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.)
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.)
The lifetime of the excited state is the reciprocal of the sum of rate constants which you are trying to find:
$$\begin{align} \tau(T_1) = \frac{1}{\sum_i k_i(T_1)} &= \pu{6 \times 10^-3 s} \\ \sum_i k_i(T_1) &= \frac{1}{\pu{6 \times 10^-3 s}} \\ &= \pu{167 s^-1} \end{align}$$
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:
$$\begin{align} k_\text{ISC} &= \Phi_\text{ISC} \cdot \sum_i k_i(T_1) \\ &= 0.1 \cdot \pu{167 s^-1} \\ &= \pu{16.7 s^-1} \end{align}$$
The lifetime of the excited state is the reciprocal of the sum of rate constants which you are trying to find:
$$\begin{align}\tau(T_1) = \frac{1}{\sum_i k_i(T_1)} &= \pu{6 \times 10^-3 s} \\\sum_i k_i(T_1) &= \frac{1}{\pu{6 \times 10^-3 s}} \\&= \pu{167 s^-1}\end{align}$$
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:
$$\begin{align}k_\text{ISC} &= \Phi_\text{ISC} \cdot \sum_i k_i(T_1) \\&= 0.1 \cdot \pu{167 s^-1} \\&= \pu{16.7 s^-1}\end{align}$$
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