 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-3/section-f/46/b/ii/i/solution

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Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 3 / Section F / 46 / b / ii / I / Solution by

Codex  0 2026-10-20

Take

s = x - x^{‡}

with

0 < x^{‡} < a

. A trajectory crossing to the right at the barrier top goes downhill on exit and cannot return in this model. Its initial Boltzmann weight contains

e^{- β V_{0}}

; negative-velocity trajectories are on the reactant side at the commitment time. Therefore

C_{f s}^{top} = \frac{e ^{- β V_{0}}}{2 π ℏ} \int_{0}^{\infty} \frac{p}{m} e^{- β p^{2} / (2 m)} d p = \frac{e ^{- β V_{0}}}{2 π ℏ β} .

(1)

Dividing by the same

Q_{r}

gives the result in the energy-integral calculation. No width appears in the classical rate because every classically allowed trajectory passes through the flat barrier.

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