Chapter 15, Problem 1P

Interpretation Introduction

Interpretation:

$\Delta {\text{G}}^0$ value should be determined for the excitation of P700 and its comparison with the energy in an Einstein of 700nm photons should be explained.

Concept introduction:

P700 is primary donor for photosystem I where P denotes the pigment and since it absorbs at 700nm hence named as P700.

In the process of photo excitation, when light is absorbed by the photosystem, one electron of chlorophyll P700 gets excited to the higher energy level.

An einstein energy is defined as the energy associated with one mole of photons of certain wavelength.

Answer to Problem 1P

Solution:

Total $\Delta G^0=164.0245kJ/mol$

$\Delta {\text{G}}^0$ obtained is almost $95.92\%$ of the Einstein energy of 700nm photons.

Given:

The raise in potential for the P700 reaction center is +0.4 to −1.3 volts.

Explanation of Solution

Step1:

Let us find change in potential first as:

$\begin{array}{c}\Delta E^0=E_{final}^0-E_{Initial}^0\\ =-1.3V-(+0.4V)\\ =-1.7V\end{array}$

Now, the free energy change is calculated as follows:

$\Delta G^0=-nF\Delta E^0$

Where, n is number of electrons involved, F is Faraday's constant (96485C/mol) and E⁰ is cell potential.

Here,

$\begin{array}{c}\Delta E^0=-1.7V=-1.7J/C\\ (1V=1J/C)\end{array}$

$n=1$

Since, the P700 gets excited through only one electron, so value of "n" will be 1.

Putting all the values we get:

$\begin{array}{c}\Delta G^0=-nF\Delta E^0\\ =-1\times 96485C/mol\times -1.7J/C\\ =164024.5J/mol\\ =164.0245kJ/mol\end{array}$

Step2:

Einstein is energy of one mole of photons.

Let us find the energy of one photon as:

$E=\frac{h\times c}{\lambda }$

Where, h is Planck's constant, c is speed of light and λ is wavelength.

Putting the values:

$\begin{array}{l}h=6.626\times {10}^{-34}J.s\\ c=3\times {10}^{8}m/s\\ \lambda =700nm=7\times {10}^{-7}m\end{array}$

Hence,

$\begin{array}{c}E=\frac{6.626\times {10}^{-34}J.s\times 3\times {10}^{8}m/s}{7\times {10}^{-7}m}\\ =2.84\times {10}^{-19}J\end{array}$

Here, energy of one photon is $2.84\times {10}^{-19}J$.

Hence, an einstein energy will be energy of one mole of photon, which can be calculated as:

$\begin{array}{c}Einstein=E\times N_A\\ =2.84\times {10}^{-19}(J/photon)\times 6.022\times {10}^{23}(photons/mol)\\ =1.71\times {10}^{5}J/mol\\ =171kJ/mol\end{array}$

Step3:

Comparison of the two energies ($\Delta {\text{G}}^0$ and Einstein) is done as follows:

In the given photoexcitation, the percentage of energy conserved by the P700 is calculated as:

$\begin{array}{c}PercentofenergyconservedinP700= \frac{EnergyconservedbyP700}{Einsteinenergyofphoton}\times \text{100}\\ \text{= }\frac{164.0245kJ/mol}{171kJ/mol}\times 100\\ =95.92\%\end{array}$

Hence, $\Delta {\text{G}}^0$ obtained is almost $95.92\%$ of the Einstein energy of 700nm photons.

Conclusion

The total obtained, $\Delta G^0=164.0245kJ/mol$.

It is almost $95.92\%$ of the Einstein energy of 700nm photons.