Hi, trying to help my niece with physical chemistry but I'm not a chemist.

You've sure set yourself up for a challenge

First, you don't need any concentration information, since photophysical rate constants are intrinsic properties of molecules. The only dimensions you need worry about here are time (for lifetimes) and inverse time (for rate constants). Quantum yields are unitless.

Second, it would be helpful to start with a Jablonski diagram, so you can visualize all the radiative and nonradiative processes that you need to keep track of and how they're related. Since it would be hopeless to try to reproduce that here, I encourage you to seek one out either in the textbook or online.

Third, you should write out in equation form the definitions of the quantum yields and lifetimes. Quantum yields can be defined in a lot of ways but the one that is needed here is this: the quantum yield of a process is the ratio of the rate constant for that process divided by the sum of the rate constants for all possible processes. So, the quantum yield of fluorescence Φ

_{f} is:

[tex]\Phi_f = \frac{k_f}{k_f + k_{nr} + k_{isc}}[/tex]

And the quantum yield of the (S1-T1) nonradiative intersystem crossing Φ

_{isc} is:

[tex]\Phi_{isc} = \frac{k_{isc}}{k_f + k_{nr} + k_{isc}}[/tex]

Here, k

_{f}, k

_{nr}, and k

_{isc} are the respective rate constants of fluorescence (radiative deactivation of the singlet state, S

_{1} S

_{0}), nonradiative internal conversion (S

_{1} S

_{0}), and nonradiative intersystem crossing (S

_{1} T

_{1}).

(Note here that there is some sloppiness in the problem, as there are actually two nonradiative intersystem crossing processes, so don't get them confused.)

The denominator in both quantum yield expressions is the same and is called the overall rate constant for singlet state deactivation. Since the inverse of any rate constant is the photophysical lifetime of the process, the inverse of the rate constant for singlet state deactivation is the intrinsic lifetime of the singlet state, τ

_{S1}. So therefore the quantum yield expressions can be written as:

[tex]\Phi_f = k_f \tau_{S1}[/tex]

[tex]\Phi_{isc} = k_{isc} \tau_{S1}[/tex]

Where

[tex]\tau_{S1}= (k_f + k_{nr} + k_{isc})^{-1}[/tex]

Quantum yields for phosphorescence Φ

_{p} are formulated in more or less the same way except that you must recall that the quantum yield of phosphorescence is usually understood to be the fraction of all excited molecules that phosphoresce, not the fraction of triplet states that deactivate radiatively. So although we use the same rate constant formulation, it has to be scaled by the quantum yield of intersystem crossing to account for the fact that not all excited molecules end up in the triplet state.

I.e.,

[tex]\Phi_p = \Phi_{isc} \frac{k_p}{k_p + k_{nr}} = \Phi_{isc} k_p \tau_t [/tex]

[tex]\Phi_{nr} = \Phi_{isc} \frac{k_{nr}}{k_p + k_{nr}} = \Phi_{isc} k_{nr} \tau_t [/tex]

Where [tex]\tau_t[/tex] is the triplet state lifetime. Do note that, for the triplet state, k

_{nr} is also formally an intersystem crossing process.

With these equations and the provided information you should be able to calculate all the parameters. However I noted that they did not give you a quantum yield of fluorescence. I didn't run through the problem myself but I suspect this will make it impossible to determine the rate constant for fluorescence (since it won't be possible to separate radiative from S1-S0 nonradiative deactivation of the singlet state). You may be expected to assume that all excited singlet that do not cross to the excited triplet deactivate radiatively (i.e., the fluorescence yield is 3%). Generally this is not a great assumption, but you could do it now to complete the exercise and then follow up with the instructor.

Do make sure you express the lifetime values in the same time unit, otherwise you will get an erroneous result.

Hope that helps.