The question is already in the title: How many photons will an excited scintillator emit? It is probably mainly given by the energy of the impinging particle and the bandgap energy. But also probably not that simple?

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    $\begingroup$ It is not that simple, but frankly there is not a good theory that allows one to calculate it in the first place. Generally, conversion efficiencies are measured, not predicted well. $\endgroup$ – Jon Custer Aug 16 '19 at 16:36
  • $\begingroup$ is there a rule of thumb? I read something like N = E_initial_photon / E_emitted_photon so this would be a linear relationship between the impinging particle and emitted photons. $\endgroup$ – Ben Aug 16 '19 at 17:37
  • $\begingroup$ At a fixed particle and particle energy, you should get (within statistics) the same number of visible photons out per particle that interacts with the scintillator. Change the particle or energy, and that ratio may change. A priori predictions are fuzzy at best. $\endgroup$ – Jon Custer Aug 16 '19 at 19:05
  • $\begingroup$ The use of "a scintillator" in the title is unusual. Experimenters are generally interested in macroscopic lengths of scintillating material in which case geometry and composition matter. But it feels like you may be interested in a single scintillating molecule or other localized target. Could you clarify? $\endgroup$ – dmckee --- ex-moderator kitten Aug 16 '19 at 21:26
  • $\begingroup$ @JonCuster That would be an answer if you fleshed it out with a few examples and a mention of energy/electron-hole-pair production. $\endgroup$ – Bill N Aug 17 '19 at 13:23

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