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

The operator ${\hat{J}}_z$ describes the $z$ component of angular momentum, whereas ${\hat{J}}^2$ describes its total squared magnitude. For a spherical harmonic $Y_{JM}$ the corresponding eigenvalues areThus the magnitude is $\hslash \sqrt{J(J+1)}$, not $J\hslash$. For ordinary orbital rotation $J=0,1,2,\dots$ and $M=-J,-J+1,\dots ,J$. The simultaneous values are possible because these two operators commute.

Normalization of a wavefunction on the unit sphere uses solid angle $d\Omega =\sin \theta d\theta d\varphi$, not the flat measure $d\theta d\varphi$. Since $|e^{-i\varphi }|=1$,where $u=\cos \theta$. ThereforeWith the usual phase convention this is the spherical harmonic $Y_{1,-1}$. An arbitrary global unit-modulus factor would not change normalization or any observable.

The azimuthal derivative is ${\partial }_{\varphi }\psi =-i\psi$, so ${\hat{J}}_z\psi =(-i\hslash )(-i)\psi =-\hslash \psi$. For the polar part of the angular Laplacian,Their sum is $-2\sin \theta e^{-i\varphi }=-2\psi$, hence ${\hat{J}}^2\psi =2{\hslash }^2\psi$. Comparing these two eigenvalues with $M\hslash$ and $J(J+1){\hslash }^2$ gives $J=1$, $M=-1$. The sign of $M$ follows from both minus signs in the ${\hat{J}}_z$ calculation.

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.