Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2021 / Paper 2A / Section A / 13 / Solution

Natural broadening is about $8\times 10^2$ times narrower than the thermal Doppler width and is negligible for this comparison. The supplied Doppler broadening formula is a half width at half maximum. Using the integer molecular mass $m=28m_{\mathrm{u}}$,For an excited-state population lifetime $\tau$, the radiating amplitude decays as $e^{-t/(2\tau )}$. Its Lorentzian line shape has natural linewidthEquivalently, compare full widths $2.11\mathrm{GHz}$ and $2.65\mathrm{MHz}$. Using the same linewidth convention in both expressions matters more than the small isotope-mass correction. This estimate assumes that the lower state's lifetime is long and leaves any pressure broadening as a separate contribution.