 Solution

Excited $\mathrm{O}({}^{1}D)$ has two competing fates. Collisional deactivation returns ground-state oxygen, which reforms ozone rapidly; reaction with water instead gives net ozone loss and two OH radicals. The chemical branching fraction for the latter isThus only about 4.92% of these ozone photolyses directly lead to the specified net chemical loss; treating every photodissociation as permanent removal would overestimate the sink. The corresponding first-order rate is $J_1{f}_w=9.8361\times 10^{-7}{\mathrm{s}}^{-1}$. Converting the given hydroperoxyl mixing ratio to a concentration gives $[\mathrm{H}{\mathrm{O}}_2]=20\times 10^{-12}[\mathrm{M}]=4.4\times 10^8\mathrm{c}{\mathrm{m}}^{-3}$, so $k_4[\mathrm{H}{\mathrm{O}}_2]=9.6800\times 10^{-7}{\mathrm{s}}^{-1}$.

For a well-mixed marine atmospheric boundary layer of depth $H=2\times 10^5\mathrm{cm}$, the downward surface flux is $v_{\mathrm{dep}}[{\mathrm{O}}_3]$. Dividing by the layer volume per unit area gives a first-order dry deposition rate $v_{\mathrm{dep}}/H=7.5\times 10^{-7}{\mathrm{s}}^{-1}$. The air density cancels on expressing the entire balance in ozone mixing ratio $\chi$:At steady state, with $P_{\chi }=0.25/3600\mathrm{ppb}{\mathrm{s}}^{-1}$,The effective removal time is about 4.28 days. This calculation assumes a uniform layer of the diagnosed depth, fixed chemical conditions and a local time-independent balance.