Photoelectric effect vs Compton scattering

mike168

I understand that in photoelectric effect, the energy of the whole photon is absorbed, freeing an electron. I don't understand why in the case of Compton scattering, the higher energy photon lost part of its energy instead of transferring the whole of its energy to the electron as in photoelectric effect? Does it mean that there is a limit that the electron could absorb so much energy and no more?
Thanks.

ZapperZ

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.

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.

sophiecentaur

For a free electron, there is a very small energy gap (only 'just' finite) between its states so any photon can interact with it. Even low frequency Radio Waves can interact with free electrons (in the ionosphere, for instance) but we usually analyse that effect in terms of how the electrons are affected by the Field rather than the Photons. There must (?) be an alternative way to approach it, though.

mike168

Hi guys
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.

ZapperZ

What is the energy gap of a free electron?

Zz.

Michio Cuckoo

well, in the photoelectric effect, the photons dont transfer all their energy, which is why the liberated electrons have a range of energies.

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.

jtbell

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.

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=8Vu...ctions&f=false

mike168

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.
increasing intensity increases the current

Thanks jtbell, this clarifies very much

vin300

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.

ZapperZ

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!

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.

truesearch

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

ZapperZ

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.

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.

Dickfore

Zero. Just substitute [itex]\theta = 0[/itex] (forward scattering) in the Compton formula.

ZapperZ

But zero isn't "just finite", which is what I was asking in that post.

Zz.

Dickfore

I don't understand. What do you mean by "just finite", and where in your post did you ask this?

ZapperZ

Look at post #3 that I was responding to.

Zz.

Dickfore

Well, if you meant that there is no gap for Compton scattering, then you were right.