Considering the photoswitching of azobenzene, I was told the following argument: Considering that the photoexcited state of azobenzene has a about 50% to 50% chance of yielding either the (E)- or (Z)-isomer, the maximum (Z)/(E) ratio achievable should be 1:1.
However, I have seen UV-VIS spectra where the (Z)-isomer seems to be in excess. So I figured that the above argument must be wrong. I would analyze the situation in the following way:
In this picture, the fact that the photoexcited state has about equal probability to yield (E) or (Z), i.e. $k_{-1} = k_2$.
So wouldn't the overall (Z) to (E) ratio be given by the expression $\dfrac{k_1\cdot k_2}{k_{-1}\cdot k_{-2}}$$\frac{k_1\cdot k_2}{k_{-1}\cdot k_{-2}}$?
I would then argue that in many cases, the cis isomer exhibits lower absorption at the wavelength of excitation, meaning that $k_{-2} << k_{1}$, which would explain that in many cases, an excess of the cis isomer is obtained. By this reasoning, the initial argument would be completely wrong, as in principle an infinite excess of the cis isomer could be obtained if $k_{-2}$ is sufficiently smaller than $k_{1}$.
Is this analysis correct or am I overlooking something?
