This question has been asked many times on this site before with different doubts in each question but I am conflicted with the vast amount of sources stating contradictory things. There are many questions on this site that state "saturation current does not increase with frequency but if kinetic energy increases then velocity of electrons increases and so current should increase". All answers to these questions explain why current cannot increase with speed of electron but do not correct the statement "saturation current does not increase with frequency". Does that mean it is true?

In a lecture I saw on the MIT open courseware site, the professor said current is directly proportional to intensity, but did not mention anything about frequency. In other papers/lectures of different universities, I saw only the current dependency with intensity and not frequency anywhere (an MIT "photoelectric effect notes" pdf said the number of photoelectrons ejected is strictly dependent only on intensity).

It does seem somewhat contradictory when you consider that photoelectrons ejected per second are proportional to photons hitting the surface per second and intensity of light = number of photons in light per second * energy of light (proportional to frequency). So if the frequency increases and intensity stays the same, photons in light should decrease and thus photoelectrons ejected should decrease and so saturation photocurrent should decrease? But then some sources state that increasing frequency increases probability to eject electrons and so both the factors (decrease in number of ejected electrons due to decrease in number of photons and increase in number of ejected electrons due to increase in frequency) cancel out and saturation current stays the same?

Let us consider we somehow got rid of all practical difficulties and consider an ideal situation (like energy loss and such). Then does changing frequency (but keeping intensity constant) change the photo/saturation current?

  • $\begingroup$ Does this previous Q&A ( physics.stackexchange.com/questions/401526/… ) answer your question? $\endgroup$ Commented Oct 30, 2023 at 5:28
  • $\begingroup$ I don't think I am able to draw a conclusion for the question "is photocurrent dependent on frequency" from the answer (which does not clearly state a conclusion and rather just explains the process of photoemission by using some terms of quantum mechanics(which I am not well versed in)) $\endgroup$
    – Ham Lemon
    Commented Oct 30, 2023 at 10:57

1 Answer 1


Your conclusion is correct. In an ideal situation, if we increase the frequency at constant intensity past the threshold where all photons extract electrons, the number of emitted electrons and, consequently, the current may decrease. The reason is that the intensity is proportional to the number of photons times their frequency. By increasing the frequency while holding the intensity constant, the number of photons must decrease.

  • $\begingroup$ If we now include the "probability of ejecting electrons increases due to increase in frequency", then can these both effects (increase in photoelectrons due to frequency increase and decreases in photoelectrons due to decrease in photons) cancel each other out? I am asking this because this also felt like a convincing explanation given here $\endgroup$
    – Ham Lemon
    Commented Nov 2, 2023 at 6:01
  • $\begingroup$ @HamLemon There may be compensation only below the threshold where all photons extract electrons, Not above that threshold. At that point, there is a one-to-one correspondence between the number of photons and the number of extracted electrons. $\endgroup$ Commented Nov 2, 2023 at 6:49
  • $\begingroup$ Oh ok, I get it now, so photocurrent may be compensated and may stay constant for varying frequency and constant intensity but saturation current will vary. Is my conclusion correct? $\endgroup$
    – Ham Lemon
    Commented Nov 2, 2023 at 6:51
  • $\begingroup$ @HamLemon Your conclusion is correct. $\endgroup$ Commented Nov 2, 2023 at 6:54
  • $\begingroup$ Ok, now I clearly understand the concept. Thank you for bearing with me and patiently explaining things to me🙏🏼 $\endgroup$
    – Ham Lemon
    Commented Nov 2, 2023 at 7:00

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