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_bⁿ in the nucleus.

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

Correct Options:
1. E_b^p − E_bⁿ is proportional to Z(Z − 1) where Z is the atomic number of the nucleus.
2. E_b^p − E_bⁿ is proportional to A^−1/3 where A is the mass number of the nucleus.
3. E_b^p increases if the nucleus undergoes a beta decay emitting a positron.

Detailed Explanation:

1. Coulomb Self-Energy of the Nucleus:
Assuming nuclear forces are charge-independent, the binding energy difference between protons and neutrons arises purely due to the pairwise electrostatic (Coulomb) repulsion among protons inside the nucleus.
If Z protons are uniformly distributed in a sphere of radius $R$, the total electrostatic potential energy $U_c$ stored in the nucleus due to pairwise interactions of $Z$ protons is given by:

$U_c=\frac{3}{5}\frac{Z(Z-1)e^2}{4\pi {\epsilon }_{0}R}$

2. Difference in Binding Energy (E_b^p − E_bⁿ):
The removal of a proton decreases the Coulomb repulsion energy of the nucleus, whereas the removal of a neutron does not directly change the electrostatic energy.
Therefore, the binding energy of a proton $E_b^p$ is lower than that of a neutron $E_b^n$ by an amount equal to the change in electrostatic energy when a proton is added/removed:

$E_b^p-E_b^n\∝-\frac{\ⅆU_c}{\ⅆZ}\∝-\frac{Z(Z-1)}{R}$

Since nuclear radius $R=R_0{A}^{1/3}$, we have:

$E_b^p-E_b^n\∝\frac{Z(Z-1)}{A^{1/3}}=Z(Z-1)A^{-1/3}$

Hence:
• $E_b^p-E_b^n$ is proportional to $Z(Z-1)$.
• $E_b^p-E_b^n$ is proportional to $A^{-1/3}$.

3. Effect of Positron (β⁺) Decay:
During positron emission (${\beta }^{+}$ decay), a proton converts into a neutron:

$p\→n+e^{+}+{\nu }_e$

This process decreases the atomic number $Z$ to $Z-1$ while keeping the mass number $A$ constant. As the number of protons decreases, the total Coulomb repulsive energy $U_c$ inside the nucleus decreases. Consequently, the remaining protons become more tightly bound, which means $E_b^p$ increases.