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:
$$\ce{A + hv -&gt; A\mathrm{*}}$$
$$\ce{A\mathrm{*} -&gt; A + hv   \tag{k1}}$$
$$\ce{A\mathrm{*} + B -&gt; A + B\mathrm{*} \tag{k2}}$$
$$\ce{B\mathrm{*} -&gt; B + hv      \tag{k3}}$$
$$\ce{B\mathrm{*} -&gt; C           \tag{k4}}$$

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.