As far as I understand, the incident photon interacts by photoelectric, Compton scattering or pair production with the electrons of the crystal (NaI(Tl) in our case). The electron that emerges from these interactions generates number of photons in the UV or visible which are proportional to gamma ray energy. Then, the light produced can be read out with photodiodes, photomultpliers, avalanche photodiodes, silicon photomultipliers, and other photo sensors, right?.

So, how does those electrons are converted into UV or visible photons?

  • 1
    $\begingroup$ In some organic scintillators it is the direct action of the passing charge's electric field on the molecules that excites molecular-modes of the scintillator and there aren't necessarily any ionization events at all. In crystalline scintillators it may be related to the band-gap, but don't quote me. $\endgroup$ Sep 7 '15 at 2:12

In the case of sodium iodide, the thing that emits the photons is the dopant - usually Tl (thallium) for sodium iodide - which is why the "full name" is usually written as NaI:Tl.

Other scintillators may have other dopants (Cesium is quite popular - for example in LaBr3:Cs, a very fast and bright scintillator with a short decay time). The mechanism for energy transfer and emission of light is quite complex - with the probability depending on part on the environment of the dopant. In the case of LaBr3:Cs the environment is "always the same" because the Cs displaces a La atom - so the dopant is always surrounded by three identical halogen ions. This makes the photon yield "always the same" and contributes to the excellent energy resolution of that particular class of scintillator (LaCl3:Cs is another one like that).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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