I recently came across a doubt in the photoelectric effects topic, and It wasn't cleared even after researching,

The doubt is, that when we increase the frequency of incident rays, current is unaffected, as the number of electrons is the same, and Kinetic energy is increased.

But if KE increases, hence velocity increases and current( charge per unit time) Should increase as more electrons pass a given point in unit time??

This problem is discussed before on this website, but I was still confused.

Can you please suggest the problem here..

  • 1
    $\begingroup$ As you noted, there's several questions on this site that address this. If you want an answer to your specific question, you should be more precise as to what you don't understand. "Still confused" is too vague. $\endgroup$
    – joseph h
    Dec 20, 2022 at 22:49
  • $\begingroup$ The confusion is that all the electrons wouldn't strike the detector plate together, there must be some gap between first photoelectron and last photoelectron that left the metalic plate, so hence, if they were moving at a higher velocity, this gap will get covered more quickly and charge per unit time should increase. Although the effect would we small, but still my textbook and teachers said it is independent of frequency. Can this be because the effect is almost negligibly small?? $\endgroup$ Dec 20, 2022 at 23:21
  • $\begingroup$ The charge per unit time received is the same as the charge per unit time emitted. It doesn't matter how much time it takes for the electrons to travel: if they are slower, there are merely more in transit on average. $\endgroup$
    – John Doty
    Dec 20, 2022 at 23:30
  • $\begingroup$ If 1000 people per day embark on ships from Europe to America, 1000 people per day arrive in America. If 1000 people per day embark on planes, they get there faster, but still, 1000 people per day arrive. $\endgroup$
    – John Doty
    Dec 20, 2022 at 23:33
  • $\begingroup$ Okk sir, that analogy did explained the concept really well. Thanks alot ☺️ $\endgroup$ Dec 21, 2022 at 1:35

2 Answers 2


If you're generating, say, 10 photoelectrons per second, it doesn't matter how fast they move to the anode. As long as the anode collects them all, it'll receive 10 electrons per second.

  • $\begingroup$ But shouldn't this be true if all the electrons touch the plate at the same time, here electrons will collide the detector plate one after another, hence velocity should matter?? $\endgroup$ Dec 20, 2022 at 23:13
  • $\begingroup$ Is it like, I am in my mind exaggerating this " one after another" too much, and the distance between the first and last electron is very very negligibly small?? $\endgroup$ Dec 20, 2022 at 23:16
  • $\begingroup$ If the light intensity is constant, the photocathode emits electrons at random, generally not all at the same time. $\endgroup$
    – John Doty
    Dec 20, 2022 at 23:26

"when we increase the frequency of incident rays, current is unaffected, as the number of electrons is the same"

How do you suppose the frequency of the incident light is increased? Presumably by using a different light source. But there is no reason to suppose that this light source will eject electrons from the surface at the same rate as the previous one. It may send out a different number of photons per second and these photons, being more energetic, will be likely to eject more electrons per 1000000 photons..

If by some chance the number of electrons ejected per second is the same with the higher frequency source, then see John Doty's answer.

  • $\begingroup$ Tuneable light sources do exist. Although GY didn't explicitly specify, I think he meant changing the frequency of the light source while maintaining the same number of photons/second, i.e. increasing the energy/photon. $\endgroup$
    – AndyW
    Dec 21, 2022 at 17:53
  • $\begingroup$ @AndyW (a) I did consider tuneable light sources, but guessed that the variation of frequency possible would be too small. I might well be wrong here. (b) I think you're right about what GY meant, but I wanted to draw attention to the assumption not being a natural one to make. On balance, though, I'm not very proud of my 'answer'. $\endgroup$ Dec 21, 2022 at 18:08

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