Past exam of the chemistry A course of the University of Cambridge / 2025 / Paper 2 / A6 / c

Subtracting the $v=0$ energy in the Morse oscillator givesThe first three relative wavenumbers are $0$, $213$, and $424{\mathrm{cm}}^{-1}$. Thus, with ${\theta }_1=213/0.695\mathrm{K}$ and ${\theta }_2=424/0.695\mathrm{K}$,This is a three-level partition function, as stipulated, rather than an unrestricted sum of the anharmonic formula to arbitrarily large $v$.

Differentiating each Boltzmann factor yieldsThe derivative is positive: warming increases occupation of both excited energy levels. At $100\mathrm{K}$,

For one mole, $Nk=R$, so the three-level Morse oscillator result isFor the full quantum harmonic oscillator with the given wavenumber,The specified anharmonic calculation is about $1.7\%$ higher. Anharmonicity lowers the first excitation spacings and increases excited-state occupation. Both results remain small because $100\mathrm{K}$ is well below the vibrational temperature. The comparison also mixes a three-level anharmonic truncation with an infinite harmonic sum; omitted higher Morse levels would make a further small positive correction at this temperature.