Under constant light intensity frequency, do electrons emitted from a metal surface travel at different speeds?

Question

I was playing around with this PheT simulation: https://phet.colorado.edu/en/simulation/photoelectric

Under a certain threshold wavelength and an intensity at 20%, the electrons were being emitted from the Sodium surface. Without changing anything, the elctrons did not share the same velocity. Why?

Right now, I am under the assumption that the atoms in which the electrons are emitted from are all the same in terms of energy level configuration. For example, in order for an electron to jump from n=1 to n=2 it would require a photon containing 3eV (making it up btw) of energy, the electron ignores all photons containing a different amount of energy.

Now, also under the assumption that the frequency of the incoming photon packets remains constant throughout, wouldn't it make sense for the emitted electrons to be all traveling at the same velocity?

TL;DR: If an electron requires a discrete amount of energy to jump, and if photon packets are shone at a metal surface without changing the frequency(they all have the same energy), why do some electrons travel faster than others when emitted as if they were able to receive more energy?

Answer 1

Photo-emission is not a simple one step process. The incident photon excites an electron, but the momentum of the initial photoelectron is in the same direction as the incident photon i.e. down into the bulk of the metal. For the electron to be emitted as a photo-electron it has either to backscatter off another electron in the metal or it has to transfer its energy to some other electron so that electron can head back out towards the surface.

As you can imagine these are rather unlikely events, and that's why only one photo-electron is emitted for every $10^5$ to $10^6$ photons. The initial quantum yield is almost 100% i.e. almost every photon excites an electron heading down into the metal but only 0.001% to 0.0001% of these initial electrons manage to backscatter with enough energy to leave the metal.

This also explains why the energy of the emitted electrons has a continuous range. The backscattering is a random process and some electrons will backscatter with almost all their initial energy while others will backscatter with only a small energy.

Answer 2

There are several factors which determine the energy acquired by a photo-electron:

1. the structure of the crystal surface determines the work function, which means that (111) surface is different from (100); only a perfect crystal has a single surface structure.

2. There is always some bandwidth to the optical source, though it may be quite small. One should expect that a proportional bandwidth will be inherited by the photo-electrons.

3. If the energy of the striking photons exceeds the work function for photo-emission, this additional energy is available to the photo-electrons, and their individual histories determine how it is apportioned.

For case (3) the photo-electron may have originated below the immediate surface, and lost some of the acquired energy on the way to the surface. Only if it still has enough energy will it escape.

4. If the surface is illuminated continually there will be a build-up of surface charge, which will tend to inhibit photo-emission, but which will also increase the variation in the energies of the emitted photo-electrons; they will interact with these electrons.

Avoid case (4) by applying a strong voltage, accelerating the emitted photo-electrons away from the photo-cathode. This also serves to reduce the work function slightly, and gets all of the photo-electrons heading in the same direction. When applied in a vacuum, this is the beginnings of an electron gun.

In experiments I conducted, generation of phot-electrons by transmission through ultrathin polycrystalline gold films, the energy spread was a fraction of an eV; this was measured indirectly in my case, but others have measured it directly. So its possible, with work and expense, to limit the enegy spread of the photo-electrons.