Say a photon hits some atom.

What determines whether there will be a photoelectric effect (photon is absorbed, electron is released) or whether there will be a Compton scattering (the photon is scattered at some angle, and the electron is released with another direction)?


6 Answers 6


For a given system that the electron is in, the primary determinant is the energy of the photon. As @DJBunk points out, this is a quantum mechanical process, so the "choice" is fundamentally random. A given interaction will occur with a probability proportional to its cross section. Figure 1 of this lecture shows how the cross section for each possible process varies with photon energy. This plot is for the interaction of photons with electrons in copper. At low energies, the photoelectric effect is the dominant effect. From about 200 keV to about 10 MeV, Compton scattering is the dominant effect. Above 10 MeV, the dominant effect is pair production. At a given photon energy, the relative probability of two processes would be the ratio of their cross sections.

The dependence of each cross section on photon energy should be similar in form for any system; the exact numbers will vary from system to system. Table 2 of that lecture gives the dependence on the atomic number, for example.

  • $\begingroup$ @Coling McFaul - Honestly I actually don't think you answered this question either - Bruno is asking why one process happens over another and you are simply replacing this with a function of energy, but now the question is what determines the process as a function of energy? $\endgroup$
    – DJBunk
    Jul 8, 2012 at 15:57
  • $\begingroup$ @DJBunk, I believe that I have answered it. I just made an edit, adding your comments from your answer about randomness. I hope that clarifies my answer. $\endgroup$ Jul 9, 2012 at 20:23
  • $\begingroup$ "Above 10 MeV, the dominant effect is pair production." - This means production of a positron, correct? This is amazing. The positron will leave with less than speed of light which makes me say: when a photon is slowed down by an electron its rest mass is not zero any more. Some hint or reference on pari production I'd consider useful. The electron is "left" unchanged? $\endgroup$ Oct 21, 2022 at 15:25

It is the energy of the incident photon.

At low Energy (Long wavelengths as in Radio waves and Microwaves, visible light) Rayleigh scattering would dominant, in which the scattered photon would have the same wavelength as the incident photon.

At higher energy (shorter wavelengths as in Ultraviolet spectrum), the electron would absorb the energy of the incident photon (assuming the incident photon frequency is above the cutoff frequency) and the electron would be ejected (no photon is scattered here, it has been absorbed by the electron)...Photoelectric Effect would dominate.

At extremely high energies (X-ray and Gamma rays), the incident photon energy is too high for the electron to absorb and maintain Energy-Momentum conservation (and that is why Photoelectric is observed on bounded electrons whereas Compton Scattering is observed on "Free" Electrons). So the electron would be ejected, Since the incident photon has transferred some of its energy to the electron which it collides with, it will be scattered but now it has less energy. Compton Scattering would dominate.

  • $\begingroup$ hi, @user-1, I still do not understand why for x ray which has extremely high energy, it could not have photoneletron effect, the electron seems could just absorb x ray and ejected out of the atom, is this right ? $\endgroup$
    – guangcun
    Jun 25, 2021 at 3:22
  • $\begingroup$ Hi @guangcun, No, because it is too much energy for the little electron to absorb, otherwise, it will break the energy-momentum conservation of the electron i.e. the electron would have to travel at a speed larger than the speed of light if it is to absorb the full x-ray photon energy. That is not what God intended :) $\endgroup$
    – user -1
    Jun 28, 2021 at 4:33
  • $\begingroup$ "... Rayleigh scattering would dominant, in which the scattered photon would have the same wavelength ..." - I wonder if in that case, as there is change of direction of the scattered photon, thus change of (radial) momentum, of the photon, "the electron would absorb the energy of the incident photon", i.e. I wonder if this holds true for Rayleigh scattering as well. Do Raleigh electrons that scatter photons change their directional impuls (are they scattered, too...)? Is there no exchange of impluse? $\endgroup$ Oct 21, 2022 at 16:53

When a photon interacts with an atom, a variety of processes can occur. You mention the photoelectric effect and Compton scattering (non-resonant inelastic scattering), but you can also have elastic scattering or resonant inelastic scattering (if the incident photon energy is tuned to an atomic transition energy). This list is still by no means exhaustive.

For each of these processes, it is possible to calculate (or measure) a cross-section that determines the relative frequency at which an event occurs given a large number of photons incident on the atom.

Now, to get to your actual question. You ask what determines which event occurs. This is a fundamental question in quantum mechanics, and often is called the "measurement problem". Consider a universe consisting of only a photon flying towards an atom. If we were to run time forward until long after the photon would have reached the atom, the system will be in a superposition of states including all possible processes (with the correct weighting to give the relative probabilities). It isn't until the system interacts with a larger ("classical") measurement device that one of the many processes is selected ("the wavefunction collapses"). According to the usual interpretation of quantum mechanics, which branch occurs is simply determined by the probability, with nothing in particular causing the selection.

Of course, there are conceptual difficulties around where the boundary between "classical" and "quantum" systems should be. You might find it interesting to read about "decoherence" as one possible mechanism for apparent wavefunction collapse.


The photoelectric effect couples two electron states whose frequency difference is equal to the frequency of the incident light. The Compton effect couples two electron states whose standing wavelength is equal to the wavelength of the incident light.

I explain how the wavelength interaction explains the Compton Effect in this blogpost.


As far as i am able to comprehend your question, the answer to it lies,according me, as we all know electrons are present in atom arranged in a set of orbitals(not orbits) for an electron to move to the next shell or orbital ,it requires a specific amount of energy(which depends upon the particular atom) .in physics,mainly in quantum nothing is half way done thus when an photon strikes ,for it change its energy level it must carry that energy(equivalent to the threshold),if does than the photoelectrc works -if not,the electron would push it back!..i would like to add that this all is related to the band theory so for a better understanding read it once. All it says is dat there are certain gaps between the bands(orbitals arranged together) therefore u need a greater amount of energy to make it jump. i hope i clarified u ! And i really apologize for my immature gesture, i will surely obey your instructions

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    $\begingroup$ Hi Jayanti, and welcome to Physics Stackexchange. Just a friendly reminder from our help center: Grammar, punctuation, and spelling are very much appreciated here, more so than elsewhere on the internet. $\endgroup$
    – user10851
    Sep 9, 2013 at 2:10

All we can ascribe to a process like you are describing ($\gamma + p \rightarrow products$) is a probability. The sum over all process will give us %100. I couldn't find a plot of anything like the process you described, but I did find this plot of Higgs decay products as a example of the probabilistic nature of quantum mechanics. On the bottom axis is the possible higgs mass (now determined to be pretty close to 125 GeV) and on the vertical axis it says given the Higgs mass, how likely it will decay to those products. enter image description here

Keep in mind calculating probabilities for hands dealt in a poker game is pretty much the same thing:

enter image description here

All we can do is calculate the probability for a single outcome, and all the possible outcomes add up to %100.

  • 1
    $\begingroup$ I suppose I should explain my downvote. This is completely irrelevant to the question. There does exist a plot that describes the cross-section for different interactions as a function of photon energy. It has nothing to do with the Higgs. $\endgroup$ Jul 8, 2012 at 2:55
  • $\begingroup$ @ColinMcFaul - Thanks for the explanation of your downvote. I probably didn't do a great job explaining it but my purpose of the higgs plot was to give an example of the probabilistic outcomes in quantum mechanics, and particle physics in particular. I couldn't find a plot of the form that you posted so I posted the closest type of plot I could find. $\endgroup$
    – DJBunk
    Jul 8, 2012 at 14:36

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