3 ms·
The amount of photons in the scintillator is proportional to the energy deposited. This is described by the Bethe-Bloch equation [1]. Pages 3-6 of this guide [2
by kaybe 10y ago
The amount of photons in the scintillator is proportional to the energy deposited. This is described by the Bethe-Bloch equation [1]. Pages 3-6 of this guide [2] have some plots and explanations on what that looks like for different particles.
How good that works in practice depends on the energy- and time resolution of the scintillator. Energy resolution depends on how different the light emission is for different energy depositions. Time resolution depends on how quickly the scintillator is back to a dark state (there are cascading effects from the deposited energy and there is a corresponding decay time). Anorganic material (eg NaI) is better in energy resolution, organic material (eg. plastic/polystyrene) is better in time resolution.
High energy resolution is good if you eg. want to look at a spectrum of nuclear fission products to find details of the fission process. High time resolution is good because it is easier to resolve different particles and you have a higher probability of a 'pure' signal.
[1] https://en.wikipedia.org/wiki/Bethe_formula https://en.wikipedia.org/wiki/Bethe_formula
[2] http://home.fnal.gov/~group/WORK/muonDetection.pdf http://home.fnal.gov/~group/WORK/muonDetection.pdf
- pif 10y ago> The amount of photons in the scintillator is proportional to the energy deposited. Exactly: it is proportional to the energy deposited in the scintillator! Which, for a cosmic ray, can be a tiny fraction of the total energy. In order to measure the kinetic energy of a particle, you need to stop it. In that case, your detector is called a calorimeter.