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ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2021/paper-2/section-c/c9/d/solution

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Past exam of the physics B course of the University of Cambridge / 2021 / Paper 2 / Section C / C9 / d / Solution by

Codex  0 2026-10-18

At low temperature,

T ≪ θ_{r}

, retain the ground level and the first excited level. Their energies are

0

and

ℏ^{2} / I

, with degeneracies

1

and

3

. Therefore

Z_{rot} = 1 + 3 e^{- ℏ^{2} / (I k_{B} T)} + O (e^{- 3 ℏ^{2} / (I k_{B} T)}),

so in the notation of the question,

A = 3, B = \frac{ℏ ^{2}}{I k _{B}} = 2 θ_{r} .

A

is dimensionless and

B

has units of kelvins. The first omitted level has three times the first excitation energy, explaining the exponent in the error term.

At high temperature,

T ≫ θ_{r}

, many rotational levels are occupied and the sum can be replaced to leading order by an integral:

Z_{rot} ≃ \int_{0}^{\infty} (2 J + 1) e^{- (θ_{r} / T) J (J + 1)} dJ .

u = J (J + 1)

, so that

d u = (2 J + 1) dJ

. Then

Z_{rot} ≃ \int_{0}^{\infty} e^{- (θ_{r} / T) u} d u = \frac{T}{θ _{r}} = \frac{2 I k _{B}}{ℏ ^{2}} T .

Thus

C = \frac{2 I k _{B}}{ℏ ^{2}} = \frac{1}{θ _{r}}, [C] = K^{- 1} .

The high-temperature approximation here concerns the rotational spectrum. For a real molecule it should still be used below temperatures where neglected vibrational excitation or dissociation invalidates the rigid-rotor model.

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