= Solution
Identify the transition-state separation with contact, $R=r_A+r_B$, take distinct atoms so $\sigma=1$, and let $E_A=N_A\Delta\varepsilon_0^{\ddagger}$. Then $\pi R^2=\rho$ and $\Delta\varepsilon_0^{\ddagger}/kT=E_A/RT$, yielding
$$
\boxed{k_{\rm 2nd}=\rho\sqrt{\frac{8kT}{\pi\mu}}e^{-E_A/RT}}.
$$
The velocity factor is the mean relative speed from the <Maxwell-Boltzmann distribution>. Thus this simple <transition state theory> model reproduces the stated <collision theory> expression. Molecular orientation constraints, quantum transmission and reaction-surface recrossing can invalidate the equality outside these simplifying assumptions.
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