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?

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$E_{b}^{p} - E_{b}^{n}$ is positive.

$E_{b}^{p} - E_{b}^{n}$ is proportional to $Z$ $(Z - 1)$ , where $Z$ is the atomic number of the nucleus.

$E_{b}^{p}$ increases, if the nucleus undergoes a beta decay emitting a positron.

$E_{b}^{p} - E_{b}^{n}$ is proportional to $A^{\frac{- 1}{3}}$ , where $A$ is the mass number of the nucleus.

answer is A, B, D.

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Detailed Solution

Total binding energy (without considering repulsions),
$E_{b} = [Z m_{p} + (A - Z) m_{n} - m_{x}] c^{2}$
Where, $_{Z}^{A} X$ is the nuclei under consideration. Now, considering repulsion: Number of poton pairs $=^{Z} C_{2}$
$\Rightarrow$ This repulsion energy $\infty \frac{Z (Z - 1)}{2} \times \frac{1}{4 π \in_{0}} \frac{e^{2}}{R}$
Where, R is the radius of the nucleus.
$\Rightarrow E_{b}^{p} - E_{b}^{n} \infty Z (Z - 1)$
$∴$ 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 $β +$ decay, number of protons decrease
$\Rightarrow$ repulsion would decrease
$\Rightarrow$ $E_{b}^{p}$ increases