Why escape peak positions are calculated using emission energy and not absorbing energy?

Question

For a X-ray monocromatic source, escape peaks energy positions are described by the difference between the incident energy and the fluorescence ($K_{\alpha}$ for example), like $E_{Escape Peak} = E_0 - K_{\alpha}$.

My context is using hybrid CdTe detectors to create energy spectrums.

I don't think I fully comprehend the escape peaks mechanism, but so far I understand the emission energy is subtracted from the incident energy promoting a smaller cloud of charge to be detected that will result in the smaller peak being the escape peak. However it would make much more sense for me if $E_{Escape Peak}$ was calculated with the difference with the electron binding energy in the absortion process like $E_{Escape Peak} = E_0 - K_{1s}$.

Answer 1

An odd fact about ionizing radiation is that the average number of "ions" (possibly electrons or holes) produced in a uniform material is almost exactly proportional to the deposited energy, regardless of how that energy cascades down to ions and heat. The constant of proportionality depends on the material. For photoelectric absorption in a semiconductor, the initial absorption event makes a photoelectron and an excited atom. The excited atom de-excites by emitting fluorescent photons and/or Auger electrons. These, in turn produce more photons, excited atoms, and electrons of lower energy. The initial photoelectron does the same. Eventually the cascade terminates when all of the incident photon's energy has the form of electron-hole pairs and heat.

That is, unless some escapes. If a photon or electron leaves the sensitive volume of the detector, the energy it leaves with doesn't contribute to the cascade, so you get fewer electron-hole pairs than expected. Escaping electrons can have a range of energies, so they produce a "tail" in the spectrum to the low energy side of a spectral peak. Escaping fluorescent photons have specific energies, though. An escaping $K_\alpha$ photon takes away a specific energy. Thus, $K_\alpha$ escape produces an event with deposited energy $E_0-K_\alpha$.