I'm doing a course in Nuclear Physics but I think it focuses more on the experimental side so not everything is that rigorous. In the notes, lecturer gives an argument for why we should use electrons to study the structure of a nucleus. It goes as follows:
Electrons are the most accurate and precise probe of nuclear sizes:
fundamental point-like particles
leptons - family of particles that do not experience the strong force
Therefore, they predominantly interact via the well understood electromagnetic force. If Δp is the momentum transferred to the recoiling nucleus then the spatial precision we can study the nucleus, Δr is given by Heisenberg's Uncertainty Relation: ∆𝒑∆𝒓~ħ hence ∆𝒑~ħ/∆𝒓.
I understand the two points he makes about point particles and leptons. Where I get lost is how he can relate the momentum transferred to the uncertainty in the momentum and then also the relation between the uncertainty in the position of the nucleus (because that's what delta r actually is) to the "structure" of the nucleus (which I'm guessing he means what it's made of i.e protons and neutrons).
Furthermore, we had a question in our tutorial where we were asked to "calculate the minimum beam energy required to study the sub-structure of the proton down to a spatial resolution of 0.05 fm using an electron beam". He works this out in a similar fashion:
"We need to use the position-momentum uncertainty principle ΔpΔx~ℏ where Δp ~pmin and Δx to be the spatial resolution required
p_min~ℏ/Δx~ℏc/0.05c~(197 "MeV" ⁄c)/0.05
Now since electron is highly relativistic:E≃pc~200⁄0.05 MeV~4.0GeV".
I don't see how delta p could be the minimum momentum.