As you can see in this link: https://www.researchgate.net/figure/With-very-low-photon-energies-the-photoelectric-effect-is-dominant-The-Compton-effect_fig4_221928565

enter image description here

Probability wise, the photoelectric effect is typically observed when the energy of the incident light is lower than what is required for both Compton scattering and pair production, but higher than the energy needed for Rayleigh scattering.

Additionally, the likelihood of the photoelectric effect increases when the matter that interacts with light possesses a large atomic number and high atomic/electronic density. Examples of such materials include Lead, Tungsten, and even more so, Uranium.

  • In general, loosely bonded electrons and free electrons are targets of Compton Scattering although the Compton Effect requires more energy to occur than the Photoelectric Effect, Inner Shell electrons let's say K-level electrons are targets of the Photoelectric effect.

How is this possible? Normally higher energy should have most probably equated more to interacting with inner shells as we see in the photoelectric effect but here we are now saying that although the Photoelectric effect requires lower energies than Compton Effect it interacts with inner shells and - visa versa - While Compton Scattering requires higher energies that photoelectric effect it just happens most probably on the valance/surface level free electrons.

For more information:

EDIT: Wikipedia: https://en.wikipedia.org/wiki/Photoelectric_effect

Headline: "Competing processes and photoemission cross-section"


1 Answer 1


What we see for x-ray photoelectric cross sections is a sawtooth pattern, where the cross section generally decreases toward higher energy, but with sharp increases at "edges" where the energy is exactly at the threshold to unbind an electron from a particular shell.

The way we capture this in the theory is that energy and momentum must balance in the interaction. Thus, a free electron cannot absorb a photon. It can only scatter photons (Compton scattering).

On the other hand, a bound electron can transfer some momentum to its atom in the process of interacting with a photon. It can thus absorb the photon (photoelectric effect). The more tightly the electron is bound, the better this works, assuming the photon has sufficient energy to unbind it. The more momentum the photon carries, the worse this works.

  • $\begingroup$ Much appreciate it bro :) $\endgroup$ Commented Jun 8, 2023 at 12:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.