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?

Total binding energy (without considering repulsions),
$E_b=\left[Z{m}_p+\left(A-Z\right)m_n-m_x\right]c^2$ 
Where,${ }_Z^{A}X$   is the nuclei under consideration. Now, considering repulsion: Number of poton pairs ${=}^Z{C}_2$  
$\Rightarrow$This repulsion energy  $\infty \frac{Z\left(Z-1\right)}{2}\times \frac{1}{4\pi {\in }_0}\frac{e^2}{R}$
Where, R is the radius of the nucleus. 
$\Rightarrow E_b^p-E_b^n\infty Z\left(Z-1\right)$ 
$\therefore$  There will be no repulsion term for neutrons.
Also, since  $R=R_0{A}^{\frac{1}{3}}$
$\Rightarrow E_b^p-E_b^n\infty A^{\frac{1}{3}}$ 
Because of repulsion among protons,  $E_b^p<E_b^n$
Since, in $\beta +$  decay, number of protons decrease
$\Rightarrow$  repulsion would decrease
$\Rightarrow$$E_b^p$    increases