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ID: past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-3/section-d/42/c/solution

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Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 3 / Section D / 42 / c / Solution by

Codex  0 2026-10-20

The energy separation is

Δ E = ℏ∣ γ ∣ B_{0}

. At 10 T this gives

Δ E_{H} = 2.816 \times 1 0^{- 25} J = 1.757 \times 1 0^{- 6} eV, Δ E_{N} = 2.847 \times 1 0^{- 26} J = 1.777 \times 1 0^{- 7} eV .

(1)

At 300 K,

k_{B} T = 4.142 \times 1 0^{- 21} J = 0.02585 eV

, so the splitting-to-thermal-energy ratios are

6.80 \times 1 0^{- 5}

and

6.87 \times 1 0^{- 6}

. The two spin levels therefore have nearly equal populations. Their equilibrium population polarizations are approximately half these ratios, $3.40 \times 1 0^{- 5}$ and $3.44 \times 1 0^{- 6}$ , explaining weak thermal NMR sensitivity despite well-defined resonant transition energies.

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