Where do holes go after photoelectric effect?

[free_electron]

Once electrons are ejected via photoelectric effect, the eventual recombination of the holes they leave behind is not discussed much (maybe not even at all) in textbooks. Also important, where they recombine as well as how long after the e-h pair formation.

 

[DaveC426913]

I don't know, but my thinking is:

If the photelectric cell is not hooked up to a circuit then it would build up a charge; it will be ionized. Eventually, it will bleed off its charge as stray electrons fill the holes.

The idea behind a photoelectric cell is that you give the electrons a place to go - and come from - by hooking it up in a circuit, say, a lightbulb or simply a voltemeter.

 

[ZapperZ]

Once electrons are ejected via photoelectric effect, the eventual recombination of the holes they leave behind is not discussed much (maybe not even at all) in textbooks. Also important, where they recombine as well as how long after the e-h pair formation.

The reason why this is not discussed in ORDINARY photoelectric effect is because these are done on metals with highly mobile charge carriers. The electron reservoir is replienished so fast, this is not an issue that one has to be concerned with. Those of us who work in photoemission spectroscopy do not normally deals with such a thing unless we deal with more exotic material in which the electron-hole pair can actually contribute to the broadening of the quasiparticle lifetime. This do not play a major role in many cases, and certainly a non-issue in simple photoelectric effect.

Zz.

 

[free_electron]

Dear DaveC, Zz,

Thanks for your responses. I also agree a charge will build up that eventually goes away. If i had a thin oxide on the metal, and the oxide was absorbing radiation to eject the electrons, but the metal is hooked up in the circuit as mentioned, we can expect no charging problem, since there is a supply of electrons from the metal?

 

[free_electron]

e-h separation in dielectric

I now have a complication. What if I had a photon (e.g., 20-100 eV) whose energy greatly exceeded a dielectric's bandgap (e.g., 9 eV for SiO2, 5.7 eV for HfO2) incident on the dielectric surface. I expect some e-h pair generation, but what happens next? It seems the electron and hole are too energetic to bind to form an exciton. Yet if they separate too far how do they get back together? I assume the hole won't go very far, but would the electron, if ejected, turn around?

 

[Gokul43201]

Any of a host of things can happen, and many them will happen. First, there's simply the possibility of photoelectron emission if your excitation energy exceeds the binding energy (~band gap + electron affinity). Additionally, you can create hot electron-hole pairs, which in turn may be energetic enough to result in production of secondary electron-hole pairs, or in auger emission, or in making secondary excitons.

Since you typically do not shoot an individual photon at the sample, you usually create a large number of electron hole-pairs, so each electron does not have to recombine with its co-created hole.

 

[ZapperZ]

I now have a complication. What if I had a photon (e.g., 20-100 eV) whose energy greatly exceeded a dielectric's bandgap (e.g., 9 eV for SiO2, 5.7 eV for HfO2) incident on the dielectric surface. I expect some e-h pair generation, but what happens next? It seems the electron and hole are too energetic to bind to form an exciton. Yet if they separate too far how do they get back together? I assume the hole won't go very far, but would the electron, if ejected, turn around?

By definition, once an electron has gone beyond the work function of the material, even in a semiconductor, then the electron is now free and no longer are attracted to the parent material. So I don't understand your question.

In the case of exciton, the typical scenario is when the electron is excited into a state in the gap region of the semiconductor. The state is one of the "Rydberg" states for an electron with a positive central potential.

Maybe a search for the Spicer 3-step photoemission model might be useful.

Zz.

 

[free_electron]

Thanks Gokul and Zz (again) for your help.

I know it's a little silly to dwell on the picture of an individual photon exciting a single electron-hole pair. It is just for the simplistic thought experiment.

I'll check out the Spicer 3-step model. Thanks again.

 

[Modey3]

This is a good topic. My question is this. Is the work function of the material the maximum energy allowed in the Rydberg energy states of the electron-hole pair. I would imagine that the Rydberg energy state of the electron-hole pair is constantly changing according to statistical processes (similar to thermo fluctuations in kinetic processes) If the fluctuating energy exceeds the work function then the electron is emmited. Is this reasoning correct? Thanks

Best Regards

Modey3

 

[ZapperZ]

This is a good topic. My question is this. Is the work function of the material the maximum energy allowed in the Rydberg energy states of the electron-hole pair. I would imagine that the Rydberg energy state of the electron-hole pair is constantly changing according to statistical processes (similar to thermo fluctuations in kinetic processes) If the fluctuating energy exceeds the work function then the electron is emmited. Is this reasoning correct? Thanks

Best Regards

Modey3

You are now mixing up two different things under two different scenarios.

The exciton energy states have nothing to do with the energy states/band of the bulk material. If you create energy states of the exciton, it does not directly "obey" the work function of the material, because you have essentially an "isolated atom" in there. So the work function of the material really has no bearing because that system is a collective, many-body effect for the whole semiconductor.

Zz.

 

[Modey3]

Zapper,

Thanks for the reply. I would think that the electron needs to be disassociated from the exciton prior to ejection from the surface. Is there an energy associated with the disassociation that must be supplied prior to emission? If so wouldn't this could imply either two things: that there is a "cut-off" energy for the Rydberg energy states or the energy required to disassociate the exciton separates it without any intermediate transitions.

On a experimental note, it would be interesting to see how XPS peaks shift in quantum dots or wells compared to the bulk.

Best Regards

Modey3

 

[Gokul43201]

The exciton states can be thought of as a perturbation of the conduction electron states. If you supply the difference in energy between the conduction band minimum and the excitonic ground state (the n=1 Rydberg state for the exciton) to an exciton (essentially, this is like the "ionization potential" for the exciton), you then make an unbound, conduction electron (as well as a free hole). Additionally, if you supply what is usually called the electron affinity, you will get photoelectron emission.