The question

If one exposes a scintillator to a gamma source, one will observe two features for each characteristic energy in the source: a photopeak, due to complete absorption of photons, and a Compton spectrum, due to scattering of the photons by electrons. The geometry of the scattering process imposes a maximum energy for the re-emission: the well-known Compton edge, discussed on Wiki and here.

But in the labs that I've done, there appears to also be a low-energy cutoff, below which the Compton spectrum is not visible. For example, below is a spectrum of Cs-137 produced via a germanium scintillator, and plotted by Spectrum Technologies' USX (UCS-30) software. What I want to know is:

What causes this low-energy cutoff?

A blue-on-white log-plot of a spectrum, indicating counts against channels.  The counts rise from 0 to around 4000 by channel 17, then decrease slightly until a discontinuous jump by channel 60, labeled "???".  The counts follow a "U"-shape to around channel 165, where they precipitously drop off; this is labeled "Compton edge."  Around channel "234", there is sudden spike in counts, labeled "Photopeak", followed by a permanent drop-off to O(10) counts within 20 channels; the count numbers continue to decrease until around 500, where they reach 0 permanently. Here, channels are unit internal to the software; from calibration with known photopeaks, we have roughly $E=(2.734\text{ keV})n+21.9\text{ keV}$ where $n$ is the channel number and $E$ the corresponding energy.

Some calculations

In units where the energy of the incident gammas is $1$, the energy of a Compton scattering emission is given by Compton's formula $$E=1-\left(1+\frac{1-\cos{\theta}}{E_e}\right)^{-1}$$ where $E$ is the energy of the emitted photon, $E_e$ is the rest energy of the electron, and $\theta$ the change in the angle of the electron's velocity over the course of the scattering event. [Edit: This seems to have cause some confusion. By measuring energies units of "incident gamma energy," all energies henceforth are really the ratio of that energy to the energy scale of the incident gamma.] Thus the energy scale of each low-energy cutoff has a corresponding scattering angle. For the $661.6$ keV Cs-137 photopeak above, the cutoff corresponds to about $46^{\circ}$; for Ba-133 ($1.1732$ MeV) it's $26^{\circ}$, and for Na-22 ($511$ keV) it's $60^{\circ}$. Because these are very different numbers, I don't think the cause is geometric (like, say, the angular acceptance of the detector).

The cutoff isn't intrinsic to the quantum mechanics of scattering either. The probability of each scattering angle is given by the Klein-Nishina formula $$P\propto\left(1+\frac{1-\cos{\theta}}{E_e}\right)^{-2}\left(\frac{1}{1+\frac{1-\cos{\theta}}{E_e}}+\frac{1-\cos{\theta}}{E_e}+\cos^2{\theta}\right)$$ From this equation, one would expect the intensity of the Compton spectrum should be a cubic in the energy observed: $$I\propto(1-E)^3+(E_e^2+2E_e)(1-E)^2+(1-2E_e-2E_e^2)(1-E)+E_e^2$$ Obviously, cubics do not exhibit discontinuities.


Your question marked peak - ??? - corresponds to the secondary photon energy for a 180 degree backscatter Compton event. The Compton edge you have marked corresponds to the initial maximum energy transfer to an electron in the detector. The secondary photon then may or may not escape the detector. That low energy bump (and edge) is the secondary photon interaction, via photoelectric interaction, in the detector but outside the processing electronics time window for a single energy "count."

Edit: You have a problem with the formula you are using. Because the $(1-\cos\theta)$ term is multiplied by the ratio of the primary photon energy to the electron mass energy, then added to a "$1$", you can't simply plug in $1$ for the primary photon energy. You need to do some accurate algebra to extract $E_{\gamma}$. That's why your angle numbers don't work.

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  • $\begingroup$ Do you mean that the "???" peak is the photopeak from the photons re-emitted following Compton scattering? $\endgroup$ – Jacob Manaker May 25 '19 at 20:12
  • $\begingroup$ Yes, that's what I said. See the edit about your calculations. $\endgroup$ – Bill N May 25 '19 at 20:21
  • $\begingroup$ I don't agree. The "Compton edge" peak is the photopeak from photons re-emitted following scattering. The curve between the two markings is fit well by the cubic in the problem statement ($\chi^2/N=3.37$ with $E_e$ as a fixed parameter and assuming Poissonian uncertainty); it's just cut off at those two edges. $\endgroup$ – Jacob Manaker May 25 '19 at 20:29
  • $\begingroup$ The $1$ and the ratio are both dimensionless, so setting the primary photon energy to be my energy unit shouldn't affect the results. $\endgroup$ – Jacob Manaker May 25 '19 at 20:30
  • $\begingroup$ No. The Compton edge corresponds to the energy of the maximum electron energy gained in an event. That means the secondary photon is scattered at 180 degrees. The secondary photon then has a low energy and may or may not escape. The region between the Compton edge and the backscatter peak is the region where secondary photons scattered at medium angles are displayed. And your algebra is wrong! You cannot throw a 1 in there. Do the math with a real energy and see if you get corresponding results. Plus, I did the calculation for Cs-137 and it matches the spectrum. $\endgroup$ – Bill N May 25 '19 at 20:38

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