Gamma Spectrometry Forum

Hi Everyone

I am rather curious about something. I have found the following very interesting document:

https://indico.cern.ch/event/774281/con ... cindet.pdf

There is a chart on Page 11 giving some data for several scintilators. According to this chart, NaI(Tl) produces 38 photons of light per keV of detected gamma radiation. The max peak wavelength of the 38 light photons is 415nm which is approximately equivalent to 0.003 keV. If we multiply this by 38, we get 0.114 keV. So my question is what happens to the remaining 0.886 keV?

My guess is that it escapes as "heat" in the form of infrared radiation. Am I correct?

Thank you very much.

Kind regards

Tim

Tim,

You ask a good question, I don't know the answer, but I think your guess is pretty good, someone else reading this might have something to add.

Steven

Hi Steven

Thank you very much indeed for your very kind reply.

I look forward to receiving a reply from someone elese.

Kind regards

Tim

Tim,

This is refered to as scintillator efficiency, not to be confused with detector efficiency.
As you said, the rest of the energy is going into wasted heat, and we are trying to make scintillators as efficient as possible.
But it's all a very statistical process, where a photon hits an electron, how close the nearest dopant hole is etc.

Hi GigaBecquerel

Thank you very much for another informative reply.

Am I correct in guessing that the wasted heat is in the form of infrared radiation?

Thank you very much.

Kind regards

Tim