Past exam of the particle and nuclear physics course of the University of Cambridge / 2015 / Paper 1 / 4 / ii

The mirror nuclei have interchanged neutron and proton numbers. Neglect their strong-interaction binding difference and describe them with the same uniformly charged sharp sphere of radius $R$. The nuclear Coulomb energy isChanging one neutron into a proton decreases the nucleon rest-mass sum by $m_n-m_p=1.294\mathrm{MeV}/c^2$, but raises the Coulomb energy. The diagram gives $(M_F-M_O)c^2=2.25\mathrm{MeV}$ for nuclear masses, soWith the standard electromagnetic conversion $\alpha \hslash c\simeq 1.44\mathrm{MeV}\mathrm{fm}$, the uniform-sphere Coulomb displacement energy estimate isThis $R$ is the uniform sphere's outer radius. If “charge radius” means the rms nuclear charge radius, the same distribution givesThe numerical result is an estimate within the charge-density model, not a model-independent radius measurement. A discrete-proton correction replaces $Z^2$ by $Z(Z-1)$, giving $\Delta [Z(Z-1)]=16$ and $R\simeq 3.90\mathrm{fm}$. The main answer above follows the literal continuum charged-sphere model. Strong isospin-breaking, nonuniform density and changes in the valence-nucleon distribution also limit the extraction.