# Compton Effect vs. Photoelectric effect

1. Sep 24, 2010

### jordan41590

Ok I'm doing some gamma ray spectroscopy and finding that the two main effects occurring are Photoelectric and Compton. I want to fully hash out similarities and differences for these two effects so I know exactly what's going on. As a starting point, I know that in the photoelectric effect the photon's energy is fully absorbed, ejecting an electron and ultimately emitting a new photon when another electron drops down, whereas with compton the first photon scatters. But is the main difference the amount of energy lost by the photon, or is there a more fundamental difference?

2. Sep 24, 2010

### tom.stoer

The main difference is that the photoelectric effect deals with electrons bound in matter, whereas the Compton effect deals with free electrons. So in the photoelectric effect the matter can "absorb" energy and momentum whereas in the Compton effect only two particles are involved.

If the photon has energy and momentum (E, p) = (E, E/c) you can't explain why you observe electrons with (E-W, -k) w/o taking into account something that absorbs energy and momentum in order to guarantuee energy-momentum conservation.

Microscopically both effects are identical (electron-photon scattering), but whereas in the Compton effect there is only this fundamental process, in the photoelectric effect additional the the additonal interactions must not be neglected.

3. Sep 24, 2010

### zonde

It seems that you have mixed up http://en.wikipedia.org/wiki/Compton_scattering" [Broken].

Last edited by a moderator: May 4, 2017
4. Sep 24, 2010

### tom.stoer

No. Thompson scattering is elastic scattering of classical electromagnetic radiation,
whereas both the photoelectric effect and the Compton effect and quantum mechanical processes. Thompson scattering is the classical limit of Compton scattering

Last edited by a moderator: May 4, 2017
5. Sep 24, 2010

### zonde

Compton effect deals with electrons bound in matter - that's what I'm saying.
You was saying it deals with free electrons.
"Part of the energy of the X/gamma ray is transferred to a scattering electron, which recoils and is ejected from its atom (which becomes ionized), and the rest of the energy is taken by the scattered, "degraded" photon."

6. Sep 24, 2010

### Staff: Mentor

I would amend tom.stoer's statement to read that "the Compton effect deals with almost-free electrons" or better, "the Compton effect deals with electrons whose binding energies are much smaller than the incident photon energy."

As you probably know, the basic analysis of the Compton effect assumes that the electron is "free" initially. This is of course not actually true for an electron bound in an atom! Nevertheless, the binding energy of the "outermost" electrons is on the order of a few eV, whereas for Compton scattering we use at least X-ray photons (a few tens of keV) or gamma-ray photons (hundreds of keV or even more). In those situations the electrons are "approximately free" as far as the incoming photons are concerned.

7. Sep 24, 2010

### tom.stoer

exactly

8. Sep 24, 2010

### Bob S

Neodymium-YAG laser photon beams at SLAC have been scattered off of "free" electrons in 20-GeV electron beams to produce Compton-backscattered multi-GeV photons. So there is no requirement that the electrons be bound. Because the recoil electron is free, the incident photon has to have significantly more energy than the target electron binding energy. See

http://www.slac.stanford.edu/pubs/beamline/25/3/25-3-melissinos.pdf

Bob S

9. Sep 24, 2010

### tom.stoer

Wikipedia can be misleading or sometimes even wrong.

Compton scattering deals with free or quasi-free electrons. It means the QED scattering process where one single photon interacts with one single electron. Whether the electrons are bound in atoms, molecules or metal doesn't matter, as long as the binding energy is small compared to the photon's energy. Please have a look at the calculation of the scattering angle, cross section etc. You never use anything else but one incoming photon and one incoming electron, you never calculate final state interactions of the electron with an atom or ion or something else.

The only reason that you use electrons bound in atoms is that it's difficult to have enough free electrons sitting together and waiting for the next photon to come ...

10. Sep 25, 2010

### zonde

Tanks a lot for your clarification. It started to feel like we are going go in circles.

So would it be more correct to say that the main difference between photoelectric effect and Compton effect is difference in energy scale?

11. Sep 25, 2010

### zonde

Question was about Compton scattering but this is Compton-backscattering or inverse Compton scattering.
Why do you want to mix them together?

12. Sep 25, 2010

### tom.stoer

No, not really.

If you have a free electron then the Compton scattering formulas apply for arbitrary small energies. But if an electron is bound in matter and if binding energy and photon energy are comparable, then one has to take binding effects into account.

13. Sep 25, 2010

### Bob S

NdYAG laser photon scattering off of high energy electron beams is regular Compton scattering of photons off of free electrons, just Lorentz-transformed from the (free) electron rest frame into the laboratory rest frame. A Compton backscatter peak is observed with Cs137 gamma rays in sodium iodide, for example.

The photoelectric effect (=deep core photoejection) requires that all of the incident photon energy be absorbed by the ejected electron, which in turn requires that the electron be bound, not free. The photoelectric effect cross section (per nucleus) varies as ~Z4/E3 where E is the photon energy, so it drops off very fast at energies above the atomic K-edge energy. The Compton scattering (also called Klein-Nishina) cross section (per nucleus) is roughly proportional to ~Z/E at high energies. Compton scattering may (but is not required to) occur off of lightly-bound electrons.

Bob S

14. Sep 25, 2010

### ZapperZ

Staff Emeritus
You continue to make this mistake after being repeatedly corrected.

The standard photoelectric effect is not equal to core level photoemission. You need to get out of this mindset, because you are giving incorrect information. The standard photoelectric effect (such as the one done by Millikan and millions others in undergraduate physics labs) are done with typically a discharge tube with photons of a few eV, and using metals as the photocathode. This means that the photoelectrons are from the conduction band, not the core level.

If you do a "photoelectric effect" experiment on core level, you will not get the Einstein relation. This is why this is NOT the standard photoelectric effect. If you want to talk about core level photoemission, then wait until that subject comes up. Do not bring it up when the question is on the photoelectric effect, because it is totally irrelevant!

Zz.