in photoelectric effect? I know that when light is incident on certain metallic surfaces, electrons are emitted from the surface. And I know that the maximum velocity can be calculated. But why does the textbooks use 'maximum velocity' rather than 'velocity'? My assumption is that those emitted photoelectrons don't have the same velocity. If it is the case, then what causes the phenomenon? And is it possible to derive the velocity distribution of these photoelectrons by simple experiments?
 A: Photo-emission from a metal surface is a multistep process. When the photon hits the metal surface it excites a photoelectron with a very high quantum yield, but that photoelectron is travelling in the same direction as the photon i.e. down into the metal. For an electron to be emitted from the surface the initial photoelectron has either to ricochet back out of the metal, or more likely to transfer energy to other electrons so one of the other electrons has enough energy to leave the surface. This process is essentially random and consequently has a very low quantum yield, so the overall quantum yield for photoemission is around $10^{-5}$ to $10^{-6}$.
The scattering of the initial photoelectron occurs via inelastic collisions, so whether it's the original electron scattered backwards or other electrons scattered by the initial photoelectron, the energy of the electron leaving the surface is less that the energy of the photon. The balance of the energy goes into lattice vibrations of the metal i.e. heat.
Offhand I don't know the energy distribution of the photoelectrons, but it is a well studied phenomenon.
