UPSC Combined Geo-Scientist, Mains · 2024 · Mains Paper II
Q1(o). Fluorescence rate constant of tryptophan
Question
The fluorescence quantum yield and observed fluorescence lifetime of tryptophan in water are ϕF,0 = 0.20 and τ₀ = 2.6 ns, respectively. Calculate the fluorescence rate constant, kF.
Answer in brief
kF = ϕF/τ₀ = 7.7 × 10⁷ s⁻¹ (radiative lifetime 13 ns).
Key points
- S₁ decays by all paths together: τ₀ = 1/(kF + kIC + kISC + …), ϕF = kF/(kF + kIC + kISC + …) = kF τ₀.
- kF = ϕF/τ₀ = 0.20/(2.6 × 10⁻⁹ s) = 7.7 × 10⁷ s⁻¹.
- Radiative (natural) lifetime τrad = 1/kF = 13 ns; non-radiative knr = (1 − ϕF)/τ₀ = 3.1 × 10⁸ s⁻¹.
- Tryptophan fluorescence (λem ≈ 350 nm) is the intrinsic probe of protein folding.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Coversfluorescence quantum yieldfluorescence lifetimeradiative rate constanttryptophannon-radiative decay
All parts of Question 1
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