We all know that the amount of current flowing b/w the plates is independent of the frequency. If I were to ask why doesn't it depend then you'd probably say that it depends on the number of electrons rather than the frequency or you might show me a current vs frequency graph.

But here is my theoretical argument. Electric current is defined as the amount of charge flowing through a point/space/wire per unit time. Let I be current due to photo electrons ejected by photons of some frequency. Now if I would increase the frequency, the kinetic energy of the ejected electron would increases, therefore it will move faster and therefore contribute more current (charge will through the same point many times per unit time) . Thereby, giving a net current greater than I.

What is wrong with my argument?

After thinking a lot, I came up with a possible reason. However fast the electrons may move, but when they hit the metal plate, they would lose their velocity and start moving in drift velocity. Now there is a problem with this argument. If this were to be true then I should expect to see a big chunk of extra electrons getting stuck at the anode.

If my reasoning was correct, then why can't I see a big charge buildup at the anode.

  • $\begingroup$ If the intensity of light is kept constant then the statement "We all know that the amount of current flowing b/w the plates is independent of the frequency" is possibly not correct. $\endgroup$
    – Farcher
    Commented Nov 15, 2018 at 10:29

2 Answers 2


One way to look at current is "the total number of electrons passing a particular plane per unit time, multiplied by their charge".

How fast they are going doesn't matter - if they are going faster, they will appear to be further apart.

A given amount of light (above the critical frequency) will knock a given number of photo-electrons into space. That is the number of electrons that flows - and regardless of their velocity, they will give rise to the same current.

If you had a constant space density of charge, then making that cloud of electrons move faster would increase the current. But that is not what you have here - you have a fixed number, not a fixed density.

Imagine cars sitting in a traffic jam. Perhaps the four lane highway reduced to a single lane. And let's imagine one car per second passes a given point. Maybe the car is doing 10 km/h. Now we look five km "downstream". The road is 4 lanes wide again, and cars are going 120 km/h. How many cars per second do you see passing you? Of course it is one car per second - that's the number that was going through the narrow point. So although the cars are now going much faster, the road is still transporting the same number of cars. The cars are much further apart. If somehow everyone in the traffic jam (with the cars bumper-to-bumper) figured out how to drive really fast at the same time, the number of cars per second (the "current") would be much greater.

  • 1
    $\begingroup$ You want to tell that however fast the electrons (like the 4 lane road - between the plates) go they will ultimately end up waiting near the anode (like traffic jam in single lane? ). $\endgroup$
    – Yashas
    Commented Jul 8, 2015 at 9:23
  • $\begingroup$ @YashasSamaga - yes that's exactly it. Or even better than a traffic jam, like a "metered ramp" that you see on some highways - the light is green just long enough to let one car (electron) at a time on to the highway (you see those here in the US some times to help keep traffic flowing on the highway). The photons that knock electrons loose are like the traffic light, letting only so may electrons per second participate in conduction. Of course when electrons "bunch up" they repel each other so they will ultimately not all sit "waiting near the anode". $\endgroup$
    – Floris
    Commented Jul 8, 2015 at 10:31
  • $\begingroup$ Then wouldn't charge start piling up at the anode? I would either expect the charge build up and go to infinity as time passes or it is going to be a infinite sum which will converge at some number. And this charge buildup would produce an electric field which would oppose the photo-electrons. (Assuming that the metal plate is infinity large that it won't run out of electrons - maybe this is a bad assumption) $\endgroup$
    – Yashas
    Commented Jul 8, 2015 at 15:52
  • $\begingroup$ What is the force that makes the electrons move? It is the electric field. Any momentum of the electrons (because energy of photons was greater than work function) will not result in polarization - only in (a tiny bit of) heating. Electrons repel each other - they move under the influence of a field. After flying across the gap they will lose their energy the moment they hit the anode. No bunching, no "running out of electrons". Your intuition is off in this one, I'm afraid. $\endgroup$
    – Floris
    Commented Jul 8, 2015 at 18:30
  • $\begingroup$ Thanks, I understood, the momentum of the electrons are lost when they hit the plates, therefore the velocity won't matter. Oops, the electrons would come back because the plates are connected by wire. But if light was just flashed on some metal plate, then after some time we start seeing less number of electrons. $\endgroup$
    – Yashas
    Commented Jul 9, 2015 at 0:34

I think that it depends upon the situation.

See if you talk about saturation photocurrent, then NO, the photocurrent will not depend upon the frequency because all the photoelectrons that are emitted do reach the other plate, and contribute into photocurrent, any increase in frequency will not change the number of photoelectrons reaching the other plate (becuase all the possible photoelectrons at that intensity have already been emitted), so photocurrent will not change.

But if it is not about saturation photocurrent, when a photon strikes and emits a photoelectron, the photoelectron may lose it's Kinetic energy due to collisions and even though it was removed, it may not be able to reach the other plate and will not contribute to the photocurrent. In this case increasing the frequency may increase the chances of the photoelectron to come out of the collisions and reach the other plate.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.