Calculating rate constant of intersystem crossing from Jablonski diagram I am to calculate the rate constant for $T_1 \to S_0$ intersystem crossing, with reference to the Jablonski diagram attached below. I know this should not be difficult, but for some reason I cannot put the number together to get the correct result.

I have that

$$
\Phi_\text{ISC} = \frac{k_\text{ISC}}{\sum_i k_i}
$$

where the denominator represents a sum of all decay pathways for the $T_1$ state. However, how can this sum be calculated if we do not know the rate constants to begin with?
The lifetime of the excited state is the reciprocal of the sum of rate constants which you are trying to find:

$$\begin{align}
\tau(T_1) = \frac{1}{\sum_i k_i(T_1)} &amp;= \pu{6 \times 10^-3 s} \\
\sum_i k_i(T_1) &amp;= \frac{1}{\pu{6 \times 10^-3 s}} \\
&amp;= \pu{167 s^-1}
\end{align}$$

The two lines going from the $T_1 \to S_0$ state represent intersystem crossing (ISC) and phosphorescence. Squiggly lines represent nonradiative transitions, of which ISC is one, so the quantum yield for ISC is 10%. (Straight lines represent radiative transitions, in this case phosphorescence.)

Therefore:

$$\begin{align}
k_\text{ISC} &amp;= \Phi_\text{ISC} \cdot \sum_i k_i(T_1) \\
&amp;= 0.1 \cdot \pu{167 s^-1} \\
&amp;= \pu{16.7 s^-1}
\end{align}$$