# Accumulation of electrons inside vacuum tube during photoelectric effect

I'm thinking of the general photoelectric effect experiment set-up.

We know that after certain value of potential difference between Cathode and Anode, the photocurrent of the circuit will reach plateau and stay near constant even when you increase the potential (due to saturation current of diode).

When the diode reach its saturation current, wouldn't electrons still be able to travel up to certain point of the Anode as long as the light is shined to the Cathode (Assuming Cathode hasn't reached enough net positive charge to emit more electrons)? What would happen to the electrons inside this vacuum tube then? Would it kind of sit around in some region? What would happen as more electron gets accumulated inside this tube?

• where is the diode in your diagram, do you mean the tube with the anode and cathode? why should it have a saturation current? Unless the vacuum stops being a vacuum? Commented Oct 10, 2022 at 5:51
• @annav Yes, I meant the anode and cathode. So from my researching, I got that increasing the potential difference doesn't increase the photocurrent because of saturation current (maybe this was wrong). Commented Oct 10, 2022 at 6:02
• as far as I know saturation currents are in semiconductors that have carriers both ways :en.wikipedia.org/wiki/Saturation_current Commented Oct 10, 2022 at 6:28
• @annav Okay, why does increasing the voltage not change the photocurrent from certain value? Commented Oct 10, 2022 at 6:30
• Are you sure? the effect is measured the opposite way "Finding the opposing voltage it took to stop all the electrons gave a measure of the maximum kinetic energy of the electrons in electron volts. " see the plot hyperphysics.phy-astr.gsu.edu/hbase/mod2.html . Also en.wikipedia.org/wiki/… . Commented Oct 10, 2022 at 10:32

You seem to mis-understand saturation current cause. Imagine you want a photocurrent to be about $$2~C/s$$, by raising a voltage even more, but given light intensity can knock-out from a photosensitive material only $$1~C/s$$. Where additional $$1~\text{Coulomb}$$ will come from ? So answer is simple, - voltage can only speed-up electrons at maximum rate which is the same rate at which light intensity can pluck electrons out of cathode.
• Also, there's a nice paper in nature communications journal which analyzes photocurrent switching effect and finds-out relation between a photocurrent and a light flux, which is : $$i = i_0\left( {1 - {\mathrm{{e}}}^{ - k\phi }} \right) .$$ Hope that helps. Commented Oct 10, 2022 at 10:59
• How will voltage increase number of electrons if light didn't supply them more from cathode? Remember, everything starts from the action of light,- $\phi_0$ flux of light say can hit $N_0$ electrons per second out of cathode- and that's it, you can't get more even you raise voltage to the sky. But raising light intensity from $\phi_0 \to \phi_1$, where $\phi_1 \gt \phi_0$, helps, because now light frees more electrons $N_1 \gt N_0$ per second which can be speed-up by applied voltage. Commented Oct 10, 2022 at 14:49