= Solution
Write $M=m_A+m_B$ and $\mu=m_Am_B/M$. An atom has no rotational or vibrational internal factor under the stated electronic assumption. The transition-state <moment of inertia> is $I=\mu R^2$, so the high-temperature <rigid rotor> result is
$$
q_{\rm rot,TS}=\frac{8\pi^2\mu R^2kT}{\sigma h^2}.
$$
The translational factors therefore give
$$
\frac{f'_{\rm TS}}{f_Af_B}
=\left(\frac{h^2}{2\pi\mu kT}\right)^{3/2}
\frac{8\pi^2\mu R^2kT}{\sigma h^2}.
$$
Substituting into the <transition state theory> expression and simplifying,
$$
\boxed{k_{\rm 2nd}=\frac{\pi R^2}{\sigma}
\sqrt{\frac{8kT}{\pi\mu}}\,
\exp\left(-\frac{\Delta\varepsilon_0^{\ddagger}}{kT}\right)}.
$$
For distinct atoms, $\sigma=1$. This is a number-density rate coefficient, with units of <volume> per time. A rate law in molar concentrations uses $N_Ak_{\rm 2nd}$. The derivation assumes a classical high-temperature rotor at the dividing surface and a transmission coefficient of one.
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