A system of f ive identical, non -interacting particles with mass m and spin
3
2 is
confined to a one-dimensional potential well of length L. If the lowest energy of the
system is N
𝜋2\hbar2
2mL2, the value of N (in integer) is _____
Official Answer Key:
8 to 8
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.63)**
**1. Statistical Quantum Mechanics of Fermions**:
- System of $5$ identical non-interacting particles of mass $m$ and spin $s = 3/2$ in a 1D potential well of length $L$.
- Orbital single-particle energy eigenvalues:
$E_n = n^2 E_1, \quad \text{where } E_1 = \frac{\pi^2 \hbar^2}{2mL^2}$
**2. Pauli Exclusion Principle & Ground State Filling**:
- Spin degeneracy for spin $s = 3/2$:
$g_s = 2s + 1 = 2\left(\frac{3}{2}\right) + 1 = 4 \text{ spin states per spatial level}$
- Because the particles have half-integer spin ($s = 3/2$), they are **fermions** obeying the Pauli exclusion principle. At most $4$ particles can occupy any single spatial level $n$.
- Ground state configuration for $5$ particles:
- Level $n = 1$: Accommodates $4$ particles $\implies E(n=1) = 4 \times (1^2 E_1) = 4 E_1$.
- Level $n = 2$: Accommodates remaining $1$ particle $\implies E(n=2) = 1 \times (2^2 E_1) = 4 E_1$.
- Total ground state energy of the 5-particle system:
$E_{\text{total}} = 4 E_1 + 4 E_1 = 8 E_1 = 8 \left(\frac{\pi^2 \hbar^2}{2mL^2}\right)$
- Comparing with $N \frac{\pi^2 \hbar^2}{2mL^2}$ yields:
$N = 8$
**3. Official Verification**:
The verified official IIT answer is **8** (Numerical range: `8 to 8`).
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