 Solution

Let brackets denote number densities and impose a local photochemical steady state. The atomic-oxygen and ozone balances areSubtracting gives $J_1[{\mathrm{O}}_2]=K_4[\mathrm{O}][{\mathrm{O}}_3]$. If photodissociation is the dominant direct ozone loss, $J_3[{\mathrm{O}}_3]\gg K_4[\mathrm{O}][{\mathrm{O}}_3]$, the second balance becomes $K_2[\mathrm{O}][{\mathrm{O}}_2][\mathrm{M}]\simeq J_3[{\mathrm{O}}_3]$. Eliminating atomic oxygen yields the Chapman ozone equilibriumHigh in the atmosphere ultraviolet photons make $J_1$ large but the third-body density is small; low down, $[\mathrm{M}]$ is large but O2-dissociating ultraviolet radiation has been absorbed. Their product peaks at intermediate altitude, producing an ozone layer.