QB_PH_SEC9_018: Nuclear Properties & Nuclear Forces

In the semi-empirical mass formula, the pairing energy term $\delta(A, Z)$ is

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Pairing Interaction**: Short-range attractive nuclear forces strongly favor pairing identical nucleons ($p-p$ and $n-n$) with antiparallel spins into singlet $J = 0$ configurations. 2. **Binding Energy Term Sign Convention**: $B(A, Z) = a_v A - a_s A^{2/3} - a_c \frac{Z^2}{A^{1/3}} - a_{\text{asym}} \frac{(A - 2Z)^2}{A} + \delta(A, Z)$ - **Even $Z$, Even $N$ (Even-Even)**: Both protons and neutrons are fully paired $\implies \delta = +\frac{a_p}{\sqrt{A}} > 0$. - **Odd $A$ (Even-Odd or Odd-Even)**: One unpaired nucleon $\implies \delta = 0$. - **Odd $Z$, Odd $N$ (Odd-Odd)**: Two unpaired nucleons $\implies \delta = -\frac{a_p}{\sqrt{A}} < 0$. 3. **Conclusion**: Positive for even-even, zero for odd-$A$, negative for odd-odd. Option (A) is correct.