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ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-4/59/b/v/solution

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Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 4 / 59 / b / v / Solution by

Codex  0 2026-10-20

With the supplied sum, the second-order approximation is

E_{1} ≃ \frac{h ^{2}}{8 m a ^{2}} - \frac{2048 m a ^{4} q ^{2} ε ^{2}}{h ^{2} π ^{4}} (4.12 \times 1 0^{- 3}) .

(1)

For an electron in the

1 μ m

box,

E_{1}^{(0)} = 6.02 \times 1 0^{- 26} J

and the correction is

- 4.61 \times 1 0^{- 27} J

, giving $5.56 \times 1 0^{- 26} J = 3.47 \times 1 0^{- 7} eV$ .

For

10 μ m

, the formal result is approximately $- 4.61 \times 1 0^{- 23} J = - 2.88 \times 1 0^{- 4} eV$ . It is not a credible approximation: even the minimum electric potential energy in that box is only

- 5 μ eV

, whereas this answer is much lower. The ground-to-first-excited-state mixing estimate increases from

0.160

160

when the length grows tenfold. The shift grows as

a^{4}

while the level spacing decreases as

a^{- 2}

, so the larger box is far outside the perturbative regime.

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