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]

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]

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]

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.

 

[Michio Cuckoo]

well, in the photoelectric effect, the photons don't 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]

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.

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

 

[mike168]

well, in the photoelectric effect, the photons don't 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.

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

Thanks jtbell, this clarifies very much

 

[vin300]

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]

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.

 

[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]

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.

 

[Dickfore]

What is the energy gap of a free electron?

Zz.

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

 

[ZapperZ]

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.

 

[Dickfore]

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]

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.

 

[Dickfore]

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]

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]

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]

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]

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]

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]

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]

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.

 

[Dickfore]

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?

 

[sophiecentaur]

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.

Would you also say that "main body Physics" insists that a gas is a continuum because it uses the gas laws? Those gas laws came from the statistics of large numbers of discrete particles and we all live with that.
They are only models, remember.

 

[truesearch]

I would say that it is misleading to introduce 'work function' as part of the explanation of electrical conduction in metals.

 

[ZapperZ]

Would you also say that "main body Physics" insists that a gas is a continuum because it uses the gas laws? Those gas laws came from the statistics of large numbers of discrete particles and we all live with that.
They are only models, remember.

Then derive using your model the DOS for free electron gas and show that it has a "just finite gap" at the Fermi energy. So far, you have made claims with ZERO materials to support them.

Zz.

 

[sophiecentaur]

Why are you being so precious about this? The support for my statement is that calculus is used for that model of 'yours'.
Btw, do you know what the word 'calculus' means?

 

[Dickfore]

I would say that it is misleading to introduce 'work function' as part of the explanation of electrical conduction in metals.

I would say it isn't, since we're discussing photoelectric effect. Do you know Einstein's formula for the photoeffect?

 

[truesearch]

Yes I do and that equation describes the process of ejecting electrons from a metal, not conduction of electricity through a metal.
I wonder if SC is confusing the small ( but not zero) energy gap between donor or acceptor impurities and the conduction or valence band in semiconductors with the continuous conduction band in metals ?

 

[sophiecentaur]

There is no confusion. I should be grateful if you would point out why, for simple cases of just a few charges interacting, you get integer values of quantum numbers and, hence discrete 'levels'. And yet, increasing the number involved, changes that 'in principle'. Of course, you can't do the same sums for large quantities and naturally go to statistics and calculus to get an answer.
Tell me where the 'gear change' happens. Is it for 10 atoms, 1000, 100,000?
We are surely just looking at two ends of a range of situations. It would be daft to consider the actual value of energy gaps when a band model is more suitable just as it would be daft to assume a continuum for just a few atoms. we are not dealing with reality. We are applying the most convenient model to describe things. How is it more than that?

 

[Dickfore]

There is no confusion. I should be grateful if you would point out why, for simple cases of just a few charges interacting, you get integer values of quantum numbers and, hence discrete 'levels'. And yet, increasing the number involved, changes that 'in principle'. Of course, you can't do the same sums for large quantities and naturally go to statistics and calculus to get an answer.
Tell me where the 'gear change' happens. Is it for 10 atoms, 1000, 100,000?
We are surely just looking at two ends of a range of situations. It would be daft to consider the actual value of energy gaps when a band model is more suitable just as it would be daft to assume a continuum for just a few atoms. we are not dealing with reality. We are applying the most convenient model to describe things. How is it more than that?

Now I see that you had made a conceptual mistake. You claim that even a macroscopic body has a discrete spectrum, but the levels are very finely spaced.

While technically this is true, it is as useless as the idealization of an infinite crystal, which gives continuous bands exactly.

The problem comes when you try to define your system! Namely, you need to make the system isolated, so that it does not exchange particles nor energy with the environment, to really calculate energy eigenvalues. But, every system is essentially open. Because of the fine spacing between the levels, you can never claim that you isolated your system so well to be able to distinguish individual discrete eigenvalues.

Let us estimate the difference between subsequent energy levels. For this, assume a crysta lsystem with a cubic unit cell with side a. The total number of atoms in the system is N, so that $N = (L/a)^3$. The discrete values for the wave vector are:
$$k_{i} = \frac{2 \pi \, n_i}{L}, \ (i = x, y, z)$$
Assume the free electron model so that the energy eigenvalues are:
$$E = \frac{\hbar^2 \, k^2}{2 m}$$

Then, the uncertainty in energy due to a neighboring level is:
$$\Delta E = \frac{\hbar^2 \, \vert k_i \vert}{2 m} \, \Delta k_i$$
The order of magnitude for the wave vector corresponds to the Fermi wavevector, found from the total electron density n:
$$\frac{N_e}{a^3} = 2 \frac{1}{(2\pi)^3} \, \frac{4 \pi k^{3}_{F}}{3} = \frac{k^{3}_{F}}{3 \pi^2} \Rightarrow k_F = \frac{1}{a} \, \left( 3 \pi^2 \, N_e \right)^{\frac{1}{3}}$$
where $N_e$ is the number of electrons per atom.

The uncertainty in the wave vector is given by:
$$\Delta k_i = \frac{2 \pi}{L} = \frac{2 \pi}{a} \, N^{-\frac{1}{3}}$$

Then, we have:
$$\Delta E = \frac{\hbar^2}{2 m} \, \frac{1}{a} \, \left( 3 \pi^2 \, N_e \right)^{\frac{1}{3}} \, \frac{2 \pi}{a} \, N^{-\frac{1}{3}}$$
$$\Delta E = \frac{1}{4} \, (\frac{3}{\pi})^{1/3} \, \frac{h^2}{m \, a^2} \, \left( \frac{N_e}{N} \right)^{\frac{1}{3}}$$

For typical crystals $a \sim 5 \stackrel{o}{A}$. Take an Avogadro number of atoms, and one electron per atom ($N_e = 1$). We have:
$$\Delta E = 0.246 \times \frac{(6.626 \times 10^{-34})^{2}}{9.11 \times 10^{-31} \times (5 \times 10^{-10})^2} \times (6.022 \times 10^{23})^{\frac{-1}{3}} \, \mathrm{J} \times \frac{1 \, \mathrm{eV}}{1.602 \times 10^{-19} \, \mathrm{J}} = 2.9 \times 10^{-5} \, \mathrm{eV}$$

 

[sophiecentaur]

Now I see that you had made a conceptual mistake. You claim that even a macroscopic body has a discrete spectrum, but the levels are very finely spaced.

While technically this is true, it is as useless as the idealization of an infinite crystal, which gives continuous bands exactly.

The problem comes when you try to define your system! Namely, you need to make the system isolated, so that it does not exchange particles nor energy with the environment, to really calculate energy eigenvalues. But, every system is essentially open. Because of the fine spacing between the levels, you can never claim that you isolated your system so well to be able to distinguish individual discrete eigenvalues.

Thanks for that. I take it that this bit is the 'nub' of what you're saying. That makes excellent sense, the fact being that you only need to have your system in an environment to be 'stirring things up' enough to blur out any discreteness of levels that a simple model would suggest.
I was thinking that the 'line spreading' effect as you increase the pressure in a gas (due to the additional interactions) was all that counts but the system wouldn't, even then, be isolated.

Well, at least that's something!. In a real situation, there would be quite a small upper limit to the number of atoms for what I suggested to be actually true. You have accounted for the 'gear change'.

 

[vela]

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.

If you analyze the absorption of a photon by a free electron, you'll find you can't conserve both energy and momentum, so with Compton scattering, the scattered photon has to be there. With the photoelectric effect, the rest of the atom absorbs the extra momentum, so the electron can effectively take off with all the energy provided by the photon.