 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/section-b/8/c/solution

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Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Section B / 8 / c / Solution by

Codex  0 2026-10-18

For

E ≫ V_{0}

, the two wavenumbers nearly coincide:

k_{2} = k_{1} 1 - \frac{V _{0}}{E} ≃ k_{1} (1 - \frac{V _{0}}{2 E}) .

(1)

Consequently

B / A ≃ V_{0} / (4 E) \to 0

and

C / A \to 1

. To leading order the complete wavefunction is

Ψ (x, t) ≃ A exp [i (k_{1} x - \frac{Et}{ℏ})] .

(2)

The small fractional change in kinetic energy scarcely changes the propagation wavenumber, so reflection vanishes and the particle behaves approximately as a free right-moving plane wave. A more accurate phase at large positive

x

retains

k_{2} x

on the transmitted side; even a small wavenumber difference accumulates over a sufficiently long distance.

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