Bremsstrahlung in scintillators

[BBV]

The monochromatic gamma rays enter a scintillator (e.g. NaI). Then I compute the deposited energy spectrum. I need a techinque of modeling of the bremsstrahlung produced by the recoil electrons without direct modeling of the electrons trajectory. What could I do?

 

[Vanadium 50]

Can't you just use something like GEANT?

 

[Bob S]

The radiation length of a high energy gamma ray in (polystyrene) scintillator is about 42.4 cm, so it is very likely to leave the scintillator (~1/4") without any interaction. For a sodium iodide, the radiation length is only 2.6 cm, so a high energy gamma will probably interact in a 3" NaI. Above 10 MeV, all of the gamma ray energy is carried away by the electron-positron pair, which in turn create more electromagnetic showers. If you want to increase the probability of detecting a high energy gamma, put a radiation length of lead in front of the NaI. You should use an electromagnetic shower program like suggested above to model your specific geometry.

 

[Vanadium 50]

That makes absolutely no sense. The reason one uses an inorganic scintillator is to completely absorb the gamma ray to measure its energy. Putting a chunk of lead in front of it will make this impossible. And if you can't use this to measure the energy, you might as well use plastic.

 

[Bob S]

That makes absolutely no sense. The reason one uses an inorganic scintillator is to completely absorb the gamma ray to measure its energy. Putting a chunk of lead in front of it will make this impossible. And if you can't use this to measure the energy, you might as well use plastic.

If there is no lead in front of an inorganic scintillator (especially a thin one) there is a finite probability that a high energy photon will pass completely through it without any interaction, and therefore producing no ionizing radiation. The shower maximum is roughly 2-6 radiation lengths from the entrance depending on incident photon (or electron) energy. See Fig 14 on page 22 in
http://www.phys.ttu.edu/dream/ref/Profilepaper3.pdf

 

[Vanadium 50]

There is always a probability a photon will pass through without depositing any energy. So what? That's not relevant to this problem.

It's clear you have never used this sort of thing, so you might want to put your PhD aside for a moment and listen to me. People use NaI (usually NaI(Tl)) because they want to measure the energy of a gamma ray. Sticking a piece of lead in front of it causes energy to be absorbed in the lead, which makes the energy measurement in the crystal worthless. That's why, when you look at a NaI counter, it's not wrapped in lead.

NaI is used to measure energy. That's why efficiency is less important than energy resolution for applications that choose Na(I).

 

[Bob S]

A single NaI(tl) crystal is too small to contain the complete shower of a 20-GeV photon. That is why experimenters use lead-scintillator sandwiches that are often 30 or more radiation lengths thick. See Fig 14 in
http://www.phys.ttu.edu/dream/ref/Profilepaper3.pdf
[Edit] I added a thumbnail of the EM shower development curve from the above URL for 5 to 200 GeV electrons. This curve is nearly identical to the shower curve for photons. The shower maximum is at 5 to 7 radiation lengths. To identify a high-energy photon, put a veto counter in front of 1 or 2 radiation lengths of lead, followed by a NaI(Tl). For low energy photons: the attenuation length of a 10-MeV photon in NaI(Tl) is about 0.04 cm² per gram, and when combined with the density of 3.67 grams per cm³ yields an attenuation length of 6.8 cm. So a 3" by 3" sodium iodide will not even contain that.

 

[Vanadium 50]

A single NaI(tl) crystal is too small to contain the complete shower of a 20-GeV photon.

And where exactly did the OP say that he was looking at 20 GeV photons?

I'm making the assumption that the OP is using the right tool for the job. You're making the assumption that it's the wrong tool. Which makes more sense? Furthermore, you're giving advice such that if it turns out the OP is not some sort of incompetent stumblebum, and he actually is using the correct instrumentation for his application, it will make things worse and not better.

That is why experimenters use lead-scintillator sandwiches that are often 30 or more radiation lengths thick.

Except of course, when they don't. KTeV, L3 and CMS both used crystal scintillators, CsI, BGO and PbWO4 respectively. These were/are interested in multi-GeV photons. Crystal Ball even used NaI, although it's not terribly popular any more in HEP because it is so slow.

I added a thumbnail of the EM shower development curve from the above URL for 5 to 200 GeV electrons.

200 GeV? Where on Earth does the OP say he is looking at 200 GeV photons?

 

[BBV]

Thanks for discussion and my bad I've not given any feedback.

The real problem is that usual detector's sizes are about 3" x 3" and the energies of photons are 1 - 10 MeV. In my now modeling I implement only two types of photo-atomic reactions: compton scattering and pair production. It allows to make the reasonable shape of the energy spectrum. But when the energy of the incident photons are higher than 1 MeV I have the unreal photopeak/compton ratio. This is caused by ignoring the energy losses produced by the bremsstrahlung produced by the recoil electrons. I mean I should not only track the scattered photons, I should watch what the energy escapes the detector after recoil electron loses it.

So the problem could be formulated as follows - in the center of the 3"x3" NaI detector the 5 MeV* electron is born. How much energy escapes the detector while the electron slows down? And it is important that I need to avoid the Monte Carlo modeling of the full electron trajectory.

Bob_S, according to my problem the lead container doesn't help me. The problem touches the processes inside the crystal. The energy losses are caused by the escaping it and I don't care what photons enter. And I don't consider GeVs, only the first MeVs.

Vanadium 50, we can use GEANT or MCNP to make the proper modeling but in educational purposes we need to know how the spectrum is made.