 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2024/paper-1/a1/a/ii/solution

1 = \int_{0}^{a} ∣ Ψ_{n} (x) ∣^{2} d x = ∣ C_{n} ∣^{2} \int_{0}^{a} \frac{1 - c o s ( 2 nπ x / a )}{2} d x = ∣ C_{n} ∣^{2} \frac{a}{2} .

Choosing an irrelevant overall phase to be zero gives

Ψ_{n} (x) = \frac{2}{a} sin \frac{nπ x}{a} .

The amplitude has units of inverse square root length, so

∣ Ψ_{n} ∣^{2} d x

is dimensionless.

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