The binding energy of nucleons in a nucleus can be affected by the pairwise Coulomb repulsion. Assume that all nucleons are uniformly distributed inside the nucleus. Let the binding energy of a proton be $$E_b^p$$ and the binding energy of a neutron be $$E_b^n$$ in the nucleus.

Which of the following statement(s) is(are) correct?

The extra effect that distinguishes a proton from a neutron inside a nucleus is the electrostatic (Coulomb) repulsion among protons. All other (strong-interaction) pieces of the binding energy are the same for both nucleons.

1. Coulomb energy of a uniformly charged nucleus
For a uniformly charged sphere of radius $$R$$ containing $$Z$$ protons, the total Coulomb potential energy is

$$U_C \;=\;\frac{3}{5}\,\frac{1}{4\pi\varepsilon_0}\, \frac{Z(Z-1)e^{2}}{R}$$

(the factor $$Z(Z-1)$$ counts the number of distinct proton pairs). The nuclear radius follows the empirical relation

$$R \;=\; r_0\,A^{1/3}$$

so

$$U_C \;=\;\frac{3e^{2}}{20\pi\varepsilon_0 r_0}\; Z(Z-1)\;A^{-1/3}$$

Hence the Coulomb contribution is proportional to $$Z(Z-1)$$ and also to $$A^{-1/3}$$.

2. Difference between proton and neutron binding energies
A neutron carries no charge, so its binding energy is unaffected by the Coulomb term. For a proton, the repulsive energy reduces its binding. Therefore

$$E_b^{\,p}-E_b^{\,n}\;=\;-\,k\,Z(Z-1)\,A^{-1/3}$$

where $$k=\dfrac{3e^{2}}{20\pi\varepsilon_0 r_0}$$ is a positive constant. Thus the magnitude of the difference is indeed proportional to $$Z(Z-1)$$ and to $$A^{-1/3}$$, but the sign is negative (a proton is less tightly bound than a neutron).

3. Checking each statement

Case A: $$E_b^{\,p}-E_b^{\,n} \propto Z(Z-1)$$ True — the proportionality appears directly from the expression above. Case B: $$E_b^{\,p}-E_b^{\,n} \propto A^{-1/3}$$ True — the same expression also contains the factor $$A^{-1/3}$$. Case C: $$E_b^{\,p}-E_b^{\,n}$$ is positive False — the calculated sign is negative. Case D: $$E_b^{\,p}$$ increases if the nucleus emits a positron (β+ decay) In β+ decay a proton converts to a neutron, so $$Z$$ decreases by 1. The Coulomb repulsion term becomes smaller, which reduces the penalty on the remaining protons; consequently the binding energy of a proton becomes larger. Hence the statement is true.

Therefore the correct options are:
Option A, Option B and Option D.