 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2023/paper-1/a1/a/solution

ℏ = m_{e} = e = 4 π ε_{0} = 1

, and the hydrogen atom has

\hat{H} Ψ = - \frac{1}{2 n ^{2}} Ψ, \hat{l}^{2} Ψ = l (l + 1) Ψ, \hat{l}_{z} Ψ = m Ψ,

\overset{s}{^}^{2} Ψ = \frac{3}{4} Ψ, \overset{s}{^}_{z} Ψ = m_{s} Ψ.

The allowed quantum numbers are

n = 1, 2, \dots, l = 0, 1, \dots, n - 1, m = - l, - l + 1, \dots, l, m_{s} = \pm \frac{1}{2} .

s = 1/2

. The energy eigenvalues are in hartrees; restoring units gives

ℏ^{2} l (l + 1)

and

ℏ m

for the orbital operators and the corresponding factors for spin. The spin-independent hydrogen Hamiltonian does not split the two spin projections.

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