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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/a/ii/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 / a / ii / Solution by

Codex  0 2026-10-20

The probability current is

j = \frac{ℏ}{2 mi} (ψ^{*} ψ^{'} - ψ ψ^{*'}) = \frac{ℏ}{m} Im (ψ^{*} ψ^{'}) .

(1)

For the incident and transmitted travelling waves,

j_{in} = ℏ k ∣ A ∣^{2} / m

and

j_{tr} = ℏ k ∣ F ∣^{2} / m

. Thus $P (E) = ∣ F / A ∣^{2}$ here. If the asymptotic wave numbers differed, the ratio of outgoing to incoming velocities would also be required. Current conservation gives

P + ∣ B / A ∣^{2} = 1

for a real nonabsorbing barrier.

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