# Does the photoelectric effect obey Ohm's law?

So, I've been reading about the photoelectric effect for my modern physics class, and I was confused about how Ohm's law works in relation to it. Say we have a photoelectric apparatus that simply generates some current and then applies a stopping potential V to bring the current to zero. If we increase the intensity, this increases the current. However, then V = IR, and resistance remains the same, so shouldn't voltage increase? This is against the fact that intensity and stopping potential are independent, so does this mean that Ohm's law just fails in this case? Thanks so much.

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The physical scale of these two phenomena are very different. Ohm's law works on a scale where individual electron-atom interactions are something you average over in a statistical population. –  dmckee Jul 7 at 3:03
Thanks @dmckee, that also makes a lot of sense. –  nick Jul 7 at 3:28

does this mean that Ohm's law just fails in this case

Ohm's law is not universal. The ideal resistor circuit element is defined by Ohm's law but not all circuit elements obey Ohm's law; Ohm's law only applies to ohmic devices.

Physical resistors and conductors approximately obey Ohm's law but, for example, semiconductor diodes, transistors, thyristors, solar cells, vacuum tubes, batteries, thermistors, etc. etc. etc. do not.

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This is not the exact answer to the question @Alfred. The question asks the connection between ohm's law and photoelectric effect(if any). –  Shamayeta Jul 7 at 12:54
@Shamayeta, I've addressed the question as I believe appropriate. Clearly, the OP refers to an "apparatus" through which a current circulates thus there is a current variable $I$. In addition, there is a voltage $V$ across the photoelectric device. From this context, I gather that the OP is under the impression that Ohm's law is universal and thus, should relate the circuit variables $V$ and $I$. My answer goes precisely to what I believe is the OPs misconception in that context. –  Alfred Centauri Jul 7 at 13:17