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ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2a/section-a/16/a/solution

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Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2A / Section A / 16 / a / Solution by

Codex  0 2026-10-20

Let

s = [S_{1}]

and treat the quencher concentration as a constant reservoir. Weak steady excitation, one emitting population, and the three specified processes give

\frac{d s}{d t} = I_{abs} - (k_{F} + k_{Q} [Q]) s = 0, s = \frac{I _{abs}}{k _{F} + k _{Q} [ Q ]} .

(1)

Dividing the fluorescence photon-production rate by the absorbed photon rate therefore gives

ϕ_{Q} = \frac{k _{F} s}{I _{abs}} = \frac{k _{F}}{k _{F} + k _{Q} [ Q ]} .

(2)

This assumes no saturation, photochemical loss, reabsorption, static association or additional excited-state decay, and that each counted absorbed photon reaches the stated

S_{1}

population. In this restricted scheme

ϕ_{0} = 1

and the unquenched lifetime is

τ_{0} = 1/ k_{F}

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