0
$\begingroup$

The photoelectric effect is most probably seen when the incoming light has lower energy than the energy needed for both Compton scattering and pair production to happen.

The probability of the photoelectric effect to occur also increases when the matter that light interacts with has a big atomic number and high atomic/electronic density such as Lead, Tungsten, and even better Uranium.

  • Given the energy difference between Gamma and X-Rays, and the for high probability, photoelectric effect requiring low energetic photons to occur, how is it possible that X and Gamma rays are both prevented by the photoelectric effect that is observed in those materials?

Is it that energy levels of photons that photoelectric effect has to occur on materials like Tungsten are on the high energy boundary of X-Rays and Low Energy boundary of Gamma Rays, and thus rest of the matter-photon interaction effects such as COMPTON and PAIR PRODUCTION are almost always on the GAMMA Spectrum then?

Improve this question
$\endgroup$
6
  • $\begingroup$ The photoelectric effect is usually the term for the interaction of IR to near UV light with quasi-free electrons in metals. It's low energy physics taking place in the single digit eV range. Do you mean high energy x-ray photoionization that requires keV to hundreds of keV? $\endgroup$
    – FlatterMann
    Commented Jun 8, 2023 at 20:40
  • 1
    $\begingroup$ @FlatterMann We usually call x-ray photoionization "photoelectric" if the photon is absorbed, "Compton" if it's scattered. $\endgroup$
    – John Doty
    Commented Jun 8, 2023 at 20:50
  • $\begingroup$ @FlatterMann Exactly what John Doty said. Additionally, in my OP, I am talking specifically about high probability of photoelectric effect to happen, which is low energy but X-Rays interestingly. $\endgroup$
    – medical physics
    Commented Jun 8, 2023 at 20:51
  • $\begingroup$ The low energy photoelectric effect happens in very thin layers of metals (100nm if I remember correctly), but I am not aware that x-ray photoionization has similarly large cross sections. It takes a significant amount of bulk (in the 0.25mm range for lead, I believe) to absorb 100keV photons. The absorption cross section per atom is therefor significantly smaller for x-rays than for optical photons. Back of the envelope that scales with 1/E, doesn't it? Am I missing something? $\endgroup$
    – FlatterMann
    Commented Jun 8, 2023 at 21:04
  • 1
    $\begingroup$ @FlatterMann More like 1/E^3. So steep that for low energy x-rays it is seriously difficult to get them into the detector through windows and surface layers. See amptek.com/products/x-ray-detectors/… $\endgroup$
    – John Doty
    Commented Jun 8, 2023 at 21:08

1 Answer 1

Reset to default
2
$\begingroup$

Compton scattering occurs at low photon energies, where it goes by the name "Thomson scattering". There's no rigorous distinction, and there is no threshold. However, with details depending on the material, the photoelectric cross section is much higher at low photon energies, while the Thomson cross section is nearly constant, so the photoelectric effect dominates. The pair production cross section is, of course, zero below 1.022 MeV.

Shielding employing the Compton effect is a bit problematic, since the photon isn't absorbed, but scattered to lower energy. Low background hard x-ray and gamma ray detectors often employ "active shields" which are themselves detectors. If you see a detection in you main detector within a few nanoseconds of a detection in the shield, you have high confidence that it was a gamma ray that came through the shield.

Improve this answer
$\endgroup$

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.