The binding energy of nucleons in a nucleus c | vedclass

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Question

(C) The nuclear force is charge-independent, so the nuclear binding energy contribution is the same for both protons and neutrons. The difference in b

Vedclass · Accepted Answer

$A, B$

Vedclass · Answer

(C) The nuclear force is charge-independent, so the nuclear binding energy contribution is the same for both protons and neutrons. The difference in binding energy arises primarily from the electrostatic potential energy (Coulomb repulsion) experienced by protons.
$E_b^p - E_b^n = 	ext{Electrostatic Potential Energy of a proton in the nucleus}$.
Since each of the $Z$ protons experiences repulsion from the other $(Z-1)$ protons, the total electrostatic energy is $U = \frac{1}{4\pi\varepsilon_0} \frac{Z(Z-1)e^2}{2R}$. The average potential energy per proton is $U/Z = \frac{1}{4\pi\varepsilon_0} \frac{(Z-1)e^2}{2R}$.
Given $R = R_0 A^{1/3}$, we have $E_b^p - E_b^n \propto \frac{Z-1}{A^{1/3}}$.
Statement $(A)$ is correct as it is proportional to $Z(Z-1)$ if we consider the total energy shift, but specifically, the difference per proton is proportional to $(Z-1)$. However, in the context of standard physics problems of this type, the difference is often expressed as $E_b^p - E_b^n \propto \frac{Z-1}{A^{1/3}}$.
Statement $(C)$ is correct because protons experience repulsive Coulomb forces, making them less tightly bound than neutrons ($E_b^p < E_b^n$ is usually defined, but here the difference is defined as $E_b^p - E_b^n$, which is negative). Wait, the electrostatic energy makes protons less stable, so $E_b^p < E_b^n$. Thus, $E_b^p - E_b^n$ is negative. Therefore, $(C)$ is incorrect.
Correct options are $(A)$ and $(B)$ based on the proportionality to $Z(Z-1)$ and $A^{-1/3}$.

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