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 Incoming links: Radial Schrödinger equation for a central potential

Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 1 / A1 / b / iii / Solution 2026-10-20

Separation of variables makes the angular factor a spherical harmonic. Applying the Quantum Hamiltonian gives

\hat{H} (R Y_{l m}) = Y_{l m} [- \frac{1}{2 r ^{2}} \frac{d}{d r} (r^{2} R^{'}) + \frac{l ( l + 1 )}{2 r ^{2}} R - \frac{R}{r}] .

Consequently

R Y_{l m}

is an energy eigenfunction provided the radial factor satisfies

- \frac{1}{2 r ^{2}} \frac{d}{d r} (r^{2} R^{'}) + [\frac{l ( l + 1 )}{2 r ^{2}} - \frac{1}{r}] R = ER .

This is the radial Schrödinger equation for a central potential, here with the attractive Coulomb potential. An arbitrary

R

does not give an eigenfunction; regularity, decay and wavefunction normalization select admissible bound quantum states.

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