= Solution
For a transition starting at $J$ and ending at $J+2$, the molecule gains
$$
\Delta\widetilde\varepsilon
=\widetilde B\big[(J+2)(J+3)-J(J+1)\big]
=\widetilde B(4J+6).
$$
<Energy> conservation therefore gives the scattered-light <wavenumbers>
$$
\boxed{\widetilde\nu_{\rm S}(J)=\widetilde\nu_L-(4J+6)\widetilde B,\qquad J=0,1,2,\ldots.}
$$
The first shifts below the laser are \b[$6\widetilde B,10\widetilde B,14\widetilde B,18\widetilde B$]. Adjacent Stokes lines are separated by \b[$4\widetilde B$], while the first line is $6\widetilde B$ from the Rayleigh line. The figure uses positive <Raman shift> $\widetilde\nu_L-\widetilde\nu_{\rm S}$; increasing shift means decreasing scattered-light wavenumber.
\Image[/past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2023/paper-1-rotational-raman.png]
{title=Original line-position sketches for the diatomic and BF3 rotational Raman series; stick heights carry no intensity information}
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