Photoelectric effect

[My name aint bob]

hey everyone a quick question, can a photoelectric current flow without a potential difference of any kind provided from a battery? with light still shining on a material releasing electrons though?

[Lasercookie]

From my understanding, yes. Electrons will still be ejected and yes we will still have a photoelectric current.

The battery in the photoelectric effect is there to provide a charge to the anode/cathode (I can't remember which is which.... I probably should know the difference between the two by now). We can give a positive charge to the metal plate that's on the other side (the one that we're not shining light upon) and that will attract the ejected electrons towards it and complete the circuit etc.

If you gave that other metal plate a negative charge, then it would repel the ejected electrons (and might stop the current all together).

But if we didn't have that battery, and hence just two metal plates, the vacuum and so on connected in a circuit really, would the ejected electrons have enough energy to make it to the other side?

Perhaps a very rough way of showing it is this:

We know that stopping voltage will always be negative, since if we were applying a positive voltage we wouldn't be slowing down the electrons, we'd be speeding them up. The stopping voltage is such that it has a sufficient energy to stop the most energetic electrons from reaching the other metal plate.

Note that negatives here represent the direction of the battery and hence current.





So will stopping voltage ever be zero (e.g. no battery)? Yes, if the energy of the electrons was zero. That would only be the case if we didn't have electrons being ejected. Removing the battery won't affect that, since we still have the light being shone on it and that's what provides the energy to eject the electron. Note that that light has to be of sufficient energy in the first place.

That didn't really show that the electrons would still have enough energy to reach the other side though. My way of understanding that it would is that it's in a vacuum, and if the voltage of the battery is zero then there are no opposing forces acting upon that electron and hence by Newton's first law / inertia it would keep travelling until it did eventually reach the other side. I think that's kind of flawed though, because I'm not sure if we should be thinking about the motion of electrons in terms of Newton's laws. 

I think this is a better approach of convincing you. You could also consider what if your premise is true, that a zero voltage across the plates was enough to stop a current being registered, what would this imply? It'd imply that the stopping voltage would be equal to zero. You know this isn't true. The stopping voltages you're given aren't usually zero. Sometimes it's -2 V, for example. This means that if you applied -1V we'd still have some electrons hitting the other plate and hence a current registered. You could take a look at those graphs of stopping voltage vs current here too, everybody loves graphs.

Have a a source: "If one starts with zero voltage across the plates, then some current will flow when the light is turned on". That was just some random result from google (this one: http://spiff.rit.edu/classes/phys314/lectures/photoe/photoe.html), you could search for more too.

Take a look at that link, the most interesting thing about that page is that it includes diagrams from Lenard's original paper, which is pretty cool. That page doesn't seem to be a translation of that original paper though, might be able to find one elsewhere. I'm rambling too much though, that was irrelevant to what you were asking. I'll stop here then.