0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Max Feb 27 '18 at 15:59
  • $\begingroup$ 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... $\endgroup$ – Zhe Feb 27 '18 at 16:41
0
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ So then how would I show that Φenergy transfer = Φf(B) + ΦC? Would I need to work out the equations for both the Φf(B) and ΦC and then show they are mathematically equal to Φenergy transfer. Also what specifically is the Φenergy transfer referring to? So for eaxmple you can show that the Φf(B) = ΦET + k3/(k3 +k4) $\endgroup$ – Henry Cooper Feb 28 '18 at 21:30
  • $\begingroup$ No the argument is in para 2 and I reword it as follows. There are only two pathways for B* to go to, one is B* fluorescence to ground state B and the other is to C. These yields must add to be 1 as there is nothing else B* can do according to your scheme. As B* is produced from A* it follows that the yield you want can only be that of the amount of B* produced (since the sum of decays of B* is 1) and this is the yield of energy transfer to form B*. $\endgroup$ – porphyrin Feb 28 '18 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.