In $P_2\xrightleftharpoons[k_1]{I_a}2Q\xrightarrow{k_2}R$, $I_a$ is absorbed light intensity and the first-step quantum yield is one. The steady-state Q concentration is ______.
Photodissociation produces 2Q at a rate 2Ia, because Ia already counts absorbed photons and the quantum yield is one. Both recombination channels consume Q in pair events, at combined rate 2(k1+k2)[Q]². Balance production and loss to obtain **[Q]=√[Ia/(k1+k2)], A**. Multiplying Ia by [P2] again would double-count absorption.