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: Understanding the Concept:
Each of the \(N\) classical (distinguishable, non-interacting) particles can sit in either of two energy levels, \(\varepsilon_1\) or \(\varepsilon_2 = \varepsilon_1+\Delta\). At temperature \(T\), a single particle's average energy comes from the ordinary canonical (Boltzmann) weighting of these two levels, this is the classic "two-level system."

Step 2: Key Formula or Approach:
The single particle partition function is \(z = e^{-\beta\varepsilon_1}+e^{-\beta\varepsilon_2}\), with \(\beta=1/(k_BT)\). The average energy per particle is \(\langle\varepsilon\rangle = -\dfrac{\partial \ln z}{\partial\beta}\), and the total internal energy is \(U(T) = N\langle\varepsilon\rangle\).

Step 3: Detailed Explanation:
Factor out \(e^{-\beta\varepsilon_1}\): \(z = e^{-\beta\varepsilon_1}(1+e^{-\beta\Delta})\), so \(\ln z = -\beta\varepsilon_1+\ln(1+e^{-\beta\Delta})\).
\[ \langle\varepsilon\rangle = -\frac{\partial\ln z}{\partial\beta} = \varepsilon_1+\frac{\Delta e^{-\beta\Delta}}{1+e^{-\beta\Delta}} = \varepsilon_1+\frac{\Delta}{e^{\beta\Delta}+1} \]
\[ U(T) = N\left[\varepsilon_1+\frac{\Delta}{e^{\Delta/k_BT}+1}\right] \]
Now look at the two extreme temperatures. As \(T\to0\), \(\beta\to\infty\) so \(e^{\beta\Delta}\to\infty\) and the correction term vanishes, giving \(U\to N\varepsilon_1\): every particle sits frozen in the lower level.
As \(T\to\infty\), \(\beta\to0\) so \(e^{\beta\Delta}\to1\) and the correction term saturates at \(\Delta/2\), giving \(U\to N(\varepsilon_1+\Delta/2)\): both levels become equally populated and \(U\) stops changing.
In between, \(U\) rises smoothly from the low plateau to the high plateau as \(k_BT\) becomes comparable to \(\Delta\), tracing out an S-shaped (sigmoid) curve that flattens at both ends.

Step 4: Why the other options are wrong.
Option (B) keeps rising without any upper limit, but \(U\) is physically bounded above by \(N(\varepsilon_1+\Delta/2)\) once the two levels are equally populated, it cannot increase forever.
Option (C) shows \(U\) decreasing with \(T\), but heating a two-level system can only push more particles into the higher level, never fewer, so the internal energy cannot fall as \(T\) rises.
Option (D) also grows without bound and skips the flat low-temperature plateau altogether.

Final Answer:
\(U(T)\) starts flat near \(N\varepsilon_1\), rises through an S-shaped transition, and levels off at \(N(\varepsilon_1+\Delta/2)\), this is exactly the shape in graph (A). \[ \boxed{U(T):\ N\varepsilon_1 \to N\left(\varepsilon_1+\frac{\Delta}{2}\right),\ \text{sigmoid shape, option (A)}} \]