# Photoelectric effect vs Compton scattering

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.

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ZapperZ
Staff Emeritus
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.
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
Gold Member
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.

Hi guys
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
Staff Emeritus
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.
What is the energy gap of a free electron?

Zz.

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
Mentor
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)
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.

well, in the photoelectric effect, the photons dont transfer all their energy, which is why the liberated electrons have a range of energies.
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.
2. Increasing intensity of light has no effect, but frequency does
increasing intensity increases the current

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.

Thanks jtbell, this clarifies very much

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.
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
Staff Emeritus
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.
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.

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

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ZapperZ
Staff Emeritus
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.
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.

What is the energy gap of a free electron?

Zz.
Zero. Just substitute $\theta = 0$ (forward scattering) in the Compton formula.

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

Zz.

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

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

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

Zz.

Look at post #3 that I was responding to.

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

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

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 I asked for the nature of this gap.

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

Zz.

sophiecentaur
Gold Member
I only said small gap because someone would have pointed out that the band does not actually have zero gaps - there are just a huge number of quantum numbers involved - effectively a continuum.

ZapperZ
Staff Emeritus
I only said small gap because someone would have pointed out that the band does not actually have zero gaps - there are just a huge number of quantum numbers involved - effectively a continuum.
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.

sophiecentaur
Gold Member
This is getting philosophical. My point is that a finite number of particles will have a finite set of interactions. If somebody's model introduces an integral rather than a sum then that' s fair enough and it's clearly the sensible approach. But if two copper atoms have discrete levels then so do 2e23 atoms. If no, when do you make the switch.
Remember it's all models.

ZapperZ
Staff Emeritus
This is getting philosophical. My point is that a finite number of particles will have a finite set of interactions. If somebody's model introduces an integral rather than a sum then that' s fair enough and it's clearly the sensible approach. But if two copper atoms have discrete levels then so do 2e23 atoms. If no, when do you make the switch.
Remember it's all models.
So instead you decided to contradict the prevailing solid state models and introduce your own model that has not been verified?

Zz.

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

Zz.
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.)
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?

ZapperZ
Staff Emeritus
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.)
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?
But this is many-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.

The definition of a conductor here is that there is no gap at 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.

But this is many-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.

The definition of a conductor here is that there is no gap at 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.
ZapperZ, have you ever heard of a work function of a metal?