Question on the photoelectric effect?

[mklaben15]

For the work function, is it taken into consideration that the metal has oxidized, giving it a different energy to release the electron? Or is it so small that it is negligible?

 

[ZapperZ]

For the work function, is it taken into consideration that the metal has oxidized, giving it a different energy to release the electron? Or is it so small that it is negligible?

Depends on the experiment.

If you are simply looking at the value of the stopping potential, then no, the oxide layer isn't important. This is because metal oxide tends to be insulators, will have a higher "work function" than metals. And depending on the type of oxide, the typical light source used in a standard Photoelectric Effect experiment will have a penetration depth longer than the oxide layer, so the electrons from the metal underneath can escape. Since these electrons will have higher KE than the electrons from the metal oxide (common metals used in this experiment tend to have smaller work function), then the stopping potential is dictated by the metal's photoelectrons, not from the oxide layer.

In photoemission spectroscopy experiment (a more complex type of photoelectric effect), the oxide layer definitely will affect the result. It is why samples are either cleaned, cleaved, or transferred in ultra-high vacuum environment before the experiment.

Zz.