Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2021 / Paper 2A / Section B / 22 / b / iii

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- Solution iii

Solution

The steady-state approximation for the excited singlet gives

0 = I_{abs} - (k_{F} + k_{ISC}) [S_{1}],

and that for the triplet gives

0 = k_{ISC} [S_{1}] - (k_{P} + k_{D} + k_{Q} [Q]) [T_{1}] .

Combining these with the quantum yield definition gives the more general expression

ϕ_{P} = \frac{k _{ISC}}{k _{F} + k _{ISC}} \frac{k _{P}}{k _{P} + k _{D} + k _{Q} [ Q ]} .

Under the supplied limit

k_{ISC} ≫ k_{F}

, this becomes

ϕ_{P} ≃ \frac{k _{P}}{k _{P} + k _{D} + k _{Q} [ Q ]} .

The approximation is the near-unit triplet formation yield, not an assumption that phosphorescence is the only triplet loss process.

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