# Photoelectric effect question

1. Jul 13, 2014

When a fixed intensity and frequency Em radiation strike a metal plate , is the outgoing electron/s with a higher energy if the plate is charged to some potential (volts) than if the plate is simply neutral?

2. Jul 13, 2014

### Staff: Mentor

Yes. If you charge the plate with a negative voltage the electrons will be accelerated to a higher velocity and have a higher kinetic energy than if it is neutral.

3. Jul 14, 2014

so if my photon bombardment is big enough in intensity and powerful enough in frequency then the outgoing current of electrons of a negatively charged plate would have the eV of the original charge the plate had + the energy of the incoming photon?

4. Jul 14, 2014

### Staff: Mentor

The intensity and frequency don't matter (not much at least, as long as the frequency is high enough to eject an electron). A single photon will eject a single electron, whose energy is the energy of the photon plus any extra energy from the charged plate, minus the work function (the energy needed to remove the electron from the metal).

5. Jul 14, 2014

### sophiecentaur

The frequency of the photons will affect the speed (kinetic energy) of the ejected electrons and that can be determined by measuring the so-called Stopping Potential (look it up) of the photoelectrons. The graphs in this link show how any energy beyond the threshold (work function) of the metal goes into the KE of the emitted photo electron. It is always worth while noting that the graph shows the Maximum energy. There will be electrons with lower energies, too (You could imagine some of them having been 'buried' a small distance below the surface, although this a slightly naive model)

6. Jul 14, 2014

What would happen if I had two plates with a Potential difference across them set in vacuum, with some distance between the plates so that electric breakdown doesnt occur.
If I would shine a laser on one of the plates , and the laser frequency would be high enough to cause significant photoelectron emmision from the charged plates or from one of them , would there now be a current between the plates and the Potential difference would gte short circuited?

Also , if we take a isolated plate , how long can the electron be emitted from the plate , even though there are huge amount of electrons in a small metal plate doesnt the plate run out of them after some time of continious emission?

7. Jul 14, 2014

### sophiecentaur

If the 'other' plate were held at a positive potential with respect to the emitting plate, a current would flow (one electron per photon. The electrons would have their original KE plus the eV due to the acceleration across the plates.
If the other plate were at a high enough negative potential, no electrons would reach it.

How much have you read about this topic, apart from what we have been telling you? There are hundreds of Google hits and there has to be one which will be at a level appropriate to your present knowledge. Simple Q and A is not always the best way to acquire knowledge.

8. Jul 14, 2014

### Staff: Mentor

Of course. I meant that the intensity and frequency don't matter in response to Salvador's 2nd post. As long as the frequency is enough to eject an electron, you will have the effect he mentions.

9. Jul 14, 2014

The internet does have alot about the photoelectric effect but not so much about the specific maters that i'm into.
I do understand the effect itself though.

another question , if I charged the vacuum seperated plates to say 10kv, and then used a very low intensity but high frequency laser on the positive plate , would the current between the plates would be proportional to the intensity of the incoming photons ? or would there be a current runaway once a current has formed between the plates?

I do understand about frequency and the kinetic energy of the outgoing electron which in the case of the charged plate would be the incoming KE of the photon + the potential of the electron from the plate in eV right ?

10. Jul 14, 2014

### UltrafastPED

Your laser should strike the negatively charged plate; then when an electron is ejected it will be swept away to the positive plate. The energy of the electrons will be 10 keV plus any excess kinetic energy from the photo-emission.

Usually you would match the material of anode with the photon energy (as determined by the wavelength of the laser light) so that each photon is sufficient to overcome the work function of the metal, but no greater.

So for example, the work function for gold is just over 4 eV, so laser light in the deep UV, around 260 nm (=4.5 eV) would work well. For a 780 nm wavelength laser such as the Ti-sapphire ultrafast lasers, this would require a set of BBO crystals; the first for doubling from 780 to 390 nm (1.5 eV -> 3.0 eV), and then a second crystal to shift from 390 to 260 nm (3.0 eV -> 4.5 eV).

When you say "high frequency laser" I'm assuming you mean a high-repetition rate pulsed laser. Then as each pulse liberates some electrons, they will be swept away from the cathode before they can build up and interfere with the next set of electrons. This requires some careful analysis of electron pulse propagation, as well as proper photo-electron gun design.

OK ... to answer your question: yes, the current will be proportional to the laser intensity, as long as the intensity is not too high; if too high you will initiate non-linear effects, such as multi-photon absorption, or melting/vaporization of the target.

11. Jul 14, 2014

but if the laser radiation causes the photoelectron emission and forms a small current , shouldn't the current get bigger because the plates have a PD across them and now because of the photoelectron current the charges on the plates have a path through whioch they can discharge, so shouldn't the current then be the photoelectric current + whatever current the power supply of the plates can generate ?

12. Jul 14, 2014

### Staff: Mentor

No, as the electrons will only be excited enough to cross the gap between plates by the laser photons. So the current will be equal to the number of photons hitting the negative plate. Increasing the voltage between the plates only increases the energy each electron has, it doesn't increase the number of them.

13. Jul 14, 2014

Ok so if I have 10kv across the plates and a power supply capable of 10kv 1amp then how big of a current will flow between the plates if I switch on the laser and have the power supply already switched on to the plates?

14. Jul 14, 2014

### sophiecentaur

You have to ask yourself "where will all these extra electrons come from, for the extra current?"
Unless the PD between the plates is high enough to tear electrons and ions from the surface of the metal, there will only be the number of electrons equal to the number of photons arriving.
The breakdown voltage between two plates in a good vacuum is many tens of kV but if you put enough volts across the gap, you can get an arc and the photoelectric effect would not be relevant.

It is a shame (from the experimenter's point of view, only) that so few metals will yield photoelectrons for visible wavelengths. It only works for alkali metals, I believe and they are so reactive that you need to do the experiment in a vacuum with a recently cleaned metal surface.

15. Jul 14, 2014

### Staff: Mentor

That depends entirely on the intensity of the laser. The 10kv power supply can support a current UP TO 1 amp, but it is very unlikely that your laser will be ejecting that many electrons.

Think about it this way. If you turn off the laser, there is no path between the plates for current to flow, so you have an open. Turning on the laser ejects UP TO one electron per absorbed photon, and it is these electrons that form the current flow. Whether the plates are charged to 10 volts or 10 kv, doesn't matter since the voltage doesn't eject electrons.

16. Jul 15, 2014

I understand that the voltage is not the cause for electron flow in the photoelectric effect , I was rather thinking that once the photoelectrons form the current all the other electrons now can use that current to get to the other plate what they would normally want to do but seems like this is not the case.

17. Jul 15, 2014

### Staff: Mentor

No, that is indeed not the case. The vacuum between the plates is an excellent insulator, so each individual electron must be excited out of the metal by a photon. They cannot use other electrons as a conduction path. In fact, the electrons would resist this since they are all negatively charged.

18. Jul 15, 2014

### sophiecentaur

Go along with their mates, you mean?
Fact is that, for any electron to leave the surface, it needs a photon to supply the work function energy. If you think (classically and very approximately) of the sort of field that would be needed by considering the, say 4eV work function of an alkali metal. This involves an equivalent Potential of 4V across a very thin layer (say, a micron), which is a field of 4MV/metre. This is the basis of QM; Energy exists in peaks (quanta) and not a continuum, which took a lot of accepting when the idea first came up. (And it still does!)

Last edited: Jul 15, 2014
19. Jul 15, 2014

### Chronos

Last edited: Jul 15, 2014
20. Jul 15, 2014

### Staff: Mentor

Sure, but this requires an extremely large voltage between the plates. Much higher than the example given, and much higher than what would ever be used in a photoelectric experiment.