Effect of Pairwise Coulomb Repulsion on Nucleon Binding Energy

To determine the correct statements, we analyze the effect of pairwise Coulomb repulsion on the binding energy of nucleons.

1. Electrostatic Energy of the Nucleus

Assuming that nucleons are uniformly distributed inside a spherical nucleus of radius $R = R_0 A^{1/3}$, the pairwise Coulomb repulsion energy among the $Z$ protons is given by: $U_C = \frac{3}{5} \frac{1}{4\pi\varepsilon_0} \frac{Z(Z-1)e^2}{R} = C \frac{Z(Z-1)}{A^{1/3}}$ where $C = \frac{3e^2}{20\pi\varepsilon_0 R_0}$ is a positive constant.

2. Binding Energy Expression

The total binding energy of a nucleus with atomic number $Z$ and mass number $A$ can be written as: $B(Z, A) = B_{\text{strong}}(Z, A) - U_C(Z, A)$ where $B_{\text{strong}}(Z, A)$ is the contribution from the strong nuclear force, which is charge-independent (i.e., symmetric with respect to protons and neutrons).

The binding energy (separation energy) of a proton $E_b^p$ and a neutron $E_b^n$ in the nucleus $(Z, A)$ are given by: $E_b^p = B(Z, A) - B(Z-1, A-1)$ $E_b^n = B(Z, A) - B(Z, A-1)$

3. Difference Between $E_b^p$ and $E_b^n$

Subtracting the neutron binding energy from the proton binding energy: $E_b^p - E_b^n = B(Z, A-1) - B(Z-1, A-1)$

Since the nuclear force is charge-independent, $B_{\text{strong}}(Z, A-1) = B_{\text{strong}}(Z-1, A-1)$. Therefore, the difference arises solely from the Coulomb potential energy: $E_b^p - E_b^n = -\left[ U_C(Z, A-1) - U_C(Z-1, A-1) \right]$

Substituting the expression for $U_C$: $U_C(Z, A-1) - U_C(Z-1, A-1) = \frac{C}{(A-1)^{1/3}} \left[ Z(Z-1) - (Z-1)(Z-2) \right]$ $= \frac{C}{(A-1)^{1/3}} (Z-1)\left[ Z - (Z-2) \right] = \frac{2C(Z-1)}{(A-1)^{1/3}}$

Thus, we obtain: $E_b^p - E_b^n = -\frac{2C(Z-1)}{(A-1)^{1/3}} \propto -\frac{Z-1}{A^{1/3}}$

4. Analysis of the Options

• Option A: $E_b^p - E_b^n$ is proportional to $(Z-1)$, not $Z(Z-1)$. Thus, Statement A is incorrect.
• Option B: $E_b^p - E_b^n$ is proportional to $A^{-\frac{1}{3}}$. Thus, Statement B is correct.
• Option C: Since $Z \ge 1$ and $C > 0$, $E_b^p - E_b^n = -\frac{2C(Z-1)}{(A-1)^{1/3}} \le 0$, which is negative. Thus, Statement C is incorrect.
• Option D: In positron ($\beta^+$) decay, a proton turns into a neutron ($Z \to Z-1$ while $A$ remains constant). The binding energy of a proton in the daughter nucleus $(Z-1, A)$ becomes: $E_b^{p \prime} = E_{\text{strong}} - \frac{2C(Z-2)}{A^{1/3}}$ Since $Z-2 < Z-1$, the negative Coulomb repulsion term decreases in magnitude, leading to $E_b^{p \prime} > E_b^p$. Thus, $E_b^p$ increases. Hence, Statement D is correct.

Conclusion

The correct options are B and D.