# Why doesn't saturation current in the photoelectric effect depend on the frequency of light absorbed ny the metal emitter?

If current I is given by I = nAev, where n is the number of electrons per unit volume, A is the area, e is the charge of an electron and v is the velocity of the electron, it means that current increases with increase in velocity of the electron which increases with the frequency of light incident on the metal emitter. Why then doesn't saturation current increase with increase in frequency?

In the context of the photoelectric effect it is probably not a good idea to use the equation $I = nAev$ because the speed of the electrons is not constant across the gap.
It is better to use $I=Ne$ where $N$ is the number of photoelectrons collected per second by the plate remote from the plate emitting the photoelectrons and $I$ is the current.

The saturation current occurs when all the emitted photoelectrons are collected.

The saturation current can change with changing frequency and so a lot of the graphs in textbooks are not correct.

The intensity of light is proportional to the number of photons arriving per second times the frequency of a photon.

If you increase the frequency of the light keeping the intensity the same then the number of photons hitting the surface per second decreases which might well have an effect on the number of photoelectrons being collected per second (photoelectric current).

If you have all the incoming photons of high enough energy for each of them to release a photoelectron and then increase the frequency whilst keeping the intensity the same the saturation current will decrease.

The Phet photoelectric emission simulation can be used to show such behaviour.

If the frequency is kept constant then the saturation current does increase as the intensity is increased.

• How can we justify your third para? "If you increase the frequency of the light keeping the intensity the same then the number of photons hitting the surface per second decreases" – samjoe Mar 14 '17 at 11:11
• @samjoe I have added a sentence to explain this. "The intensity of light is proportional to the number of photons arriving per second times the frequency of a photon." – Farcher Mar 14 '17 at 11:14
• Your answer is very correct and consistent! Thanks for explanation. So the graph is indeed incorrect. – samjoe Mar 14 '17 at 11:20

In the relation, $I=neV_dA$, the velocity is the drift velocity of the electrons inside the conducting material.

The saturation current depends on the intensity of the light,for a given frequency. More photons correspond to more electrons. One photon knock out one electron and if you increase the number of photons (or increase the intensity) you increase the current.

If you increase the frequency of the photon, you increase the electron's kinetic energy outside the surface. When the increase you increase the frequency you increase stopping potential but the saturation current remains the same for a given intensity. At zero potential, the value of photoelectric current is higher for the photon whose frequency is greater than the other frequency of other photons. But the saturation current doesn't depend on the frequency. When the Photo-electons collide with the anode plate they lose some of their kinetic energy. More energy implies loss of more energy.

Well in the equation $I=neAV$ with the increase in $V$, $I$ increases only if n, e, A does not decrease . Now just think if you increase the drift velocity of the electron then they would move from their position and the no of electrons per unit volume (n) should decrease. Thus the resultant effect is that I does not change.

Also you want to increase the velocity of the electron keeping $n$ constant so that would mean that the electrons are randomly moving in a given region only and hence their contribution to the net current is zero.