I am going to carrying out an investigation in the lab on gamma ray spectroscopy and to help me understand the graphs that will be displayed to me, I am researching on the types of interactions that happen within the crystal.

As I know so far there a three

  1. Photoelectric absorption
  2. Compton Scattering
  3. Pair Production

The issue I am having is with the photoelectric absorption and how light (optical photons are being generated) , the book I am ref from is 'Radiation Detection and Measurement, G.F Knoll, pg 49-50'. Also the crystal I am using for my experiment is a sodium/Thallium crystal.

So here my understand of how the photoelectric interaction is occurring in the crystal.

As the gamma photo enters the crystal it is 100% absorbed by an atom to which is comes into contact with and a photo electron is ejected, this is normally from the K-shell of the atom, so for my lab the gamma photo should have and energy > than the binding energy of the atom.

So when the photo electron is ejected a gap is left within the K-shell so a surrounding electron the fill it place and a characteristic x-ray is emitted and through phosphorylation and collisions with the photo electron optical light is generated by energy dissipation.

Is this along the correct line to what is actual happening within the crystal with regards to the photoelectric interaction.

  • $\begingroup$ That is the gist, but all sorts of things happen. The gamma ray could be hit an electron and transfer part of its energy. A lower energy gamma ray could then be scattered out of the detector. The excited atom could emit an x-ray that escapes the detector. -- and so on... $\endgroup$
    – MaxW
    Sep 24, 2018 at 16:14
  • $\begingroup$ @MaxW: The processes you're describing are not the photoelectric effect. The OP specifically asked about the photoelectric effect. $\endgroup$
    – user4552
    Sep 24, 2018 at 21:22
  • $\begingroup$ A high-energy photoelectron can also go near a nucleus and emit Bremsstrahlung radiation which may or may not escape from the crystal. $\endgroup$
    – MaxW
    Sep 25, 2018 at 0:33
  • $\begingroup$ So you can have, both characteristic and Bremsstrahlung radiation, I did think this but all lit that I have read never really mention it, to which I wounder that maybe it not very likely to happen within the crystal. $\endgroup$
    – james2018
    Sep 25, 2018 at 10:02

1 Answer 1


No. There are several steps involved. They all involve electrons and photons and atoms but with very different energies and numbers. Some accounts blur over the distinctions, but it's really quite clear.

First stage: a single high energy photon (gamma ray) interacts in the crystal. It may give all its energy to an atomic electron (photoelectric effect) or just some of it (Compton effect) or even, if its energy is large enough, have all its energy converted to an electron and a positron. So this electron (and maybe positron) has an energy of hundreds of keV. (The binding energy of the target electron is small on this scale).

Second stage: this electron (/positron) with its energy moves through the NaI crystal. It passes the crystal atoms and the field from its moving charge excites their atomic electrons to higher levels: as it does so it loses energy and slows down, eventually stopping. It travels a mm or so, which is a long way in terms of atoms, so it leaves lots of excited electrons behind it. These drop down to their original level and many of them give out photons as they do so which are in the visible region with typically eV energies.

Third stage: these photons bounce around the crystal and many of them make it to the photomultiplier tube window, and make electrons by the photoelectric effect. These are amplified by the dynodes and the current pulse is recorded.

The size of the pulse is proportional to the number of electrons at the last dynode. Which is proportional to the number at the first dynode (by a multiplication factor of $10^{12}$ or so). Which is proportional to the number of ~eV photons striking the photocathode (with a photoefficiency of ~10%). Which is proportional to the number of photons produced by the track. Which is proportional to the energy of the electron. Which is the energy of the gamma ray (for PE reactions) or a random fraction of it (for Compton Scattering).

With pair production, both electron and positron have tracks and excite atoms, the photomultiplier counts the photons from the two tracks together.

  • $\begingroup$ The OP specifically asked about the photoelectric effect, so your material about pair production and Compton scattering is not relevant. The OP also didn't ask about things like the operation of the photomultiplier tube. If you cut all that material out of your answer, then it seems to me that your answer contains nothing that wasn't already stated by the OP. $\endgroup$
    – user4552
    Sep 24, 2018 at 21:25

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