The GATE Grind

GATE 2025 PH – Question 42

Thermodynamics and Statistical Mechanics · macrostates and microstates · 2 marks · Multiple choice

Consider a two-level system with energies $+\epsilon,-\epsilon$ and populations $N_+,N_-$. The total energy is $E$ and particle number is $N=N_++N_-$. In the thermodynamic limit, the inverse absolute temperature is (given $\ln N!\simeq N\ln N-N$)

  1. $\frac{k_B}{2\epsilon}\ln\frac{N-E/\epsilon}{N+E/\epsilon}$
  2. $\frac{k_B}{\epsilon}\ln N$
  3. $\frac{k_B}{2\epsilon}\ln N$
  4. $\frac{k_B}{\epsilon}\ln\frac{N-E/\epsilon}{N+E/\epsilon}$

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Correct answer: (A) $\frac{k_B}{2\epsilon}\ln\frac{N-E/\epsilon}{N+E/\epsilon}$

Explanation

The system has $N_+$ particles at energy $+\epsilon$ and $N_-$ at $-\epsilon$.

**Energy and number:**
$$N=N_++N_-,\qquad E=\epsilon(N_+-N_-)\;\Rightarrow\;N_\pm=\frac12\left(N\pm\frac E\epsilon\right).$$

**Entropy** (counting configurations): $S=k_B\ln\dfrac{N!}{N_+!N_-!}$, which with Stirling's approximation is
$$S=k_B\left[N\ln N-N_+\ln N_+-N_-\ln N_-\right].$$

**Temperature:** $\dfrac1T=\dfrac{\partial S}{\partial E}$. Using $\dfrac{\partial N_\pm}{\partial E}=\pm\dfrac1{2\epsilon}$:
$$\frac1T=k_B\left[-\frac{\ln N_+}{2\epsilon}+\frac{\ln N_-}{2\epsilon}\right]=\frac{k_B}{2\epsilon}\ln\frac{N_-}{N_+}=\frac{k_B}{2\epsilon}\ln\frac{N-E/\epsilon}{N+E/\epsilon}\quad(\text{option A}).$$

(For $E<0$, most particles are in the lower level and $T>0$.)