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ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2024/paper-1/a3/b/iii/solution

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Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 1 / A3 / b / iii / Solution by

Codex  0 2026-10-20

The relevant allowed absorption is

1 \to 2

. Under the stipulated population-only intensity model, the vibrational levels are nondegenerate and

\frac{I _{12}}{I _{01}} = \frac{N _{1}}{N _{0}} = exp [- \frac{E _{1} - E _{0}}{k _{B} T}] = exp [- \frac{h c ( 440.6 c m ^{- 1} )}{k _{B} T}] .

Set this ratio to

0.10

and solve:

T = \frac{h c ( 440.6 c m ^{- 1} )}{k _{B} ln 10} = \frac{( 1.4387769 K cm ) ( 440.6 c m ^{- 1} )}{ln 10} ≃ 275.3 K .

The energy in the exponential is the measured

0 \to 1

gap, not the harmonic constant

444.3 c m^{- 1}

. At this temperature,

N_{2} / N_{0} ≃ 0.0102

, so considering the first thermally excited manifold is sensible. Actual band intensity ratios also involve transition moments and rotational structure; the calculation follows the question's simplifying assumption.

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