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Forum: Processes Involving Optical Photons
Re: Question Problem in Scintillation Simulation with Nonlinear Light Yield (Xing Zhang)
Date: 16 Jan, 2012
From: Gumplinger Peter <Gumplinger Peter>

Sorry, for the late reply.

> I found that when proton energy exceeded 5.5 MeV, the proton would
> deposit its energy in many steps by ionization process. For example, the
> below list the track information of a 9.13 MeV proton. The light yield
> in every step is calculated with the deposition energy in every step.
> But due to the nonlinear light yield, the sum of the light yield in all
> steps is much less than the light yield from a 9.13 MeV proton. So is
> there any technique to make the light yield be calculated according to
> the energy of proton at the beginning in a track while not the
> deposition energy in every step?

Your question can be reworded: What does it take to induce the 9.13MeV proton to lose all of its energy in one single step? Good question, I actually don't know the answer.

> What I want to do is the 14MeV neutron response in a BC501A or EJ301
> scintillator.

I presume the protons are the result of a neutron interacting in your scintillator. Since they are not the primaries in your simulation I have one suggestion. One way to not produce the secondary proton is to increase the production range cut threshold so protons are not actually tracked in your G4 simulation and all of the energy going into proton production is dumped at the point where the neutron interacted. Please, see:

http://geant4.web.cern.ch/geant4/UserDocumentation/UsersGuides/ForApplicationDeveloper/html/ch02s04.html#sect.HowToSpecParti.RangeCuts

Since no secondary proton is produced, the energy loss is attributed to the neutron and consequently, the scintillation yield must be defined for a neutron. But since the neutron is not one of the particles in the list of possible 'scintillation by particle type' options, the default yield, "ELECTRONSCINTILLATIONYIELD", must be provided for it as a function of energy.

Let me know if that works, Peter

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