QB_PH_SEC9_025: Nuclear Properties & Nuclear Forces

For a fixed odd mass number $A = 125$, the atomic number $Z_0$ of the most stable isobar in Green's approximation of the semi-empirical mass formula is given by $Z_0 = \frac{A}{2 + 0.015 A^{2/3}}$. Calculate $Z_0$. (Round off to the nearest integer)

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Valley of Beta Stability Condition**: $\left.\frac{\partial M(A, Z)}{\partial Z}\right|_A = 0 \implies Z_0 = \frac{A}{2 + \frac{a_c}{2 a_{\text{asym}}} A^{2/3}}$ 2. **Evaluation with Standard Empirical Coefficients**: $Z_0 = \frac{A}{2 + 0.015 A^{2/3}}$ 3. **Numerical Computation for $A = 125$**: $A^{1/3} = 125^{1/3} = 5.0$ $A^{2/3} = 5.0^2 = 25.0$ $\text{Denominator} = 2 + 0.015 \times 25.0 = 2 + 0.375 = 2.375$ $Z_0 = \frac{125}{2.375} = 52.6316$ 4. **Nearest Integer (Physical $Z$)**: 53 (Iodine-125).