Approximate the atmosphere above $R_p$ as isothermal with constant gravity and <atmospheric scale height> $H$, so
$$
P(z)=P_0e^{-z/H}.
$$
Vertical transport over one scale height has <eddy mixing time>
$$
\tau_{\rm mix}\simeq\frac{H^2}{K_{zz}}.
$$
The <chemical quench level> satisfies $\tau_{\rm chem}(z_q)=\tau_{\rm mix}$. Since $\tau_{\rm chem}=\eta z$,
$$
z_q=\frac{H^2}{\eta K_{zz}},
\qquad
\boxed{P_q=P_0\exp\left(-\frac{H}{\eta K_{zz}}\right)}.
$$
Above this level, mixing is faster than reaction and freezes the deeper abundance of A. Larger $K_{zz}$ moves the quench level deeper and raises $P_q$. Important examples are <carbon monoxide–methane quenching> and <nitrogen–ammonia quenching>; phosphine destruction is another tracer of vertical quenching.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 315 / 4 / d / Solution

{title2=$\tau_{\rm mix}\simeq H^2/K_{zz}$}

The eddy mixing time over one atmospheric scale height is $\tau_{\rm mix}\simeq H^2/K_{zz}$, where $K_{zz}$ is the vertical eddy diffusion coefficient.


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