Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 1 / A3 / b

The expanded feature resolves the Q branch: each line has $\Delta J=0$ but starts from a different lower rotational level. In the simple rigid rotor band model,Because the upper rotational constant is larger, the lines extend towards higher wavenumber. Their adjacent spacings areso they crowd together at the low-frequency edge and become increasingly separated. A strictly resolved bending-band analysis may distinguish the upper parity component sampled by Q lines from that sampled by P/R lines; it is not necessary to impose identical high-resolution constants on every branch in this sketch.

The intensity envelope reflects rotational state populations and the rotational transition strength. For this perpendicular Q branch, the usual rotational line-strength factor gives the approximate dependenceAt small $J$, the increasing magnetic degeneracy makes the line intensity rise; at large $J$, Boltzmann suppression dominates. The maximum is therefore at an intermediate $J$. This is a population envelope across distinct transitions, rather than one intrinsically broad vibrational line.

Count the successive Q branch lines from the low-frequency edge, taking $J=1$ as the first allowed bend transition; the first lines are crowded and partly unresolved. The maximum near $712.55{\mathrm{cm}}^{-1}$ is approximately $J=12$. Treating $J$ as continuous in the intensity envelope from part (i),HenceUsing $hc/k_{\mathrm{B}}=1.43878\mathrm{K}\mathrm{cm}$, $B^{\prime \prime }=1.47786{\mathrm{cm}}^{-1}$ and $J_{\max }\simeq 12$ givesA useful check avoids locating the exact maximum. The plotted $J\simeq 20$ line has roughly $0.70$ of the $J\simeq 12$ peak height. Thusgives about $659\mathrm{K}$. An appropriate graphical estimate is $T\simeq 6.6\times 10^2\mathrm{K}$, with precision limited by line counting, line overlap and reading the plotted intensities. There is no basis for reporting a temperature to the nearest kelvin from this figure.