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Apparently, the answer to this is that the saturation current is reached when all the ejected electrons are able to reach the anode. But if I keep the cathode illuminated with light of frequency above the threshold frequency, wouldn't more electrons reach the anode in a given time interval if I increase the voltage? Since the electrons are replenished in the cathode due to the circuit, electrons should continue to be ejected and the current should increase without bounds as I keep on increasing the voltage.

Does this have something to do with the experimental procedure? Like, is the cathode irradiated with pulses and not continuously illuminated?

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Akshit
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In a typical photoemission experiment one electron is emitted for every $10^5$ to $10^6$ photons that strike the metal surface. This ratio is called the quantum efficiency and its exact value depends on the metal. However in any one experiment the ratio can generally be taken as constant.

If you have a constant light intensity then the number of photons per second hitting the surface is constant, so the number of electrons emitted per second is constant. Let's call this number $N$. As you increase the voltage you capture a bigger and bigger fraction of these $N$ electrons, but the limit is when you capture them all in which case the current will be:

$$ I_\text{sat} = N\,e $$

because if the charge of each electron is $e$ and we have $N$ of them per second the total charge per second is $N\,e$.

This is the saturation current. The current can't get any bigger than this because $N$ is determined by the light intensity not the applied voltage.

John Rennie
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  • Suppose we had 10 V and 10^4 electrons were incident per second. Now we increase the voltage to 20. Then we will have more than 10^4 electrons incident on the cathode per second since they reach the cathode faster. – Akshit Sep 09 '16 at 10:16
  • @Akshit: when you double the voltage the electrons that are in transit between the metal surface and your collector will accelerate, so you get a brief spike in the current. But if the metal is emitting $10^4$ electrons per second then once steady state is attained the collector can't collect more than $10^4$ electrons per second because you can't summon up extra electrons from nowhere. – John Rennie Sep 09 '16 at 10:19
  • The circuit's complete, right? So those 10^4 electrons moving out of the cathode per second can be replenished by the electrons which come back into the cathode through the circuit – Akshit Sep 09 '16 at 10:20
  • Yes, but the limiting factor is the light intensity. No more than $10^4$ electrons per second can be emitted from the metal surface unless you increase the light intensity. – John Rennie Sep 09 '16 at 10:22
  • Look at it like this: suppose the electrons are ejected with velocity 1 m/s. Then at 10 V, in one second, 10^4 electrons reach the cathode. When the voltage is 20 V, it takes less than 1 second for the 10^4 electrons to reach the cathode, which makes the current larger than before. – Akshit Sep 09 '16 at 10:30
  • Am I missing something here? Does this have something to do with the steady state you described? – Akshit Sep 09 '16 at 10:30
  • @Akshit: if $10^4$ electrons per second are emitted then on average an electron is emitted every 0.1ms. That electron takes a time $t$ to reach the collector so there is a time lag between the electron being emitted and it being received at the collector. However we still have an electron emitted at the metal every 0.1ms and an electron received at the collector every 0.1ms. If you increase the voltage you decrease the time taken to reach the collector, but this does is reduce the lag between emission and collection. You still have an emission every 0.1ms and a collection every 0.1ms. – John Rennie Sep 09 '16 at 10:40
  • So what I've understood is this: 1. To have a saturation current, we need to have discrete packets like photons and not continuous like in the classical theory, and 2. On doubling the intensity, the saturation current doubles. – Akshit Sep 10 '16 at 01:51
  • Your point (2) is certainly correct. I'd have to think about point (1). – John Rennie Sep 10 '16 at 04:32
  • If you have $N$ electrons, each with charge $e$, there is a current on the LHS but a charge on the RHS of your equation. – Jasper Oct 17 '18 at 20:58