Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2A / Section B / 21 / a / ii / Solution

The $K=0$ lines give two equations independent of ${\tilde{D}}_{JK}$:Solving yields $\tilde{B}\simeq 0.25022\mathrm{c}{\mathrm{m}}^{-1}$ and ${\tilde{D}}_J\simeq 2.1\times 10^{-7}\mathrm{c}{\mathrm{m}}^{-1}$. The small distortion is obtained by subtracting close numbers, so its precision is considerably worse than that of $\tilde{B}$.

For components in a fixed group,For example the $J=2,K=2$ displacement gives ${\tilde{D}}_{JK}=0.00008/24$, while $J=3,K=3$ gives $0.00024/72$. Both give ${\tilde{D}}_{JK}\simeq 3.3\times 10^{-6}\mathrm{c}{\mathrm{m}}^{-1}$; the smaller gaps are consistent after rounding. The pure rotational transitions have $\Delta K=0$, so the $(\tilde{A}-\tilde{B})K^2$ and ${\tilde{D}}_K{K}^4$ terms cancel. Neither $\tilde{A}$ nor ${\tilde{D}}_K$ can be determined from these zero-field lines.