Question

In 2 N H\(_2\)SO\(_4\), an organic compound shows fluorescence with quantum yield $\phi_f = 0.42$ and fluorescence rate constant $k_f = 5.25 \times 10^{7}\ \text{s}^{-1}$. The observed fluorescence lifetime (correct to 1 decimal place) is ............ ns.

Hint:

Fluorescence lifetime depends on total decay rate, not only on fluorescence rate constant.

Answer 1

Correct Answer: 7.9

Step 1: Use quantum yield formula.
\[ \phi_f = \frac{k_f}{k_f + k_{nr}} \] Step 2: Rearranging for total decay rate.
\[ k_f + k_{nr} = \frac{k_f}{\phi_f} \] Step 3: Substitute values.
\[ k_f + k_{nr} = \frac{5.25 \times 10^{7}}{0.42} \] \[ k_f + k_{nr} = 1.25 \times 10^{8}\ \text{s}^{-1} \] Step 4: Fluorescence lifetime.
\[ \tau = \frac{1}{k_f + k_{nr}} = \frac{1}{1.25 \times 10^{8}} \] \[ \tau = 8.0 \times 10^{-9}\ \text{s} = 8.0\ \text{ns} \] Step 5: Conclusion.
Observed fluorescence lifetime = 8.0 ns.