Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2A / Section A / 13 / Solution

Using mass $7\mathrm{u}$ for the dominant lithium isotope gives a transition wavenumber ${\tilde{\nu }}_0=14925\mathrm{c}{\mathrm{m}}^{-1}$. Inserting the thermal speed into the stated Doppler broadening expression givesThe supplied formula is the Gaussian half width at half maximum; the corresponding full width is about $0.256\mathrm{c}{\mathrm{m}}^{-1}$. Meanwhile exponential population decay with lifetime $\tau$ produces a natural linewidth expressed as a full widthHere $c$ must be in $\mathrm{cm}{\mathrm{s}}^{-1}$ in the second expression. Equivalently the fine-structure separation is about $9.0$ GHz, the Doppler half width about $3.83$ GHz, and the natural linewidth expressed as a full width about $5.89$ MHz. The natural linewidth is therefore negligible compared with thermal broadening; low density also makes collision broadening less important.

The doublet is resolvable in principle, but only marginally against the Doppler width. The separation exceeds one Doppler full width, so two components can survive, although their profiles overlap strongly. The $J=3/2$ and $J=1/2$ upper levels have different statistical weights, so the weaker member need not make an equally conspicuous maximum. A spectral resolving power $R\gtrsim {\tilde{\nu }}_0/0.3\simeq 5\times 10^4$ is the basic instrumental requirement; a width well below $0.1\mathrm{c}{\mathrm{m}}^{-1}$ and adequate signal would make the test more convincing. An arbitrarily fine spectrometer cannot remove the physical Doppler overlap. This conclusion uses the stipulated thermal, optically thin picture; additional photospheric velocity fields or optical-depth effects would require more information.