Discussion
The \(g_{N}\) value for \(F^{19}\) nucleus is 5.256. If it is placed in a magnetic field of strength 1.0 T, the relative population in two spin states at 300 K will be, (Given \(\mu_{N} = 5.05 \times 10^{-27}\) JT\(^{-1}\) and \(k = 1.38 \times 10^{-23}\) JK\(^{-1}\)) :
Answer: A
The relative population is essentially 1.0.
The energy gap between the two nuclear spin states and the thermal energy are
\[ \begin{aligned} \Delta E &= g_{N}\mu_{N}B = 5.256 \times 5.05 \times 10^{-27} \\ &= 2.65 \times 10^{-26}\ \text{J} \\ k_{B}T &= 1.38 \times 10^{-23} \times 300 = 4.14 \times 10^{-21}\ \text{J} \end{aligned} \]
\[ \begin{aligned} \frac{N_{upper}}{N_{lower}} &= e^{-\Delta E/k_{B}T} \\ &= e^{-6.4 \times 10^{-6}} \approx 0.999994 \end{aligned} \]
The two populations differ by only a few parts per million, which is why NMR signals are weak. A ratio of 2.0, 2.5 or 0.5 would need a gap comparable to kT, about a million times larger than the nuclear Zeeman splitting at 1 T.
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