Cs137 - Ratio between 662 KeV peak and Compton edge

[tim.hbn]

Hi Everyone

The following are two Cs137 spectra that I have found on the Internet.

Caesium-137_Gamma_Ray_Spectrum-en.jpg (28.2 KiB) Viewed 27837 times

Cs137_Spectrum.png (14.89 KiB) Viewed 27837 times

Do you notice that the ratio between the 662 KeV peak height and the height of the Compton edge is very different in each spectrum?

Does anyone here know the reason for these differences?

Is there a way of reducing Compton scattering when doing a Cs137 spectrum?

Thank you very much.

Tim

[Sesselmann]

Tim,

Good question....

When a gamma ray enters a scintillation crystal all hell breaks loose (literally). A gamma ray from Cs-137 enters a crystal with 662 keV, and it takes as little as a few eV to dislodge an electron from the outer shell of an atom (H), so hundreds of electrons literally scatter everywhere, and as these electrons settle back into their orbits they release light and X-Ray fluorescence. Most of these x-rays are reabsorbed in the crustal and down graded to visible light, but some escape through the side of the crystal never to come back.

Lets assume an event where 200 keV of a 662 keV gamma ray is scattered or lost from the detector then the remaining energy 462 keV is converted to photons which gives one count in the compton plateau region.

Hypothetically an infinitely large crystal would have no compton scatter, but since it is impractical and cost prohibitive to have a very large crystal we have to accept compton scatter.

Scattering losses only occurr near the surface of a crystal, therefore it is proportionally larger in smaller detectors, because of the surface to volume ratio.

One other consideration for height differences in the main photopeak is resolution, obviously if you detect the same number of counts over a wider peak, then the peak will be lower.

Steven

[NuclearPhoenix]

Tim,

like Steven already said, it's probably something to do with the volume of your crystal. Because lower energies are absorbed much better than higher ones, smaller crystals are more efficient at these (lower) energies. That's the same reason why you need a bigger and bigger scintillator, the higher you want to go. Crystal efficiency is really bad at higher energies and more so for a smaller crystal. This of course also depends heavily on the material density. The higher the density, the better the efficiency, because the probability of absorption is much higher. I've attached a graph from Gordon Gilmore's great book "Practical Gamma-ray Spectroscopy" (2008).

The second spectrum was probably done using a smaller crystal than the first one or another (less dense) material, resulting in a lower efficiency at the 662 keV peak. As a result, the Compton edge is closer to the Cs-137 photopeak.

[tim.hbn]

Hi Steven and NuclearPhoenix

Thank you both very much indeed for your very informative replies.

When a gamma ray enters a scintillation crystal all hell breaks loose (literally). A gamma ray from Cs-137 enters a crystal with 662 keV, and it takes as little as a few eV to dislodge an electron from the outer shell of an atom (H), so hundreds of electrons literally scatter everywhere, and as these electrons settle back into their orbits they release light and X-Ray fluorescence. Most of these x-rays are reabsorbed in the crustal and down graded to visible light, but some escape through the side of the crystal never to come back.

It seems from what you have said here that when a gamma photon enters the crystal, there are two times when visible light is released. The first is when the electrons settle back into their orbits and the second is when the x-rays are reabsorbed in the crystal and down graded to visible light. Is this correct?

If my summarization is correct then am I correct that the first time contributes to the main photo peak and the second time contributes to the Compton scatter?

Thank you very much.

Kind regards

Timothy Scrimgeour

[Sesselmann]

It seems from what you have said here that when a gamma photon enters the crystal, there are two times when visible light is released. The first is when the electrons settle back into their orbits and the second is when the x-rays are reabsorbed in the crystal and down graded to visible light. Is this correct?

Timothy Scrimgeour

No, that's not quite correct, every gamma ray entering the crystal has the full energy 661.7 keV and if/when all of it (or as much as possible) is converted to light, you will get a count in the main photo peak. It is only gamma rays which strike close to the sides of the crystal where some energy is lost to the surrounding space, in which case some light has been lost.

Think of it in terms of (full energy) - (lost energy) = (compton)

Larger crystals or crystals with higher density loose less energy, therefore more counts in the main photo-peak.

There are some popular experiments you can do learn about compton, below is a link to a paper.

https://www.gammaspectacular.com/download/Expt_22.pdf

Steven

[tim.hbn]

Hi Steven

Thank you very much for your reply.

No, that's not quite correct, every gamma ray entering the crystal has the full energy 661.7 keV and if/when all of it (or as much as possible) is converted to light, you will get a count in the main photo peak. It is only gamma rays which strike close to the sides of the crystal where some energy is lost to the surrounding space, in which case some light has been lost.

Am I interpreting you correctly in thinking that basically, whenever a gamma photon hits an atom that is not near one of the sides of the crystal, it sets off a chain reaction which eventually converts all of the energy into visible light but if it hits an atom close to one of the sides, some of the energy leaks out of the crystal in the form of X-rays?

Do you have any idea how close it has to be to one of the sides of the crystal in order for energy leakage to happen?

Thank you very much.

Kind regards

Tim

[Sesselmann]

Do you have any idea how close it has to be to one of the sides of the crystal in order for energy leakage to happen?

This will depend on the type of crystal and the energy of the gamma, the attached efficiency chart is for NaI crystal and was clipped from a Saint.Gobain publication.

You are starting to uncover the subtle uncertainty of gamma detection.

I have been working on some code extensions to Impulse for putting the compton scatter back where it belongs, but I am not sure at this stage if the juice is worth the squeeze.

Steven

NaI efficiency

[tim.hbn]

Hi Steven

Thank you very much for your informative reply.

Just to confirm, am I correct about what I asked about the chain reaction?

Than you very much.

Kind regards

Tim

[Sesselmann]

Just to confirm, am I correct about what I asked about the chain reaction?

Yes, not sure if chain reaction is the right description as it is only the initial energy being absorbed. We have to be careful how we describe interactions between photons and electrons, because at the quantum level it's waves not particles, but if we take the naive view that a gamma ray hitting a scintillator is like the initial break in billiards, then compton is caused when one or more balls fall into a pocket on the break.

[tim.hbn]

Yes, not sure if chain reaction is the right description as it is only the initial energy being absorbed. We have to be careful how we describe interactions between photons and electrons, because at the quantum level it's waves not particles, but if we take the naive view that a gamma ray hitting a scintillator is like the initial break in billiards, then compton is caused when one or more balls fall into a pocket on the break.

Hi Steven.

Thank you very much indeed for this clarification and for your very kind patience with me. I think I am getting there.

I am guessing that in your analogy, balls falling in the pockets represent x-rays produced in the scintillator escaping from the sides of the scintillator and balls not falling in the pockets represent visible light which then gets detected by the photon counter. Is this roughly right?

Thank you very much.

Kind regards

Tim