# Photochemistry and steady state approximation

I am unsure whether to ask this in physics or chemistry as there is overlap but I am taking this in a chemistry module.

System A undergoes photophysical pathways: $$\ce{A + hv -> A\mathrm{*}}$$ $$\ce{A\mathrm{*} -> A + hv \tag{k1}}$$ $$\ce{A\mathrm{*} + B -> A + B\mathrm{*} \tag{k2}}$$ $$\ce{B\mathrm{*} -> B + hv \tag{k3}}$$ $$\ce{B\mathrm{*} -> C \tag{k4}}$$

It wants me to show that the quantum yield of energy transfer is equal to the quantum yield of fluorescene B and the quantum yield of photochemical conversion from B to C.

I am quite stuck on this problem. I assume the steady state approximation needs to be employed in order to determine quantum yields for the processes asked for. I am wondering what the general strategy would be to answer this question?

• I'd begin by drawing a tree with the possible reaction paths. Which 'quantum yields' are associated with each branch of the tree? Think about which steps are reversible and which steps aren't. Extra hint: Double funnel. – Max Feb 27 '18 at 15:59
• I think the big question here is which processes do you expect to be reversible? And if the answer is none, then it's just the forward rate, the whole way through... – Zhe Feb 27 '18 at 16:41

In general if you have to find a yield of any process x this is, $\displaystyle \phi_x = \frac{\text{rate of process x}}{\text{rate of absorption}} \equiv \frac{\text{rate const x} }{\text{sum of all other rate constants from x}}$.
In your example $\displaystyle \phi_{ET}= \frac{k_{ET}[B]}{k_f^A+k_{ET}[B]}$ where superscript $A$ refers to fluorescence from $A^*$.
To work all this out write down $d[A^*]/dt = +k_{abs}[A]- k_f^A[A^*]-k_{ET}[A^*][B]$ and use steady state.