# (What causes) the *other* Compton edge?

## 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?

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.

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.
• 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. May 25 '19 at 20:29
• The $1$ and the ratio are both dimensionless, so setting the primary photon energy to be my energy unit shouldn't affect the results. May 25 '19 at 20:30