Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.18)**
**1. Quantum Mechanical Principles & Magic Numbers in a 3D Box**:
- Energy levels of a particle in a 3D cubical box of side $L$:
$E(n_x, n_y, n_z) = \frac{\pi^2 \hbar^2}{2m L^2} \left(n_x^2 + n_y^2 + n_z^2\right), \quad n_x, n_y, n_z \in \{1, 2, 3, \dots\}$
- Nucleons have spin $s = 1/2$, so each spatial orbital accommodates $2$ identical nucleons (spin degeneracy $g_s = 2$).
- Magic numbers correspond to closed shell cumulative nucleon counts.
**2. Step-by-Step Shell Filling**:
- **Shell 1**: $(n_x, n_y, n_z) = (1, 1, 1)$
- $n_x^2 + n_y^2 + n_z^2 = 3$
- Spatial degeneracy $= 1$.
- Number of nucleons $= 1 \times 2 = 2$.
- **Cumulative count = 2**.
- **Shell 2**: $(2, 1, 1), (1, 2, 1), (1, 1, 2)$
- $n_x^2 + n_y^2 + n_z^2 = 6$
- Spatial degeneracy $= 3$.
- Number of nucleons $= 3 \times 2 = 6$.
- **Cumulative count = 2 + 6 = 8**.
- **Shell 3**: $(2, 2, 1), (2, 1, 2), (1, 2, 2)$
- $n_x^2 + n_y^2 + n_z^2 = 9$
- Spatial degeneracy $= 3$.
- Number of nucleons $= 3 \times 2 = 6$.
- **Cumulative count = 8 + 6 = 14**.
- **Shell 4**: $(3, 1, 1), (1, 3, 1), (1, 1, 3)$
- $n_x^2 + n_y^2 + n_z^2 = 11$
- Spatial degeneracy $= 3$.
- Number of nucleons $= 3 \times 2 = 6$.
- **Cumulative count = 14 + 6 = 20**.
- The first four magic numbers are **2, 8, 14, 20**.
**3. Official Verification**:
The verified official IIT Master Answer is **Option (C)** (`2, 8, 14, 20`).