The Photoelectric Effect question

[ZapperZ]

I'm going to accept your statement that the simple form of the Schroedinger equation we learned in undergrad school is inadequate to solve for the band structure in a metal. Sometimes I use the phrase "the Schroedinger equation" as a shorthand for "standard qm calculations of the electron wave function". I think this is how they solve for band structures.

What you CAN do at the undergrad level is solve for some simple physical cases like the potential well or the hydrogen atom. Both these cases show clearly how the different energy levels have different frequencies. And when you take superpositions of states with different frequencies, the resultant charge density sometimes oscillates, giving you a classical antenna. This is certainly true for such combinations as the 1s-2p state of the hydrogen atom. Maybe you've seen the applets people have made up of this and posted on the internet? It's also true for some superpositions of bound states with free states.
You can construct some pretty good ones with the potential well. Actually, in the one I'm thinking of, you don't necessarily get oscillating charge; you get surface charge waves traveling faster than the speed of light. This of course is also a valid classical antenna; it's the same kind of charge waves you get when ordinary light falls on a metal sheet at an angle.

These theoretical cases show exactly the type of behavior characteristic of the photo-electric effect experiments. Below the threshold frequency, there is no coupling between the bound states and the free states. Above the theshold frequency, the superposition of these states interacts strongly with radiation. So one state can be driven into the other state.

This is wrong. A hydrogen atom has no "work function". It has a ionization potential, but it doesn't have a continuous band that a metal has! The study of solid is not identical to the study of atom and molecules. That's why we have "solid state physics".

You have just shown here that you can't derive the work function. Do you still claim to be in possession of the ability to arrive at the photoelectric effect using QM and classical wave theory? If you do, let's see it explicitly.

Zz.

 

[monish]

You have just shown here that you can't derive the work function.

That's not what I was trying to show. It was you who said that it couldn't be derived from Schroedinger's equation, and I said I was willing to accept your assertion. I don't know if it's true; I'm pretty sure the work function comes out, as I said, from a pretty conventional q-m analysis. I accepted your claim because I thought I could show how the frequency threshold arises naturally from Schroedinger's equation without needing to do an exact calculation of the work function. That's why I used easy examples that would be accessible at the undergrad level.

Do you still claim to be in possession of the ability to arrive at the photoelectric effect using QM and classical wave theory? If you do, let's see it explicitly.

Zz.

I honestly think I did a pretty decent job of explaining the frequency threshold. Which was the main objection to the wave theory that has been raised so far. Is there any other aspect of the experiment you'd like me to try to explain?

Marty

 

[ZapperZ]

I honestly think I did a pretty decent job of explaining the frequency threshold. Which was the main objection to the wave theory that has been raised so far.

You did? I re-read what you wrote and no where is there ANY explanation of it. You used the H-atom, which isn't even close to being anywhere near to even mimic the work function or the photoelectric effect spectrum. Somehow, the fact that the conduction band is continuous, while the H-spectrum is discrete doesn't bother you at all?

I don't know how you could convince yourself that what you are doing here has any agreement with the photoelectric effect.

Zz.

 

[monish]

You did? I re-read what you wrote and no where is there ANY explanation of it. You used the H-atom, which isn't even close to being anywhere near to even mimic the work function or the photoelectric effect spectrum...I don't know how you could convince yourself that what you are doing here has any agreement with the photoelectric effect.

Well, the analogue of the work function in this case would be what I think is called the ionization potential. In the photoionization of hydrogen, you need the minimum frequency before you can drive the transition. In the photo-electric effect, similar idea.

I don't know what you mean about explaining the "photoelectric effect spectrum". If you'd care to clarify this I can try to respond.

Somehow, the fact that the conduction band is continuous, while the H-spectrum is discrete doesn't bother you at all?

Zz.

No, it never bothered me until you mentioned it just now. I'm not sure I see how this makes an essential difference in the relevant physics. In both cases, were coupling from a bound state to a free state. I just find the hydrogen atom (and the potential well) easier to calculate explicitly because the bound state wave function is readily described.

If you can explain why the photo-electric effect is qualitatively different, I'd be interested.

 

[ZapperZ]

Well, the analogue of the work function in this case would be what I think is called the ionization potential. In the photoionization of hydrogen, you need the minimum frequency before you can drive the transition. In the photo-electric effect, similar idea.

I don't know what you mean about explaining the "photoelectric effect spectrum". If you'd care to clarify this I can try to respond.

No, it never bothered me until you mentioned it just now. I'm not sure I see how this makes an essential difference in the relevant physics. In both cases, were coupling from a bound state to a free state. I just find the hydrogen atom (and the potential well) easier to calculate explicitly because the bound state wave function is readily described.

If you can explain why the photo-electric effect is qualitatively different, I'd be interested.

You know what the photoelectric effect is, don't you? Now how are you able to derive QUANTITATIVELY all of that have been observed?

For example, if you look at the energy spectrum of the photoelectrons, you'll see a continuous spectrum of energy. You don't see that in an H-atom spectrum. Not only that, take carbon atoms. Arrange it one way, you get one value of the work function, but arrange it another way, you get ANOTHER diferent value of the "work function". Why? The crystal structure dictates how the valence shell overlaps and by how much. Yet, these are still carbon atoms forming two different material, diamond and graphite. In your example, you'll see discrete spectrum of only ONE type, no matter how you arrange them, because all you care is the isolated energy spectrum. It just doesn't fit the experiment.

And oh, btw, my avatar is one such example of a photoemission spectrum that included both the energy (horizontal axis) and momentum (vertical axis) distribution of the photoelectrons. I can assure you that you DO NOT get this when you look at the ionization spectrum of isolated atoms.

If you think you can derive all of the photoelectric effect using simple QM and wave theory, please submit it for consideration for publication, or to the IR forum. I believe I have seen enough of this to qualify it as speculative, unverified personal theory, which is a violation of our guidelines. If the OP has any followup, he/she may contact me to have this thread reopened.

Zz.