 Solution

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.