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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

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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"

medical physics
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  • This seems a strange quote AFAIK. Can you give a link for these definitions of the photoelectric effect? I know this effect http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html – anna v Jun 08 '23 at 11:32
  • @annav This source is free: https://www.nrc.gov/docs/ML1122/ML11229A667.pdf https://www.researchgate.net/figure/With-very-low-photon-energies-the-photoelectric-effect-is-dominant-The-Compton-effect_fig4_221928565 – medical physics Jun 08 '23 at 11:38
  • see this question and answer https://physics.stackexchange.com/questions/206263/why-is-the-k-shell-electron-preferred-in-the-photo-electric-effect – anna v Jun 08 '23 at 11:39
  • This seems to be about x-ray and gamma-ray interactions. – John Doty Jun 08 '23 at 11:42
  • As far as physics goes it is a misrepresentation of the photoelectric effect as known in physics , but the link explains why the probability of K shell interaction is high if the energy is available. – anna v Jun 08 '23 at 11:43
  • @annav Thank you but I am searching for an explanation for both Compton and the Photoelectric effect in relation to one another comparatively as I mentioned in my OP. – medical physics Jun 08 '23 at 11:43
  • @annav please open my question, my question is also about Compton scattering. And the link you provided does not elaborate the problem specific enough as my OP do. – medical physics Jun 08 '23 at 11:47
  • "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." This is not true, if you read the hyperphysics link the photoelectric effect in physics measurements starts on the surface of metals with quite low energy photons. it is the metals. – anna v Jun 08 '23 at 11:48
  • @annav open the question thus we can discuss it because clearly there is a misunderstanding or misinformation. My sources clearly states that lower energies than compton increases the chances of photoelectric effect. – medical physics Jun 08 '23 at 11:50
  • http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/comptint.html .there is a scattered photon in the Compton scattering, it is two different reactions. Apples with oranges. – anna v Jun 08 '23 at 11:50
  • @annav the link you provided does not weigh in the problem from the perspective I provided in my OP. Here is the link: https://www.researchgate.net/figure/With-very-low-photon-energies-the-photoelectric-effect-is-dominant-The-Compton-effect_fig4_221928565 – medical physics Jun 08 '23 at 11:53
  • @annav there are areas where the probability of different matter photon interactions to happen escalates. As you can see in the link. Open the question please. – medical physics Jun 08 '23 at 11:54
  • @annav Well, it's not quite two different reactions. A photon can scatter from an inner shell electron, with momentum and energy transfers to the atom, but with an outgoing photon of lower energy. Sort of a mixture of Compton and photoelectric. The cross section isn't very high, but calibration spectra for the SDD detectors on NICER exhibit features we interpret as this process. – John Doty Jun 08 '23 at 11:57
  • @annav Please open the question I edited it. – medical physics Jun 08 '23 at 11:59
  • @JohnDoty do you want to answer the question? if so I can remove the duplicate. – anna v Jun 08 '23 at 12:00
  • @JohnDoty please do... – medical physics Jun 08 '23 at 12:01
  • @annav You opened it. Thanks a lot. – medical physics Jun 08 '23 at 12:03

1 Answers1

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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.

John Doty
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