= Solution
The Doppler relation is the affine <change of variables>
$$
v_x=c\left(\frac{\omega}{\omega_0}-1\right),
\qquad
\left|\frac{dv_x}{d\omega}\right|=\frac{c}{\omega_0}.
$$
Transforming the one-dimensional <Maxwell-Boltzmann velocity distribution> therefore gives the normalized <thermal Doppler broadening> profile
$$
\boxed{
I(\omega)=
\frac{c}{\omega_0}\sqrt{\frac{m}{2\pi k_BT}}
\exp\left[
-\frac{mc^2}{2k_BT\omega_0^2}
(\omega-\omega_0)^2
\right].}
$$
Thus
$$
\boxed{a=\frac{mc^2}{2k_BT\omega_0^2}.}
$$
The measured Gaussian line width satisfies $\sigma_\omega/\omega_0=\sqrt{k_BT/m}/c$, so the temperature follows from
$$
\boxed{T=\frac{mc^2}{k_B}
\left(\frac{\sigma_\omega}{\omega_0}\right)^2.}
$$
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