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 / 35 / c

For a linear rigid rotor, dividing the angular momentum squared eigenvalue by twice the moment of inertia givesThe quantum degeneracy counts the values $M=-J,\dots ,J$ with the same energy. It excludes nuclear-spin multiplicity and any fine or hyperfine splitting absent from this model.

Express the rotational constant as a spectroscopic wavenumber $\tilde{B}=h/(8{\pi }^{2}cI)$. The rotational absorption selection rule $\Delta J=+1$ givesSuccessive lines therefore have spacing $2\tilde{B}$, regardless of the starting $J$. Both observed spacings are $20.88\mathrm{c}{\mathrm{m}}^{-1}$, giving $\tilde{B}=10.44\mathrm{c}{\mathrm{m}}^{-1}$. The three lines are consequently the $5\to 6$, $6\to 7$ and $7\to 8$ transitions within the stated precision; the first observed line is not the $0\to 1$ line.

Using the given isotopic masses, the reduced mass isWith $\tilde{B}=1044{\mathrm{m}}^{-1}$ and $c$ in $\mathrm{m}{\mathrm{s}}^{-1}$, the bond length follows fromThe small common $0.01\mathrm{c}{\mathrm{m}}^{-1}$ offset relative to integer multiples of the spacing is immaterial at the precision of this model. A real ground-vibrational-state rotational constant determines an effective vibrationally averaged bond length; it need not equal the exact equilibrium separation.