Discussion
Consider a two-level system with energy states \(+\varepsilon\) and \(-\varepsilon\). The number of particles at \(+\varepsilon\) level is \(N_{+}\) and the number of particles at \(-\varepsilon\) 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: \(\mathrm{ln}\, N! \simeq N\mathrm{ln}\, N - N\))
Answer: A
The answer is \(\frac{k_{B}}{2\varepsilon}\ln[\frac{N-E/\varepsilon}{N+E/\varepsilon}]\).
From \(N = N_{+} + N_{-}\) and \(E = \varepsilon(N_{+} - N_{-})\), \(N_{\pm} = \frac{1}{2}(N \pm \frac{E}{\varepsilon})\). The entropy, with Stirling's formula, is
\[ \begin{aligned} S = k_{B}[&N\ln N - N_{+}\ln N_{+} \\ &- N_{-}\ln N_{-}] \end{aligned} \]
Since \(\frac{dN_{\pm}}{dE} = \pm\frac{1}{2\varepsilon}\):
\[ \begin{aligned} \frac{1}{T} = \frac{\partial S}{\partial E} &= \frac{k_{B}}{2\varepsilon}(\ln N_{-} - \ln N_{+}) \\ &= \frac{k_{B}}{2\varepsilon}\ln\frac{N - E/\varepsilon}{N + E/\varepsilon} \end{aligned} \]
Option (D) misses the factor \(\frac{1}{2}\) from \(\frac{dN_{\pm}}{dE}\), and the \(\ln N\) options do not depend on \(E\), so they cannot give a temperature that changes with energy. For \(E \gt 0\) the result is negative, the familiar negative temperature of a population inversion.
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