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Question 42
GATE PH 2025 ā¢ Q42
MCQ ā¢ 2 Marks
Section 5: Thermodynamics and Statistical Physics
Statistical Ensembles & Partition Functions
Subtopic: Microcanonical, Canonical & Grand Canonical Ensembles
Question Statement:
Consider a two -level system with energy states +š and - š. The number of
particles at +š level is N+ and the number of particles at - š level is N-. The total
energy of the system is E and the total number of particles is N = N+ + N-. In the
thermodynamic limit, the inverse of the absolute temperature of the system is
(Given: ln N! ā N ln N - N)
Options:
(A) kB
2š ln [
N-E
š
N+ E
š
]
(B) kB
š lnN
(C) kB
2š lnN
(D) kB
š ln [
N-E
š
N+ E
š
]
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.42)**
**1. Statistical Mechanics of a Two-Level System**:
- Energy states: $+\epsilon$ and $-\epsilon$.
- Total particles: $N = N_+ + N_-$
- Total energy: $E = N_+(+\epsilon) + N_-(-\epsilon) = (N_+ - N_-)\epsilon$
- Solving for populations $N_+$ and $N_-$:
$N_+ - N_- = \frac{E}{\epsilon}, \quad N_+ + N_- = N$
$N_+ = \frac{1}{2}\left(N + \frac{E}{\epsilon}\right), \qquad N_- = \frac{1}{2}\left(N - \frac{E}{\epsilon}\right)$
**2. Microcanonical Entropy & Temperature**:
- Number of microstates:
$\Omega(N, E) = \frac{N!}{N_+! N_-!}$
- Applying Stirling's approximation ($S = k_B \ln \Omega$):
$S = k_B [N \ln N - N_+ \ln N_+ - N_- \ln N_-]$
- Thermodynamic temperature definition $\frac{1}{T} = \left(\frac{\partial S}{\partial E}\right)_N$:
$\frac{1}{T} = \frac{\partial S}{\partial N_+} \frac{\partial N_+}{\partial E} + \frac{\partial S}{\partial N_-} \frac{\partial N_-}{\partial E}$
$\frac{\partial N_+}{\partial E} = \frac{1}{2\epsilon}, \qquad \frac{\partial N_-}{\partial E} = -\frac{1}{2\epsilon}$
$\frac{\partial S}{\partial N_+} = -k_B(\ln N_+ + 1), \qquad \frac{\partial S}{\partial N_-} = -k_B(\ln N_- + 1)$
- Substituting derivatives:
$\frac{1}{T} = -\frac{k_B}{2\epsilon}\ln N_+ + \frac{k_B}{2\epsilon}\ln N_- = \frac{k_B}{2\epsilon} \ln\left(\frac{N_-}{N_+}\right)$
- Expressing in terms of $N$ and $E$:
$\frac{1}{T} = \frac{k_B}{2\epsilon} \ln\left[\frac{\frac{1}{2}(N - E/\epsilon)}{\frac{1}{2}(N + E/\epsilon)}\right] = \frac{k_B}{2\epsilon} \ln\left[\frac{N - E/\epsilon}{N + E/\epsilon}\right]$
**3. Official Verification**:
The verified official IIT answer is **Option (A)** (`kB / (2š) ln[(N - E/š) / (N + E/š)]`).
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