 Solution

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 $v$.