= Solution
Let brackets denote number densities and impose a local <photochemical steady state>. The atomic-oxygen and ozone balances are
$$
2J_1[{\rm O}_2]+J_3[{\rm O}_3]
=K_2[{\rm O}][{\rm O}_2][{\rm M}]
+K_4[{\rm O}][{\rm O}_3],
$$
$$
K_2[{\rm O}][{\rm O}_2][{\rm M}]
=J_3[{\rm O}_3]+K_4[{\rm O}][{\rm O}_3].
$$
Subtracting gives $J_1[{\rm O}_2]=K_4[{\rm O}][{\rm O}_3]$. If photodissociation is the dominant direct ozone loss, $J_3[{\rm O}_3]\gg K_4[{\rm O}][{\rm O}_3]$, the second balance becomes $K_2[{\rm O}][{\rm O}_2][{\rm M}]\simeq J_3[{\rm O}_3]$. Eliminating atomic oxygen yields the <Chapman ozone equilibrium>
$$
\boxed{[{\rm O}_3]\simeq[{\rm O}_2]
\left(\frac{K_2J_1[{\rm M}]}{K_4J_3}\right)^{1/2}}.
$$
High in the atmosphere ultraviolet photons make $J_1$ large but the third-body density is small; low down, $[{\rm M}]$ is large but O2-dissociating ultraviolet radiation has been absorbed. Their product peaks at intermediate altitude, producing an <ozone layer>.
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