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?




The processes are absorb light into A, and then A\* has only two pathways, i.e. fluoresces and reacts with B to form B\*.


The B\* has *only* two pathways, it fluoresces and converts to C. *All* B\* molecules must therefore follow one of these paths, so the sum of yields of B\* (paths fluorescence + forming C) must be the same as the yield to form B\* in the first place, thus you don't need to do a calculation.


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.


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