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Thanks.

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Thanks.

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These two are very different phenomena.

Thanks.

The photoelectric effect requires the interaction of the WHOLE SOLID. In effect, the whole solid absorbed the energy and "promotes" the electron into the vacuum state.

The Compton scattering is the interaction with the electron itself, and doesn't require either the whole solid or the atom be a part of the interaction. In fact, it could occur with a free electron (photoelectric effect cannot happen in a free electron). So the photo must have a "direct collision" with the electron, which may explain why it is more likely to occur with high energy photons with more "well-defined", shorter wavelength.

Zz.

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sophiecentaur

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Thank you for your reply.

ZapperZ,

I understand that these are very different phenomena. Can you elaborate on what you mean by interaction of the whole solid - in terms of the atomic structure/latice structure? But are they mutually exclusive? If,say, I shine a beam of X-ray onto a metal, could I find BOTH 1 and 2 below:

1. X-ray photons completely passed its energy to an electron, freeing it

2. Compton scattering, with longer wavelengths photons detected (be the probability of this be however small because of the very small chance of direct hitting a bound electron)

If BOTH 1 and 2 could be present at the same time, what determines when the photon would pass part of its energy and when it pass its complete energy to an electron, assume the photon directly hit the electron in both cases.

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What is the energy gap of a free electron?

Zz.

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But why is light a particle?

1. No time delay in photoelectric emission

2. Increasing intensity of light has no effect, but frequency does

3. Classical resonance does not apply

4. In compton, the freq of the scattered light changes

5. Also, the light appears to be radiated in only one direction, classically it should be in all directions.

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jtbell

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Both processes are possible. Which one a particular photon produces is basically a random "choice." The probabilities are determined by the interaction cross-sections (or the related absorption coefficients) for photoelectric and Compton scattering, which depend on the photon energy and on the target material.But are they mutually exclusive? If,say, I shine a beam of X-ray onto a metal, could I find BOTH 1 and 2 below:

1. X-ray photons completely passed its energy to an electron, freeing it

2. Compton scattering, with longer wavelengths photons detected (be the probability of this be however small because of the very small chance of direct hitting a bound electron)

See here for a typical graph of the absorption coefficients as a function of energy. Click ahead to page 175 if necessary.

http://books.google.com/books?id=8V...toelectric and compton cross sections&f=false

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In this case the photons do transfer all their energy, the electrons have a range of energies because it is the difference between the photon energy and the ionization energy of the electrons, which varies.well, in the photoelectric effect, the photons dont transfer all their energy, which is why the liberated electrons have a range of energies.

increasing intensity increases the current2. Increasing intensity of light has no effect, but frequency does

Thanks jtbell, this clarifies very muchBoth processes are possible. Which one a particular photon produces is basically a random "choice." The probabilities are determined by the interaction cross-sections (or the related absorption coefficients) for photoelectric and Compton scattering, which depend on the photon energy and on the target material.

See here for a typical graph of the absorption coefficients as a function of energy. Click ahead to page 175 if necessary.

http://books.google.com/books?id=8V...toelectric and compton cross sections&f=false

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I never read that about the photoelectric effect. The wikipedia page also seems to suggest it is a one-to-one interaction. It says the interaction is between a photon and the "outermost electron". Either the whole energy is absorbed or the photon is reemitted. If part of the energy is absorbed and part emitted, it has another name: the compton effect.These two are very different phenomena.

The photoelectric effect requires the interaction of the WHOLE SOLID. In effect, the whole solid absorbed the energy and "promotes" the electron into the vacuum state.

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In a metal, what is the "outermost electron"? After all, the classical photoelectric effect is done on metallic surfaces! Furthermore, for the description of the photoelectric effect that I mentioned, what is contradicting the notion that the whole photon's energy is absorbed? I never contradicted that!I never read that about the photoelectric effect. The wikipedia page also seems to suggest it is a one-to-one interaction. It says the interaction is between a photon and the "outermost electron". Either the whole energy is absorbed or the photon is reemitted. If part of the energy is absorbed and part emitted, it has another name: the compton effect.

Wikipedia is confusing photoelectric effect with photoionization done on atoms/molecules. There is a reason why we give these phenomena two different names - they have subtle differences!

I can show you other places where Wikipedia got it not quite right.

Zz.

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In the photoelectric effect the whole photon is stopped in the metal. The max energy transferred to an electron equals the energy of the photon, it is misleading to say that the 'whole solid' absorbs the energy and promotes an electron. This was one of the original problems with the photo electric effect.... if the 'whole solid' absorbs the energy then it should take a long time before any single electron gained enough energy to be ejected. In fact there is almost no time delay between the absorbtion of a photon and the ejection of an electron.

I agree about Wikipedia....WHY is it taken as the first choice for an answer.... it explains nothing

I agree about Wikipedia....WHY is it taken as the first choice for an answer.... it explains nothing

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Think again. The affected electron is the CONDUCTION BAND ELECTRON. A conduction band can only exist due to the formation of a continuous band from not only the overlapping of all the individual atoms, but also the mean-field interaction of all the electrons. In other words, the entire solid.In the photoelectric effect the whole photon is stopped in the metal. The max energy transferred to an electron equals the energy of the photon, it is misleading to say that the 'whole solid' absorbs the energy and promotes an electron. This was one of the original problems with the photo electric effect.... if the 'whole solid' absorbs the energy then it should take a long time before any single electron gained enough energy to be ejected. In fact there is almost no time delay between the absorbtion of a photon and the ejection of an electron.

And when the electron is emitted, the lattice ions (not just ONE ion) has to take up the recoil momentum!

Do you want more? How about this? In the spectrum of the emitted photoelectrons, one can also detect the phonon effects of the solid via the broadening of the electron energy spectrum! It means that the electrons are decisively affected by the many-body effects of the solid!

http://arxiv.org/abs/cond-mat/9904449

Photoelectric effect involves a lot of the solid.

Zz.

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Zero. Just substitute [itex]\theta = 0[/itex] (forward scattering) in the Compton formula.What is the energy gap of a free electron?

Zz.

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But zero isn't "just finite", which is what I was asking in that post.Zero. Just substitute [itex]\theta = 0[/itex] (forward scattering) in the Compton formula.

Zz.

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I don't understand. What do you mean by "just finite", and where in your post did you ask this?But zero isn't "just finite", which is what I was asking in that post.

Zz.

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Look at post #3 that I was responding to.I don't understand. What do you mean by "just finite", and where in your post did you ask this?

Zz.

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Well, if you meant that there is no gap for Compton scattering, then you were right.Look at post #3 that I was responding to.

Zz.

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I am puzzled on why this is now my claim.Well, if you meant that there is no gap for Compton scattering, then you were right.

You note that I was responding to sophiecentaur's claim that there's a small gap for the free electron. So I asked what is this gap! Free electrons has a continuous energy state. In a metal, there is no gap that separates the "conduction band" to the "valence band". That's why

Somehow, you now think that I'm the one claiming that a free electron has some sort of a energy gap.

Zz.

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sophiecentaur

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What bands?

If you are talking about the "band structure", then you need to tell me which material you are referring to. If you are talking about the energy band structure of a typical, standard metal that we all deal with in the first 2 chapters of Ashcroft&Mermin, then I will ask again, "What gap"?

Zz.

- #21

sophiecentaur

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Remember it's all models.

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So instead you decided to contradict the prevailing solid state models and introduce your own model that has not been verified?

Remember it's all models.

Zz.

- #23

sophiecentaur

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Not at all. There is only a contradiction in your mind. I don't have access to that reference but I wouldn't mind betting that, somewhere along the line of the argument there.will be some integration. That makes an assumption of continuous variables. (I did my Maths Analysis course many years ago.)So instead you decided to contradict the prevailing solid state models and introduce your own model that has not been verified?

Zz.

There has to be an assumption that you can jump.from discrete to continuous at some stage i.e. when the numbers are big enough so that you can come up with an answer.

With whom could I be arguing about that- apart from you?

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But this isNot at all. There is only a contradiction in your mind. I don't have access to that reference but I wouldn't mind betting that, somewhere along the line of the argument there.will be some integration. That makes an assumption of continuous variables. (I did my Maths Analysis course many years ago.)

There has to be an assumption that you can jump.from discrete to continuous at some stage i.e. when the numbers are big enough so that you can come up with an answer.

With whom could I be arguing about that- apart from you?

The

Now, if you wish to argue that you can show that there is a "just finite" gap here, then I want an exact reference. It is that simple.

Zz.

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ZapperZ, have you ever heard of a work function of a metal?But this ismany-body physics! If you think you can, for example, derive the Fermi Liquid Theory via adding up such single interactions, then I want to see it.

Thedefinitionof a conductor here is that there isno gapat the Fermi level! Period!

Now, if you wish to argue that you can show that there is a "just finite" gap here, then I want an exact reference. It is that simple.

Zz.

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