Q90 · CSIR-NET Chemistry, December 2013

Paper: CSIR-NET December 2013 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Rate Laws Order · Marks: 2 · Difficulty: Medium

The fluorescence lifetime of a molecule in a solution is $5 \times 10^{-9} \mathrm{s}$. The sum of all the noradiative rate constants ( $\Sigma \mathrm{k}_{\mathrm{nr}}$ ) for the decay of excited state is $1.2 \times 10^{8} \mathrm{s}^{-1}$. The fluorescence quantum yield of the molecule is
(a)0.1
(b)0.2
(c)0.4
(d)0.6
Answer
Answer: C ✓ checked by 4AB · confidence high

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation
1/τ = k_r + k_nr = 2×10⁸ s⁻¹ ⇒ k_r = 2×10⁸ − 1.2×10⁸ = 0.8×10⁸; φ_F = k_r τ = 0.8×10⁸ × 5×10⁻⁹ = 0.4.

Study loop for Chemical Kinetics

1. Practise the PYQsPrevious-year questions, with answers

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