Percentage of photons detected by PMT or MPPC?

[tim.hbn]

Hi Everyone

This is probably a very difficult question to answer so I am only hoping for a very rough answer.

The chart on Page 11 of the following PDF file gives the number of visible photons emitted by variuous scintilators per MeV of energy of each detected gamma photon.

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

According to the chart, a LaBr3(Ce) crystal will convert a 1 MeV gamma photon into 49000 visible photons.

According to Page 5 of the following PDF, a 6 × 6 mm2 MPPC has only 18980 microcells.

https://www.onsemi.com/pdf/datasheet/mi ... ries-d.pdf

This suggests to me that the 6 × 6 mm2 MPPC would be more than saturated by the visible photons produced by a 1 MeV gamma photon if it was to receive every photon. I am therefore guessing that under the usual circumstances inside an average gamma spectrometer, only a small percentage of all the visible photons produced by the scintillation crystal actually make it to the PMT or MPPC. Am I correct? If yes, does anyone have a very rough idea what a typical value for this percentage is?

Thank you very much.

Kind regards

Tim

EDIT: I think I have just worked out why the MPPC does not get saturated. The logic behind the reason is vaguly similar to the logic behind the Birthday Problem that you can read about here:

https://en.wikipedia.org/wiki/Birthday_problem

[Sesselmann]

Tim,

You pose an interesting question, but to begin with we know how photons are emitted isotropically over a 4pi area and the pmt in most cases cover less than 1pi. of the crystal.

49000/4 = 12250 photons [edited]

When building detectors we try to wrap the other sides of the crystal with light reflective material, usually white teflon, but this only reflects a small amount of the light and I suspect most of it does escape. [Lucas claims 90% of the light reflects, I don't have evidence to the contrary.]

Then there is Compton scatter, where some of the gamma energy escapes the crystal and some doesn't. Compton scatter is a function of surface area to volume of the crystal, imagine an infinitely large crystal would catch every gamma ray fully. This is why a spectrum from a 3" detector looks very different to a spectrum from a 1" detector, typically the 3" has a flat Compton plateau with relatively few counts compared to a 1" detector.

Once the photons make it out of the crystal in the right lucky direction they encounter a mm of glass and finally the super thin (transparent) photocathode where once again the lucky photons interact with an electron knock it out of it's orbit.

The case of an MPPC or SIPM is slightly different, but its still a game of luck for a photon to trigger a pulse.

Steven

[tim.hbn]

Hi Steven

Thank you very much indeed for your very informative reply.

> "the pmt in most cases cover less than 1pi. of the crystal ... When building detectors we try to wrap the other sides of the crystal with light reflective material, usually white teflon"
When I was asking this question, I was already taking into account that the crystal has been wrapped in something white which would have increased the photon collection to much more than 1pi.

From what you say, it sounds like even something that looks very white to the human eye is possibly only reflecting somewhat less than half of the photons that are hitting it. Am I correct?

Thank you very much.

Kind regards

Tim

[GigaBecquerel]

Steven,

I think you are a bit confused about what's happening in a crystal, and how much of it.
If something coveres 1pi of a sphere it doesn't see 1/4pi of the total output, but 1/4. The surface of a sphere is 4*pi*r², and if a quarter of that is covered the seen surface is 1*pi*r², not just r².
In compton scattering not the visible light, which is created in the process of scintillation, but the gammas are scattered.
The collection efficiency is very high, the PTFE around the crystal reflects far over 90%, and about just as much will come out of the window.
The least efficient part in the chain is the photomultiplier, which only has ~25% quantum efficiency, meaning that only ~25% of all photons which hit it will result in a photoelectron.

Lukas

[Sesselmann]

Lukas is right, my calculation was wrong, I didn't take enough care with my answer, made some minor edits so as not to confuse future readers.

Steven

[tim.hbn]

Hi Lukas

Thank you very much for your very informative reply.

You state that the quantum efficiency of a PMT is ~25%. Do you know the quantum efficiency of an MPPC?

Thank you very much.

Kind regards

Tim

[GigaBecquerel]

The "Quantum efficiency" of a SiPM is called PDE (photon detection efficiency), and it's in the datasheet you listed.

[tim.hbn]

Hi Lukas

Thank you very much for that.

The top graph on Page 7 of the datasheet gives the PDE data for the MicroFC−30035−SMT. There is no explicit PDE data for the MicroFC−60035−SMT. Do you think that the PDE data for the MicroFC−60035−SMT will be approximately the same as the PDE data for the MicroFC−30035−SMT?

Thank you very much.

Kind regards

Tim