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ID: past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2013/paper-1/1/b/solution

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Past exam of the thermal and statistical physics course of the University of Cambridge / 2013 / Paper 1 / 1 / b / Solution by

Codex  0 2026-10-20

Let

Δ = ℏΩ

and choose the nondegenerate energy levels as 0 and

Δ

. For N independent identical units,

Z = (1 + e^{- β Δ})^{N}

; set N=1 for a single two-level system. Differentiating its logarithm gives

U = \frac{N Δ}{1 + e ^{Δ/ (k_{B} T)}}, C_{V} = N k_{B} \frac{x ^{2} e ^{x}}{( 1 + e ^{x} ) ^{2}}, x = \frac{Δ}{k _{B} T} .

At low temperature,

U \sim N Δ e^{- x}

and

C_{V} \sim N k_{B} x^{2} e^{- x}

: the upper level is frozen out. At high temperature, the two levels approach equal populations, giving

U \to N Δ/2

while

C_{V} \sim N k_{B} x^{2} /4 \to 0

. The energy therefore rises monotonically to a plateau, while the heat capacity has a Schottky peak:

k_{B} T_{peak} ≃ 0.417Δ, C_{V, peak} ≃ 0.439 N k_{B} .

Unequal degeneracies would change the populations and peak; the usual equal-degeneracy model is stated here. Shifting both energy levels changes the energy baseline but not heat capacity.

Figure 1.
Original equal-degeneracy two-level internal-energy and heat-capacity curves, with the high-temperature plateau and Schottky peak marked
. Details.

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