Consider a photoelectric setup as below to investigate the maximum kinetic energy electrons. enter image description here When we graph photocurrent against the voltage applied against the electrons moving, why does the photocurrent increase when the voltage increased (until it reaches saturation current).

I can understand why a negative voltage would decrease the current relative to when no voltage is applied as the lower energy electrons will be forced to turn back, however I don’t see the explanation for positive voltages. If the electron has been dislodged from the surface with ANY amount of kinetic energy, shouldn’t it be able to make the journey to the other plate? Why should a positive voltage help? enter image description here

  • $\begingroup$ More information is needed, most importantly the physical system in question. The photocurrent generated in what? $\endgroup$
    – Gilbert
    Mar 29, 2019 at 2:57
  • $\begingroup$ Oh sorry, its a pair of parallel plates with one having light shone on it. The plates have another voltage applied to it. I’ll add a diagram to clarify $\endgroup$
    – John Hon
    Mar 29, 2019 at 3:40
  • $\begingroup$ Is the chamber evacuated, or does it contain air or some other gas? $\endgroup$
    – PM 2Ring
    Mar 29, 2019 at 6:38
  • $\begingroup$ @PM2Ring Completely evacuated $\endgroup$
    – John Hon
    Mar 29, 2019 at 6:39

2 Answers 2


The photoelectrons are emitted with random angles as well as random energies and it's for these reasons that not all the electrons will reach the collecting plate.

Some photoelectrons will be emitted at angles that simply miss the collecting plate. How much this matters will depend on the geometry. Your diagram shows a plate spacing larger than the plate diameter and in this case you'd miss quite a lot of the emitted electrons, though I assume your diagram is just an illustration and in reality you'd use large area plates close together to reduce the effect. Anyhow, using a positive voltage will collect electrons that would otherwise be lost.

Also some photoelectrons will have low enough energies that they are scattered by air molecules before they can reach the collecting plate. A positive voltage collects electrons that would otherwise be scattered. I note you've mentioned in a comment that your system is evacuated so this wouldn't apply. However it will matter for any experiments that aren't done in a vacuum.

Finally I would challenge your diagram. That diagram appears to originate on this web page and it's just an illustration and not real experimental data. It isn't clear how big the increase in current with positive voltage actually is in a real experiment. Actual experimental data is hard to find but what I have found suggests the effect of positive voltage can range from relatively small to almost zero. The variation is presumably due to variations in how the experiment is done. If you use a large collecting plate in vacuum I would expect the effect of a positive voltage on the current to be very small.

  • $\begingroup$ so basically because there is a strong force it'll change the velocity such that more of them will end up on the other plate? I agree with you on challenging the diagram, it was just cause it was in my textbook $\endgroup$
    – John Hon
    Mar 29, 2019 at 10:52

It is so because as you increase the voltage the number of electrons reaching the other plate increases because more of them have the required energy to do so until a point is reached where all the electrons released from the plate on which light is shone constitute the photo current. On further increasing the voltage more electrons won't be released by the plate as the number of electrons released depends on the intensity of the incident radiation. Hence when there is no increase in the number of electrons released per unit time the photo current obviously remains the same no matter how high voltage you may apply. Just don't make your photo-electric experiment setup a discharge tube.


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