Q.55 In 2 N H₂SO₄, an organic compound shows fluorescence with quantum yield, ϕ = 0.42 and fluorescence rate constant, kf = 5.25 × 10⁷ s⁻¹. The observed fluorescence lifetime of it under the same conditions (correct to 1 decimal place) is _______ ns.

The observed fluorescence lifetime is 8.0 ns. This value is calculated using the fundamental relationship in fluorescence spectroscopy between quantum yield (ϕ), radiative rate constant (k_f), and lifetime (τ).​

Key Formula

Fluorescence quantum yield is defined as ϕ = k_f × τ, where τ is the observed fluorescence lifetime (1 / (k_f + k_nr)), k_f is the fluorescence rate constant, and k_nr represents non-radiative decay rates.​
Rearranging gives τ = ϕ / k_f.​
Here, ϕ = 0.42 and k_f = 5.25 × 10^7 s⁻¹, so τ = 0.42 / (5.25 × 10^7) = 8.0 × 10^{-9} s = 8.0 ns.​

Step-by-Step Calculation

Convert units: k_f = 5.25 × 10^7 s⁻¹.​
τ (s) = 0.42 / 5.25 × 10^7 = 8.0 × 10^{-9} s.​
τ (ns) = 8.0 × 10^{-9} × 10^9 = 8.0 ns (rounded to 1 decimal place).​
The 2 N H₂SO₄ condition specifies the measurement environment but does not alter the formula.​

Physical Interpretation

The lifetime τ reflects the average time the excited state persists before decay via fluorescence or non-radiative paths.​
A ϕ of 0.42 indicates 42% of excitations lead to fluorescence, with k_nr / k_f = (1 – ϕ)/ϕ ≈ 1.38, shortening τ from the radiative limit (1/k_f ≈ 19.0 ns).​

Fluorescence lifetime calculation in acidic media like 2N H₂SO₄ is crucial for understanding molecular photophysics, especially for organic compounds showing fluorescence with quantum yield ϕ=0.42 and fluorescence rate constant kf=5.25×10⁷ s⁻¹. This guide delivers the observed fluorescence lifetime (correct to 1 decimal place) as 8.0 ns, tailored for CSIR NET aspirants mastering spectroscopy in molecular biology and biochemistry.

Core Principles of Fluorescence Lifetime

Fluorescence lifetime (τ) measures the excited-state duration before photon emission or non-radiative decay.​
Quantum yield ϕ equals photons emitted per photon absorbed: ϕ = kf / (kf + knr).​
Thus, τ = ϕ / kf links these parameters directly.​

Detailed Solution for This Problem

Given: ϕ = 0.42 (dimensionless), kf = 5.25 × 10⁷ s⁻¹ in 2N H₂SO₄. [query]

• Compute τ = 0.42 / 5.25 × 10⁷ = 8.0 × 10^{-9} s.​

• Convert: 8.0 ns (1 ns = 10^{-9} s).​
  No options provided, but common pitfalls include inverting (kf/ϕ ≈ 1.25 × 10^8 s, τ=8 ns wrong-way) or unit errors (ps instead of ns).​

CSIR NET Exam Relevance

Such problems test Jablonski diagram applications, where acidic conditions (2N H₂SO₄) may protonate fluorophores, tuning ϕ and τ for biotech assays. [user-information]​
Practice: If ϕ=0.4, τ=5 ns, then kf=ϕ/τ=8×10^7 s⁻¹ (similar Testbook example).​

Advanced Insights for Life Sciences

In 2N H₂SO₄, fluorescence quenching via protonation competes with emission, reducing ϕ below 1.​
Radiative lifetime (1/kf ≈19 ns) exceeds observed τ=8.0 ns due to knr.​
Applications: DNA probes, enzyme kinetics—key for CSIR NET biotech/genetics sections. [user-information]

Keywords: fluorescence lifetime, quantum yield ϕ=0.42, kf=5.25×10⁷ s⁻¹, 2N H₂SO₄, observed lifetime 8.0 ns, CSIR NET spectroscopy.