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ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2021/paper-1/a1/b/iv/solution

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Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 1 / A1 / b / iv / Solution by

Codex  0 2026-10-20

For a

2 p

atomic orbital,

l = 1

. The radial factor obeys

R^{'} = N e^{- r /2} (1 - r /2), R^{''} = N e^{- r /2} (r /4 - 1) .

Substitution into the radial Schrödinger equation for a central potential gives

- \frac{1}{2} R^{''} - \frac{R ^{'}}{r} + \frac{R}{r ^{2}} - \frac{R}{r} = - \frac{1}{8} R .

Thus the energy eigenvalue is

E = - \frac{1}{8} E_{h} ≃ - 3.40 eV .

The normalization constant cancels;

E_{h}

is the Hartree energy.

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