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

The energy separation is $\Delta E=\hslash |\gamma |B_0$. At 10 T this givesAt 300 K, $k_{B}T=4.142\times 10^{-21}\mathrm{J}=0.02585\mathrm{eV}$, so the splitting-to-thermal-energy ratios are $6.80\times 10^{-5}$ and $6.87\times 10^{-6}$. The two spin levels therefore have nearly equal populations. Their equilibrium population polarizations are approximately half these ratios, $3.40\times 10^{-5}$ and $3.44\times 10^{-6}$, explaining weak thermal NMR sensitivity despite well-defined resonant transition energies.