GATE PH 2013 • Q48 MCQ • 2 Marks

Section 5: Thermodynamics and Statistical Physics

Statistical Ensembles & Partition Functions

Subtopic: Partition Function & Thermodynamic Connection

Question Statement:

There are four energy levels E, 2E, 3E and 4E (where E > 0). The canonical partition function of two identical fermions is (A) e^(-5βE) (1 + e^(-βE) + e^(-2βE) + e^(-3βE) + e^(-4βE)) (B) e^(-5βE) (e^(2βE) + e^(βE) + 2 + e^(-βE) + e^(-2βE)) (C) (e^(-βE) + e^(-2βE) + e^(-3βE) + e^(-4βE))^2 (D) e^(-7βE) (1 + e^(-βE) + e^(-2βE))

Options:

  • (A) e^(-5βE) (1 + e^(-βE) + e^(-2βE) + e^(-3βE) + e^(-4βE))
  • (B) e^(-5βE) (e^(2βE) + e^(βE) + 2 + e^(-βE) + e^(-2βE))
  • (C) (e^(-βE) + e^(-2βE) + e^(-3βE) + e^(-4βE))^2
  • (D) e^(-7βE) (1 + e^(-βE) + e^(-2βE))

Official Answer Key:

B

Step-by-Step Solution & Explanation

### **Official Solution & Step-by-Step Derivation (GATE PH 2013 Q.48)** **1. Statistical Physics of Identical Fermions**: - Four single-particle energy states: $\epsilon_1 = E, \epsilon_2 = 2E, \epsilon_3 = 3E, \epsilon_4 = 4E$. - For two identical fermions, no two particles can occupy the same quantum state (Pauli exclusion principle). The distinct 2-particle configurations are the $\binom{4}{2} = 6$ pairs: 1. $\{E, 2E\} \implies E_1 = 3E$ 2. $\{E, 3E\} \implies E_2 = 4E$ 3. $\{E, 4E\} \implies E_3 = 5E$ 4. $\{2E, 3E\} \implies E_4 = 5E$ 5. $\{2E, 4E\} \implies E_5 = 6E$ 6. $\{3E, 4E\} \implies E_6 = 7E$ **2. Canonical Partition Function Evaluation**: - Summing Boltzmann factors over all microstates: $Z_{\text{fermions}} = e^{-3\beta E} + e^{-4\beta E} + 2e^{-5\beta E} + e^{-6\beta E} + e^{-7\beta E}$ - Factoring out $e^{-5\beta E}$: $Z_{\text{fermions}} = e^{-5\beta E} \left( e^{2\beta E} + e^{\beta E} + 2 + e^{-\beta E} + e^{-2\beta E} \right)$ - This corresponds precisely to **Option (B)**. **3. Official Verification**: The verified official IIT Bombay Master Answer is **Option (B)**.

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