= Solution
The energy separation is $\Delta E=\hbar|\gamma|B_0$. At 10 T this gives
$$
\boxed{\Delta E_H=2.816\times10^{-25}\ \mathrm J=1.757\times10^{-6}\ \mathrm{eV}},\qquad
\boxed{\Delta E_N=2.847\times10^{-26}\ \mathrm J=1.777\times10^{-7}\ \mathrm{eV}}.
$$
At 300 K, $k_BT=4.142\times10^{-21}\ \mathrm J=0.02585\ \mathrm{eV}$, so the splitting-to-thermal-energy ratios are $6.80\times10^{-5}$ and $6.87\times10^{-6}$. The two <spin> levels therefore have nearly equal populations. Their equilibrium population <polarizations> are approximately half these ratios, \b[$3.40\times10^{-5}$ and $3.44\times10^{-6}$], explaining weak thermal <NMR> sensitivity despite well-defined resonant transition energies.
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