Do you mean that the "???" peak is the photopeak from the photons re-emitted following Compton scattering?

Yes, that's what I said. See the edit about your calculations.

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.

The $1$ and the ratio are both dimensionless, so setting the primary photon energy to be my energy unit shouldn't affect the results.

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.

The backscatter secondary photon for the Cs-137 peak will be 184 keV, and the Compton edge will be about 477 keV. Those would be channels 59 and 167. And the full photopeak matches the proper channel, too, based on the energy calibration you stated.

Okay. Do you have more questions?

You seem to be right that this is a backscatter peak. The numbers work out on Na too.

Okay.

But I am going to have to disagree (strongly!) on the algebra.

What are you using for the electron energy in your formula?

511 keV. (I'm plugging in numbers and it's giving the same results 184/477 keV too, FWIW.)

Actually, that's not what you asked. Sorry, electron rest energy is .77 photopeak units.

Aha! So you must use the ratio of the photopeak energy the electron mass energy as the multiplier of the (1-cos$\theta$), 661/511, or because you are dividing, 0.77. If you have a different energy primary photon, you have to get a new divisor. I think you had a misunderstanding of what the Compton edge energy came from.

Exactly.

Thanks for arguing with me over this. I definitely didn't get it until three or four comments into the chain.

Will do.