I Quasi-Elastic Scattering Peak-Shift

[deuteron]

TL;DR

Clarifying the 8 MeV shift due to the binding energy of the proton in the nucleus

We were told in my particle physics lectures that in a quasi-elastic scattering, in a diagram of scattered electron energy $E'$-counts corresponding to an electron-nucleus scattering experiment at a fixed detector angle $\theta$, we would have a peak corresponding to the electrons that have scattered elastically off of the nucleus, and then another peak that is 8 MeV shifted to lower energies due to the binding energy of the proton in the nucleus. Just to be sure, the 8 MeV shift is happening from an elastic-scattering-off-of-free-electrons-peak that is not visible in the diagram, and not from the elastic-scattering-off-of-the-nucleus-peak, since the bound proton has a much lower energy than the whole nucleus, right?

 

[osilmag]

According to binding energy equation, Binding Energy, the bound proton would have a lower energy than the total nucleus. Are you asking if your electrons are breaking apart bound protons or losing energy to bound protons?