= Solution
\b[Natural broadening is about $8\times10^2$ 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 $m=28m_{\rm u}$,
$$
\nu_0=\frac c\lambda=6.97\times10^{14}\ \mathrm{Hz},\qquad
\Delta\nu_{\rm D,HWHM}=\frac1\lambda\sqrt{\frac{2k_BT\ln2}{m}}=1.06\times10^9\ \mathrm{Hz}.
$$
For an excited-state population lifetime $\tau$, the radiating amplitude decays as $e^{-t/(2\tau)}$. Its <Lorentzian line shape> has <natural linewidth>
$$
\Delta\nu_{\rm nat,HWHM}=\frac{1}{4\pi\tau}=1.33\times10^6\ \mathrm{Hz}.
$$
Equivalently, 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.
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