Chapter 11, Problem 11.13P

Interpretation Introduction

Interpretation:

Quenching rate constant and half-life of the fluorescence has to be determined using the given data.

Concept Introduction:

According to Stern-Volmer equation, fluorescence quantum yields of a molecule in presence and absence of a quencher (Q) at a certain molar concentration can be given as,

  $\begin{array}{c}\frac{{\Phi }_{F,0}}{{\Phi }_F}\text{ = 1 + }{\tau }_0{k}_Q\left[Q\right]\\ \frac{1}{\tau }\text{ = }\frac{\text{1}}{{\tau }_0}\text{ + }{k}_Q\left[Q\right]\\ \\ where,\\ {\Phi }_{F,0}\text{ = Fluorescence quantum yield in absence of Q}\\ {\Phi }_F\text{ = Fluorescence quantum yield in presence of Q}\\ {\tau }_0\text{ = Observed fluorescence lifetime in absence of Q}\\ \tau \text{ = Observed fluorescence lifetime in presence of Q}\\ \left[Q\right]\text{ = Molar concentration }ofQ\end{array}$

Answer to Problem 11.13P

Half-life of the fluorescence is $0.172\mu s.$

Explanation of Solution

• Calculate $\left(k_Q/k_Q\right)$:

A plot was drawn between ${\text{I}}_{abs}/I_f$ and $\left[Q\right]$ using the given information.

  $\begin{array}{ccc}\left[Q\right]/mmol.d{m}^{-3}& I_F/I_{abs}& I_{abs}/I_F\\ \begin{array}{r}\hfill 1\\ \hfill 2\\ \hfill 3\\ \hfill 4\\ \hfill 5\end{array}& \begin{array}{r}\hfill 0.31\\ \hfill 0.18\\ \hfill 0.13\\ \hfill 0.10\\ \hfill 0.081\end{array}& \begin{array}{r}\hfill 3.225806\\ \hfill 5.555556\\ \hfill 7.692308\\ \hfill 10\\ \hfill 12.34568\end{array}\end{array}$

Graph obtained is shown,

The equation for the straight line of the given plot is $\frac{I_{abs}}{I_F}=\text{1 +}\left(k_Q/k_Q\right)\left[Q\right]$.

From the equation, it is clear that the slope obtained from the plot gives $\left(k_Q/k_Q\right).$

Therefore,

  $\begin{array}{c}Slope,k_Q{\text{/k}}_f\text{ = }\frac{\left(10.0-7.692308\right)}{\left(4.0-3.0\right)\left({10}^{-3}mol.d{m}^{-3}\right)}\\ \\ =2.308\times {10}^{3}mol.d{m}^{-3}{\text{ s}}^{-1}\end{array}$

$\left(k_Q/k_Q\right)$ is $2.308\times {10}^{3}mol.d{m}^{-3}{\text{ s}}^{-1}$

• Calculate $k_Q$:

A plot was drawn between $\text{1/τ}$ and $\left[Q\right]$ using the given information.

  $\begin{array}{ccc}\left[Q\right]/\left(\times {10}^{-3}mol.d{m}^{-3}\right)& \tau /ns& 1/\tau \left(\times {10}^9{s}^{-1}\right)\\ \begin{array}{r}\hfill 1\\ \hfill 2\\ \hfill 3\\ \hfill 4\\ \hfill 5\end{array}& \begin{array}{r}\hfill 76\\ \hfill 45\\ \hfill 32\\ \hfill 25\\ \hfill 20\end{array}& \begin{array}{r}\hfill 0.013158\\ \hfill 0.022222\\ \hfill 0.03125\\ \hfill 0.04\\ \hfill 0.05\end{array}\end{array}$

Graph obtained is shown,

The equation for the straight line of the given plot is $\frac{1}{\tau }\text{ = }\frac{1}{{\tau }_0}{\text{ + k}}_Q\left[Q\right]$.

From the equation, it is clear that the slope obtained from the plot is the quenching rate constant, ${\text{k}}_Q$.

Therefore,

  $\begin{array}{c}Slope,k_Q\text{ = }\frac{\left(0.03125-0.022222\right)\left({10}^9{\text{ s}}^{-1}\right)}{\left(3.0-2.0\right)\left({10}^{-3}mol.d{m}^{-3}\right)}\\ \\ =\text{ 9}\text{.028}\times {\text{10}}^{9}mol.d{m}^{-3}{\text{ s}}^{-1}\end{array}$

Quenching rate constant is $\text{9}\text{.028}\times {\text{10}}^{9}mol.d{m}^{-3}{\text{ s}}^{-1}$

• Calculate half-life of the fluorescence:

Half-life of the fluorescence is calculated as follows,

  $\begin{array}{c}{\text{t}}_{\text{1/2}}\text{ = 0}\text{.693τ}\\ \\ t_{1/2}\text{ = 0}\text{.693}\frac{1}{k_f}\\ \\ =\text{ 0}\text{.693}\times \left(\frac{k_Q/k_f}{k_Q}\right)\\ \\ =\text{ 0}\text{.693}\times \left(\frac{2.308\times {10}^{3}mol.d{m}^{-3}{\text{ s}}^{-1}}{\text{9}\text{.028}\times {\text{10}}^{9}mol.d{m}^{-3}{\text{ s}}^{-1}}\right)\\ \\ =\text{ 1}\text{.77}\times {\text{10}}^{-7}\text{ s = 0}\text{.172 }\mu \text{s}\end{array}$

Half-life of the fluorescence is $0.172\mu s.$