# Mechanics of the photoelectric effect

1. Sep 16, 2015

### D.Freya

Given an experiment of the photoelectric effect, if we keep the intensity of the shining light source constant, by increasing the frequency of the light, the number of ejected electrons from the metal surface "decreases". I understand that increasing frequency (assuming it is already higher than the threshold) will eject electrons at higher kinetic energies, thus resulting in a higher final voltage. But I am confused as to why the number of ejected electrons decreases with higher frequencies?? I don't see the relationship.

Side note: I thought about how wavelength is inversely proportional to frequency and so higher frequency --> lower wavelength and the possibility of a photon impacting an electron decreases. I also thought about how the the increased final voltage would build up faster, preventing additional electrons from being ejected. Both answers are apparently incorrect.

2. Sep 16, 2015

### Drakkith

Staff Emeritus
Higher frequency photons have more energy, so in order to keep the intensity (power) the same you have to have fewer of these higher energy photons.