Discussion
A particle can occupy either the ground state at E = 0 or an excited state at E \(\gt\) 0. At a temperature T, the probability of the particle being in the excited state is :
Answer: C
The probability is exp(−E/kBT) / (1 + exp(−E/kBT)).
By the Boltzmann distribution, each state's population is proportional to exp(−E/kBT), divided by the partition function, the sum of these factors over all states. With levels at 0 and E,
\[ \begin{aligned} q &= 1 + e^{-E/k_{B}T} \\ p_{\mathrm{exc}} &= \frac{e^{-E/k_{B}T}}{1 + e^{-E/k_{B}T}} \end{aligned} \]
Option D is the probability of the ground state, option B is the partition function itself (which exceeds 1, so it cannot be a probability), and 0 holds only at absolute zero.
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