QB_PH_SEC9_016: Nuclear Models & Decays

According to the liquid drop model, the fissility parameter condition for a heavy nucleus to undergo spontaneous symmetric fission without an activation barrier is $\frac{Z^2}{A} \ge \left(\frac{Z^2}{A}\right)_c$. What is the theoretical critical value $\left(\frac{Z^2}{A}\right)_c$ taking surface parameter $a_s = 17.8\text{ MeV}$ and Coulomb parameter $a_c = 0.71\text{ MeV}$? (Round off to the nearest integer)

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Bohr-Wheeler Instability Criterion**: Under small quadrupole deformation $\epsilon$: - Surface area increases: $\Delta E_s = +\frac{2}{5} a_s A^{2/3} \epsilon^2$ - Coulomb repulsion decreases: $\Delta E_c = -\frac{1}{5} a_c \frac{Z^2}{A^{1/3}} \epsilon^2$ 2. **Net Energy Change**: $\Delta E = \Delta E_s + \Delta E_c = \epsilon^2 \left[\frac{2}{5} a_s A^{2/3} - \frac{1}{5} a_c \frac{Z^2}{A^{1/3}}\right]$ 3. **Condition for Spontaneous Instability** ($\Delta E < 0$): $\frac{1}{5} a_c \frac{Z^2}{A^{1/3}} > \frac{2}{5} a_s A^{2/3} \implies \frac{Z^2}{A} > \frac{2 a_s}{a_c}$ 4. **Numerical Evaluation**: - $a_s = 17.8\text{ MeV}$ - $a_c = 0.71\text{ MeV}$ $\left(\frac{Z^2}{A}\right)_c = \frac{2 \times 17.8}{0.71} = \frac{35.6}{0.71} \approx 50.14$ 5. **Nearest Integer**: 50.