Question

A gas of \(N\) classical particles that can occupy energy levels \(\epsilon_1\) and \(\epsilon_2 = \epsilon_1+\Delta\) is in equilibrium with a reservoir at temperature \(T\). From the schematics shown below, choose the correct dependence of the internal energy \(U\) on \(T\).

• A. A curve that starts flat at a low value for small \(T\), rises smoothly through an S-shaped (sigmoid) transition, and saturates to a higher flat value at large \(T\).
• B. A curve that starts at zero and keeps rising with an ever-increasing slope as \(T\) grows, with no upper flat limit.
• C. A curve that starts high at small \(T\) and falls off smoothly to a lower flat value as \(T\) increases.
• D. A curve that starts at a small nonzero value and keeps rising with an ever-increasing slope as \(T\) grows, with no upper flat limit.

Hint:

A two-level system's average energy per particle is bounded between \(\epsilon_1\) (all particles in the ground level) and \(\epsilon_1+\Delta/2\) (levels equally populated), and it rises smoothly between the two.

Answer 1

The Correct Option is A

Step 1: Work with occupation fractions instead of the partition function.
Let $p_1$ and $p_2$ be the fraction of the $N$ particles sitting in levels $\epsilon_1$ and $\epsilon_2$ at temperature $T$. Detailed balance for a classical gas in contact with a reservoir gives the Boltzmann ratio
\[ \frac{p_2}{p_1} = e^{-\Delta/k_BT} \]

Step 2: Turn the ratio into normalized fractions.
Since $p_1+p_2=1$, substituting $p_2 = p_1 e^{-\Delta/k_BT}$ gives
\[ p_1 = \frac{1}{1+e^{-\Delta/k_BT}}, \qquad p_2 = \frac{e^{-\Delta/k_BT}}{1+e^{-\Delta/k_BT}} \]

Step 3: Build $U(T)$ from the populations.
The internal energy is just the population-weighted average energy times $N$:
\[ U(T) = N(p_1\epsilon_1+p_2\epsilon_2) = N\epsilon_1 + Np_2\Delta = N\epsilon_1+\frac{N\Delta\, e^{-\Delta/k_BT}}{1+e^{-\Delta/k_BT}} \]
which is the same expression as before, written from the populations rather than from $-\partial\ln z/\partial\beta$.

Step 4: Read off the low- and high-temperature behavior.
At low $T$, $e^{-\Delta/k_BT}\to0$, so $p_2\to0$: almost every particle stays in the ground level and $U\to N\epsilon_1$, a flat plateau.
At high $T$, $e^{-\Delta/k_BT}\to1$, so $p_1\to p_2\to\tfrac12$: the levels become equally populated and $U\to N(\epsilon_1+\Delta/2)$, a second flat plateau.
Between these limits $p_2$ climbs smoothly from 0 to $\tfrac12$, so $U(T)$ rises smoothly between the two plateaus, an S-shaped curve, never overshooting the upper plateau and never decreasing.

Step 5: Rule out the wrong shapes.
Curves that keep growing without bound (options B, D) contradict $p_2$ saturating at $\tfrac12$. A curve that decreases with $T$ (option C) contradicts $p_2$ only ever increasing as $T$ rises.

Final Answer:
The population fractions saturate, so $U(T)$ must saturate too, rising from $N\epsilon_1$ to $N(\epsilon_1+\Delta/2)$ in a sigmoid shape, which is graph (A). \[ \boxed{U(T):\ N\epsilon_1 \to N\left(\epsilon_1+\frac{\Delta}{2}\right),\ \text{option (A)}} \]