deuteron
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- TL;DR Summary
- 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?