# Why doesn't saturation current in the photoelectric effect depend on the frequency of light absorbed by 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 must mean that the current increases with increase in velocity of the electron which increases with the frequency of light incident on the metal emitter. Why doesn't then 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" Mar 14, 2017 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." Mar 14, 2017 at 11:14
• Your answer is very correct and consistent! Thanks for explanation. So the graph is indeed incorrect. Mar 14, 2017 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.

In photoelectric effect the electrons will eject when sufficient frequency of light is incident. The saturation current will depends on the no.of photoelectrons ejected per second.As the frequency increases the energy in the quantum packets increases but the no. Of photons that should incident on the cathode plate remain same. Maybe the kinetic energy of the electron is more in this case, but overall the no. of electrons that should hit the anode remain same.It only vary with the intensity of the light.So on increasing the frequency probability of hitting the anode will increase(as kinetic energy increases) but that effect can be negligible.

So overall with this discussion we can conclude that saturation current only depends on intensity but not on frequency.

Hope it helps 😄

I do not think that the speed of the electron is important, and the reason for this is that the electron will face a repulsion of the electrons on the anode surface, so the importance is the number of electrons that will arrive because they will almost stop