Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2B / Section B / 34 / d / ii / Solution

Let $f$ and $o$ denote the measured fundamental and first-overtone wavenumbers, and put $y=\tilde{\omega }{x}_e$. Differences of the Morse oscillator levels giveConsequently $x_e=y/\tilde{\omega }$, $D_e/(hc)={\tilde{\omega }}^2/(4y)$ and $D_0/(hc)=D_e/(hc)-(\tilde{\omega }/2-y/4)$. Substituting the two pairs of measured bands gives the following derived quantities:The conversion is $1\mathrm{c}{\mathrm{m}}^{-1}=0.0119626566\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, after multiplication by $N_{A}hc$ and consistent length units. Therefore the requested ground-state dissociation energies are $D_0(\mathrm{A})=448.4\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ and $D_0(\mathrm{B})=388.5\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The corresponding potential-well depths are $D_e(\mathrm{A})=464.1$ and $D_e(\mathrm{B})=402.2\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. These are extrapolations under the stipulated Morse oscillator model; two low-lying bands do not establish the accuracy of the real potential near dissociation.