Chapter 17, Problem 17G.1ST

Interpretation Introduction

Interpretation:

The primary quantum yield has to be determined from the given data.

Concept introduction:

Primary quantum yield: The number of photo physical or photochemical events that lead to primary products divided by the number of photons absorbed by the molecule in the same interval.

  $\varphi \text{=}\frac{\text{number}\text{of}\text{events}}{\text{number}\text{of}\text{photons}\text{absorbed}}\text{=}\frac{{\text{N}}_{\text{events}}}{{\text{N}}_{\text{abs}}}$.

Answer to Problem 17G.1ST

The primary quantum yield was $0.17.$

Explanation of Solution

The number of photochemical events is the number of decomposed molecules and is given as follows:

    ${\text{N}}_{\text{events}}={\text{N}}_{\text{decomposed}}$.

The energy absorbed by the substance is ${\text{E}}_{\text{abs}}\text{=}\text{fPt}$, where P is the incident power, t is the time of exposure, and the factor f is the fraction of incident light that is absorbed.

${\text{E}}_{\text{abs}}$ is related to the number ${\text{N}}_{\text{abs}}$ of absorbed photons through ${\text{E}}_{\text{abs}}\text{=}{\text{N}}_{abs}hv={\text{N}}_{abs}hc/\lambda$, where $hc/\lambda$ is the energy of a single photon of wavelength.

Equating ${\text{E}}_{\text{abs}}\text{=}\text{fPt}$ and ${\text{E}}_{\text{abs}}{\text{=N}}_{abs}\left(\frac{hc}{\lambda }\right)$ for the absorbed energy is,

  $\begin{array}{l}\text{fPt}\text{=}{\text{N}}_{abs}\left(\frac{hc}{\lambda }\right)\\ \\ {\text{N}}_{abs}\text{=}\frac{\text{fPt}\lambda }{hc}----------\left(1\right)\end{array}$.

Primary quantum yield: $\varphi \text{=}\frac{\text{number}\text{of}\text{events}}{\text{number}\text{of}\text{photons}\text{absorbed}}\text{=}\frac{{\text{N}}_{\text{decomposed}}}{{\text{N}}_{\text{abs}}}$.

Substituting (1) into the above relation,

  $\varphi \text{=}\frac{{\text{N}}_{\text{decomposed}}}{{\text{N}}_{\text{abs}}}=\frac{{\text{N}}_{\text{decomposed}}hc}{\text{fPt}\lambda }$.

Given values:

  $\begin{array}{l}{\text{N}}_{\text{decomposed}}\text{=}\left(\text{0}\text{.324}\text{×}{\text{10}}^{\text{-3}}\text{mol}\right)\left(\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}{\text{mol}}^{\text{-1}}\right)\text{=}\text{1}\text{.95}\text{×}{\text{10}}^{\text{20}}\text{;}\\ \\ \text{P}\text{=}\text{87}\text{.5}\text{×}{\text{10}}^{-3}\text{W}\text{=}\text{87}\text{.5}\text{×}{\text{10}}^{-3}\text{J}{\text{s}}^{\text{-1}}\text{;}\text{t}\text{=}\text{38}\text{min}\text{×}\text{60}\text{sec}\text{=}\text{2280}\text{sec,}\\ \\ \text{λ}\text{=}\text{320}\text{nm}\text{=}\text{3}\text{.20}\text{×}{\text{10}}^{\text{-7}}\text{m}\text{,}\text{and}f=0.35.\end{array}$

By substituting the known values, the primary quantum yield is calculated as follows,

  $\begin{array}{l}\varphi \text{=}\frac{{\text{N}}_{\text{decomposed}}hc}{\text{fPt}\lambda }\\ \\ \varphi \text{=}\frac{\left(\text{1}\text{.95}\text{×}{\text{10}}^{\text{20}}\right)\left(\text{6}\text{.626}\text{×}{\text{10}}^{-34}Js\right)\left(3\text{×}{\text{10}}^{8}m{s}^{-1}\right)}{\left(0.35\right)\left(\text{87}\text{.5}\text{×}{\text{10}}^{-3}\text{J}{\text{s}}^{\text{-1}}\right)\left(\text{2280}\text{sec}\right)\left(\text{3}\text{.20}\text{×}{\text{10}}^{\text{-7}}\text{m}\right)}\\ \\ =\frac{38.8\text{×}{\text{10}}^{-6}}{2.2\text{×}{\text{10}}^{-5}}=17.6\text{×}{\text{10}}^{-6+5}=0.17.\end{array}$

The primary quantum yield was $0.17.$