Confusion about Compton scattering

[tim.hbn]

EDIT: Since originally making this post, I have discovered that I have made a mistake in my mathematical reasoning.

Hi Everyone

I am trying to understand Compton scattering but it seems that different sources of information on the Internet are saying contradictory things about it and they don't seem to explain it well enough for me to understand how it effects gamma spectra. I would therefore be very grateful indeed if someone could resolve my confusion for me.

From what I have seen on the Internet, Compton scattering happens when photons collide with electrons.

It seems that the collision reduces the energy of the photon by an amount dependent on the scattering angle.

If the angle is zero, the energy reduction will also be zero.

From looking at the Compton scattering equation and doing some extra calculation, I have calculated that if the angle is 180 degrees (the maximum angle), the energy reduction should be equal to half the rest energy of an electron. This is about 255 KeV. EDIT: This is wrong.

So according to my thinking, a very small scattering angle might reduce the 662 KeV of a photon from Cs137 to perhaps 660 KeV and the angle of 180 degrees would reduce it to 407 KeV. This suggests to me that the Compton continuum for Cs137 should be between 407 KeV and 662 KeV, EDIT: This is wrong.

However, it seems that it is actually approximately between 192 keV and 475 keV.

I got the above values from the following page:

https://physicsopenlab.org/2016/02/04/c ... ttering-2/

Can anyone here explain where I am going wrong?

Thank you very much.

[Sesselmann]

Tim,

Essentially all interactions between the gamma and the scintillation crystal is scattering, when a gamma ray hits the crystal all hell breaks loose, but in most cases scattering absorbs all the energy 😱

The Compton plateau is caused by gamma interactions that take place close to the surface if a crystal, where a scattered photon escapes the crystal completely.

The backscatter peak is caused by escaped photons bouncing off your detector housing or lead shield and coming back in again.

Therefore larger crystals show less Compton and larger diameter shielding will show less backscatter.

Steven

[tim.hbn]

Hi Steven

Thank you very much for your informative reply.

I have been doing a lot more thinking about this and I believe that I have actually been thinking about this the wrong way round and I also made a maths mistake.

I think that I may now understand it better.

It now seems to me that when a gamma photon collides with an electron in the crystal thus causing Compton scattering, the deflected photon contains the energy which could potentially be lost and the electron contains the retained energy. This is because the energy in the the electron will end up in a photon again when the electron next collides with something in the crystal.

I have managed to calculate that the theoretical Compton edge for Cs137 is at 478 KeV. However, in reality, it seems a bit lower than this. I am not sure why.

I am also now thinking that the reason that there is a non-zero minimum amount of energy which can be retained during Compton scattering is because an incoming gamma photon must be deflected by an angle greater than zero in order to escape.

Do you agree with all of this?

Thank you very much.

Kind regards

Tim

[Sesselmann]

Tim,

Yes we have to give some credit to Arthur Compton (https://en.wikipedia.org/wiki/Arthur_Compton) for figuring this out...

The absolute maximum recoil energy given to the electron occurs when θ = 180◦. At this value of θ, Eγ′ = 184keV and Eγ − Eγ′ = 478 keV. This condition shows up as an edge on the spectrum. No single process (apart from total absorption) can dump greater energy in the crystal.

From the attached student experiment, curtacy of University of Sydney.

Steven

[tim.hbn]

Hi Steven

Thank you very much for your help and for the links.

Very best wishes from

Timo

[jneilson]

It now seems to me that when a gamma photon collides with an electron in the crystal thus causing Compton scattering, the deflected photon contains the energy which could potentially be lost and the electron contains the retained energy.

Indeed. I find it easiest to understand when written as
Energy transferred to detector E_T = E_incident - E_scattered

This is because the energy in the the electron will end up in a photon again when the electron next collides with something in the crystal.

For a scintillation detector, yes it's going to end up back as photons. For a semiconductor detector, it's more about the number of electron-hole pairs ionised by the recoiling electron, I don't think there's another photon involved.

I have managed to calculate that the theoretical Compton edge for Cs137 is at 478 KeV. However, in reality, it seems a bit lower than this. I am not sure why.

Your calculation seems broadly correct - I got 477.3 keV but the exact value depends on what input nuclear data you're using.
The slightly lower cutoff you're seeing in spectra (you mentioned 475keV in the OP) doesn't seem unreasonably different, and could just be a matter of electronic noise or how you're evaluating the edge - it may not always be as sharp a cutoff as the literature would suggest due to the effects of multiple Compton scatter events.

I am also now thinking that the reason that there is a non-zero minimum amount of energy which can be retained during Compton scattering is because an incoming gamma photon must be deflected by an angle greater than zero in order to escape.

Hmm, I'm not sure. I don't tend to notice a minimum Compton threshold in my spectra - there's generally a continuum all the way to zero; or at least to my lower level discriminator noise cutoff energy. Also, a scattered photon can still escape through the opposite side of the detector - it doesn't need to scatter away to leave the detector. I guess that logic would apply to a very large / thick detector volume though.

[Sparky]

If you want to observe the Compton minimum i would suggest an easy demonstration I found in an article by Dr. David Prutchi. I found it useful for understanding Compton interactions. It places a target outside the crystal to increase the number of scattered electrons that do not interact with the crystal. This has the effect of increasing number of back scattered photons detected. The Compton Peak( a.k.a. "Compton minimum" ) is caused by 180 degree back-scattered photons in or near the source. It is sort of a compliment to the Compton edge, you see the minimum scattered photon's energy instead of the maximum recoil electron's energy. The sum of the two values will equal the energy of the original, incident photon. The Calculator linked below shows the relationship quite well.

Prutchi's demonstration is very simple to perform. You just intentionally enhance the backscatter peak by placing a target (of an appropriate Z value) behind the source. I used the copper disk from the bottom of my shield. My version of his demonstration using a TG30 (Cs-137) spark tube is shown below. The calculated edge and backscatter peak energies are marked.

https://www.diyphysics.com/wp-content/u ... rutchi.pdf

https://www.sciencecalculators.org/nucl ... cattering/

[jneilson]

If you want to observe the Compton minimum i would suggest an easy demonstration I found in an article by Dr. David Prutchi. I found it useful for understanding Compton interactions. It places a target outside the crystal to increase the number of scattered electrons that do not interact with the crystal. This has the effect of increasing number of back scattered photons detected. The Compton Peak( a.k.a. "Compton minimum" ) is caused by 180 degree back-scattered photons in or near the source. It is sort of a compliment to the Compton edge, you see the minimum scattered photon's energy instead of the maximum recoil electron's energy. The sum of the two values will equal the energy of the original, incident photon. The Calculator linked below shows the relationship quite well.

Ah, yes, I'm aware of backscatter peaks; I'd just never heard them referred to as a Compton minimum - to me, I see it as the continuum extending all the way to zero, but that it happens to have backscatter features superimposed on top, rather than seeing the continuum as "ending" at the backscatter peak.

My thought process mentally separates the types of Compton events as in-detector or ex-detector - the continuum and Compton edge features mainly come from Compton scatters in-detector, where Compton scatters cause energy to be lost from what would otherwise be full energy photopeaks; whereas backscatter peaks are an ex-detector Compton scatter sending photons that overshot the detector back towards it.

It places a target outside the crystal to increase the number of scattered electrons that do not interact with the crystal.

That's an odd way to put it that I'm not sure I quite followed to begin with - yes, the recoil electrons are part of the process, but it's the energy carried by the scattered photon that has more relevance to backscatter. My understanding of what you're doing with the target is increasing the number of missed photons that scatter back towards the detector, thus increasing the number that do interact with the crystal. I guess you can say you've increased the scattered photons by increasing recoil electrons outside of the detector, though, so I guess that is right.

you see the minimum scattered photon's energy instead of the maximum recoil electron's energy. The sum of the two values will equal the energy of the original, incident photon.

That's an interesting thought I hadn't otherwise realised - the maximum recoil energy transferred and the maximum backscatter angle are both at 180 degrees, so yes they're seeing opposite sides of a similar scatter and will sum to the incident photon energy. However, the backscatter is almost an entirely different phenomenon to in-detector Compton events, so I think you're misleading yourself if you think about it as a minimum of the continuum: for an in-detector Compton scatter, the minimum energy transferred is practically zero - when a photon is only glancingly scattered (close to 0 degrees) and carries nearly its full energy out of the crystal

[Sparky]

Joseph,
I really appreciate your detailed reply. In my simple , and probably wrong, way of thinking the Compton plateau is bracketed by the Compton edge (energy of 180 degree recoil electron ) and the backscatter peak (energy of 180 degree scattered photon). I assume the values in between are from photon/electron interactions at smaller angles. I understand that such interactions can also result with lower energies than the backscatter peak, although they appear to be less probable.

I think your comment that “backscatter is almost an entirely different phenomenon to in-detector Compton events” is true. However, I am not sure my detector respects the difference, it has a backscatter peak with or without a target. My source is not very clean, my probe has an aluminum housing, there are reasons that region looks the way it does.

[jneilson]

No problem - thanks for the discussion, it's helped me solidify my own thoughts about Compton effects - I haven't really thought about the actual physics of it since leaving university.

I guess there's partly a question of terminology - if you think solely in terms of the flat plateau portion of the continuum, you could generally say that was bracketed by the edge and the backscatter peak; but I think it makes a lot more sense to consider the full continuum down to zero rather than arbitrarily limiting your interest to the flat bit - Compton scatters don't stop at the backscatter peak. There's a full range of scatters that can happen all the way down to zero energy, as can be seen in this plot

(https://web-docs.gsi.de/~stoe_exp/web_p ... /index.php)

I sketched up a quick diagram based on your spectrum to show the different types of Compton scatters that happen in detector and out of detector and how they contribute. There's a main continuum of scatters happening in detector, and then on top of that there's the backscatter from scatters outside the detector (generally only those between ~120 and 180 degrees end up scattering back into the detector)


I guess there's also forward "backscattering" if there's a scattering surface between the source and detector, which would contribute to counts on the higher energy side, but the textbooks tend to explain counts there as due to multiple scatters rather than being from forward scattering.