Q64 · CSIR NET Chemistry, June 2014
Paper: CSIR NET June 2014 · 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