Chapter 21, Problem 1P

P700* Has the Most Negative Standard Reduction Potential Found in Nature In photosystem I. P700 in its ground state has an $\Delta {\wp }_{\circ }^{\text{'}}$= +04 V. Excitation of P700 by a photon of 700-nm light alters The $\Delta {\wp }_{\circ }^{\text{'}}$of P700* to -0.6 V. What is the efficiency of energy capture in this light reaction of P700?

Interpretation Introduction

To propose:

In the P700 light reaction, energy efficiency should be determined.

Introduction:

The P700 is also known as primary donor photosystem I which is the chlorophyll that reacts at the center. This molecule associates with the photosystem I and has 700 nm the absorption spectrum.

Explanation of Solution

The equation of quantum light energy is given as,

  $E=N\frac{hc}{\lambda }$

We have,

  $\begin{array}{l}h=6.626\times {10}^{-34}\text{Jsec}\\ N=6.02\times {10}^{23}\\ c=3\times {10}^8\text{m/sec}\\ \lambda =700\text{nm}=700\times {10}^{-9}\text{ m}\end{array}$

Substituting we have,

  $E=(6.02\times {10}^{23})\frac{(6.626\times {10}^{-34})(3\times {10}^8)}{700\times {10}^{-9}}$

  $\begin{array}{l}=170950.8\text{J}\\ =17\text{1 kJ}\end{array}$

  $\Delta G^{0\text{'}}$ Is calculated as,

  $\begin{array}{l}\Delta G^{0\text{'}}=-nF\Delta E_0^{\text{'}}\\ n=1\\ F=96.48\text{5 kJ/Vmol}\end{array}$

  $\Delta E_0^{\text{'}}=\text{acceptor}\Delta E_0^{\text{'}}-\text{donor}\Delta E_0^{\text{'}}$

  $\begin{array}{l}=-0.6-(0.4)\\ =-1.0V\end{array}$

So hence,

  $\Delta G^{0\text{'}}=-1(96.485)*-1.0$

  $\Rightarrow 96.5\text{ kJ/mol}$

P700 energy capture is calculated by the change of free energy of Gibb that is divided by one 700 nm photon for P700

  $\text{Efficiency}=\frac{96.5}{171}\times 100\%$

  $\begin{array}{l}=0.564\\ =56.4\%\end{array}$