# Kinetic energy of photoelectrons

I came across the following question:

A photon of energy $$h\nu$$ is absorbed by a free electron of a metal having work function $$W. Then:

1. The electron is sure to come out

2. The electron is sure to come out with kinetic energy $$h\nu-W$$

3. Either the electron doesn't come out or it comes out with a kinetic energy $$h\nu-W$$

4. It may come out with kinetic energy less than $$h\nu-W$$

Both options 1. and 2. seem to be correct to me. However, the correct answer provided is 4.! How is this so? Isn't $$W$$ the amount of energy required for ejection of the most tightly bound electron in the metal? Then why is there a possibility that it won't come out with kinetic energy $$h\nu-W$$?

• The process happens inside a metal lattice. The electron can collide with plenty of "stuff" before it ever reaches the vacuum. – CuriousOne Aug 16 '16 at 13:11
• Or, even without colliding with anything, is heading in the wrong direction. – Jon Custer Aug 16 '16 at 13:36
• u n c e r t a i n t y :) option 4 uses the good wording for a single electron – user46925 Aug 16 '16 at 14:09
• yep, option 4 seems to be much better. – Lamichhane88 Aug 16 '16 at 14:11

Photoemission in the usual experiment is a two step process. First the incident photon creates a photoelectron in the bulk of the metal:

The quantum efficiency for this is almost $100$% i.e. almost every photon that hits the metal creates a primary photoelectron. However this photoelectron is heading in the same direction as the original photon i.e. down into the bulk of the metal. For the electron to escape from the surface it has to rattle around bouncing off other electrons and nuclei in the metal:

However this process is very inefficient. Only about $1$ in $100,000$ of the primary photoelectrons manages to bounce back towards the surface and escape. So the overall yield is around one photoelectron emitted from the surface for every hundred thousand photons.

The electron will normally lose some energy $\Delta E$ while it's bouncing around, then it loses an energy equal to the work function when it escapes from the surface, so the final electron energy will be:

$$E = h\nu - \Delta E - W$$

The work function is a constant, but the energy lost in scattering, $\Delta E$, is very variable and can range from almost zero to so large that the electron never escapes.

I won't go through your questions 1 to 4 in detail because this is obviously a homework question. However from the explanation above you should be able to work out what the answers are.

• Oh wow, I didn't even think about this jolting around of electrons before they escape! Thanks so much! – user106570 Aug 17 '16 at 1:19
• Nonsense: most of the light that impinges on a metal is reflected. And many photoelectrons are emitted directly, as shown in ARUPS (angle-resolved photoemission spectroscopy). – Pieter Jan 1 '18 at 21:17

The work function $W$ is the minimum energy of photon that is required to eject the electron from the metal in photoelectric interaction i.e. the energy of least tightly bound electron in the lattice. However if the energy of the photon is higher than W then it can eject the electrons that are more tightly bound in the metal hence photo electrons with energy less than $h\nu -W$ can also be ejected.

• is it a gaussian or a QM Born distribution ? – user46925 Aug 16 '16 at 14:23
• you can see some photoelectron spectra here – hsinghal Aug 16 '16 at 14:35
• All what a curious needs, TY – user46925 Aug 16 '16 at 15:08