Question

In the scheme \( \mathrm{P_2} \xrightleftharpoons[\;k_1\;]{\;I_a\;} 2\mathrm{Q} \xrightarrow{k_2} \mathrm{R} \), \(I_a\) represents the intensity of light absorbed. Assuming the quantum yield of the first step is one, the steady-state concentration of \(\mathrm{Q}\) is given by

• A. \(\displaystyle \sqrt{\frac{I_a}{k_1+k_2}}\)
• B. \(\displaystyle \sqrt{\frac{I_a[\mathrm{P_2}]}{k_1+k_2}}\)
• C. \(\displaystyle \frac{I_a}{k_1+k_2}\)
• D. \(\displaystyle \frac{I_a[\mathrm{P_2}]}{k_1+k_2}\)

Hint:

If \(I_a\) is the \emph{absorbed} intensity and quantum yield \(=1\), it equals the rate of creating one event of \(2\mathrm{Q}\) formation—no extra \([\mathrm{P_2}]\) factor.
When intermediates are made/consumed in pairs, steady-state often gives a square-root dependence: \([\mathrm{Q}] \propto \sqrt{\text{generation}/\text{bimolecular loss}}\).

Answer 1

The Correct Option is A

Step 1: Write formation and consumption rates of \( \mathrm{Q} \). 
Quantum yield of step 1 is 1, and \(I_a\) is the \emph{absorbed} light intensity; hence the photochemical production rate of the pair \(2\mathrm{Q}\) is simply \(I_a\) (already normalized to the amount of light absorbed by \(\mathrm{P_2}\)). Therefore, the formation rate of \(\mathrm{Q}\) is \(I_a\). 
Loss of \(\mathrm{Q}\) occurs via two \emph{bimolecular} steps that consume two \(\mathrm{Q}\) at a time: \(2\mathrm{Q}\xrightarrow{k_1}\mathrm{P_2}\) and \(2\mathrm{Q}\xrightarrow{k_2}\mathrm{R}\). The total consumption rate is \((k_1+k_2)[\mathrm{Q}]^2\). 

Step 2: Apply steady state and solve. 
At steady state, formation = consumption: \[ I_a = (k_1+k_2)\,[\mathrm{Q}]^2 \quad \Rightarrow \quad [\mathrm{Q}] = \sqrt{\frac{I_a}{k_1+k_2}}. \] This corresponds to option (A).