Chapter 14, Problem 11P

A.

Summary Introduction

To calculate: The energy of a mole of photon at 400nm (violet light), 680 (red light) nm and 800 (near-infrared light) nm.

Explanation of Solution

Calculation:

Mole of photons, N = 6×10²³ photons/mole

Here the energy of photon at any given wavelength is equal to product of one photon and Avogadro number.

Planck’s constant, h $\begin{array}{l}\text{= 6}{\text{.6×10}}^{\text{-37}}\text{kJ sec / photon}\\ \text{=1}{\text{.58×10}}^{\text{-37}}\text{kJ sec / photon}\end{array}$

Velocity of light, c = 3×10¹⁷ nm/sec

Energy of mole of photon at 400nm λ, E= $\frac{\text{Nhc}}{\text{λ}}$

$\begin{array}{c}\text{E=}\frac{{\text{6×10}}^{\text{23}}\text{photons}}{\text{mole}}\text{×}\frac{\text{1}{\text{.58×10}}^{\text{-37}}\text{kcal/sec}}{\text{photon}}\text{×}\frac{{\text{3×10}}^{\text{17}}\text{nm}}{\text{sec}}\text{×}\frac{\text{1}}{\text{400}\text{nm}}\\ \text{=71}\text{kcal/mole}\end{array}$

Energy of mole of photon at 680nm λ, E= $\frac{\text{Nhc}}{\text{λ}}$

$\begin{array}{c}\text{E=}\frac{{\text{6×10}}^{\text{23}}\text{photons}}{\text{mole}}\text{×}\frac{\text{1}{\text{.58×10}}^{\text{-37}}\text{kcal/sec}}{\text{photon}}\text{×}\frac{{\text{3×10}}^{\text{17}}\text{nm}}{\text{sec}}\text{×}\frac{\text{1}}{680\text{nm}}\\ \text{=42}\text{kcal/mole}\end{array}$

Energy of mole of photon at 800nm λ, E= $\frac{\text{Nhc}}{\text{λ}}$

$\begin{array}{c}\text{E=}\frac{{\text{6×10}}^{\text{23}}\text{photons}}{\text{mole}}\text{×}\frac{\text{1}{\text{.58×10}}^{\text{-37}}\text{kcal/sec}}{\text{photon}}\text{×}\frac{{\text{3×10}}^{\text{17}}\text{nm}}{\text{sec}}\text{×}\frac{\text{1}}{\text{800}\text{nm}}\\ \text{=36}\text{kcal/mole}\end{array}$

Conclusion

The energy of a mole of photon at 400nm (violet light), 680 (red light) nm and 800 (near-infrared light) nm is 71 kcal/mole, 42 kcal/mole and 36 kcal/mole respectively.

B.

Summary Introduction

To calculate: The number of seconds it would take for a mole of photons to strike a square meter of leaf.

Explanation of Solution

Calculation:

One square meter of leaf receives = 0.3 kcal/sec of 680 nm light

Energy of a mole of photon at 680 nm of light = 42 kcal/mole

Time taken by a square meter of leaf to receive one mole of photon,

$\text{t= }\frac{\text{Energy}\text{of}\text{light}\text{received}}{\text{Energy}\text{of}\text{photons}\text{for}\text{680nm}\text{light}}$

$\begin{array}{c}\text{t=}\frac{\text{sec}}{\text{0}\text{.3}\text{kcal}}\text{×}\frac{\text{42}\text{kcal}}{\text{mole}}\\ \text{=140}\text{sec/mole}\end{array}$

Conclusion

It takes 140 seconds for a mole of photons to strike a square meter of leaf.

C.

Summary Introduction

To calculate: The time it would take a tomato plant with a leaf area of 1 square meter to make a mole of glucose from CO₂

Explanation of Solution

Calculation:

Time taken for mole of photons to strike a square meter of leaf = 140 seconds.

No. of photons required to fix a molecule of CO₂= 8

Time taken to synthesize a mole of glucose is given as t

$\begin{array}{c}\text{t=}\frac{\text{140}\text{sec}}{\text{mole}\text{photons}}\text{×}\frac{\text{8}\text{mole}\text{photons}}{\text{mole}{\text{CO}}_{\text{2}}}\text{×}\frac{\text{6}\text{mole}{\text{CO}}_{\text{2}}}{\text{mole}\text{glucose}}\\ =6720\sec \\ =112\min \end{array}$

Conclusion

For a tomato plant with leaf area of 1 sq.meter it would take 112 minutes to make a mole of glucose from CO₂. The obtained value is an exaggeration as the incident photons are not entirely captured by the 1 sq.meter of leaf. Several factors like saturation of Photosystem, location of leaf on the stem and many more affect the overall efficiency.

D.

Summary Introduction

To calculate: The efficiency of conversion of light energy into chemical energy after photon capture.

Explanation of Solution

Calculation:

$\text{Efficiency=}\frac{\text{Time}\text{taken}\text{to}\text{generate}\text{a}\text{mole}\text{of}\text{glucose}}{\text{No}\text{.}\text{of}\text{photons}\text{required}\text{to}\text{fix}\text{a}\text{molecule}\text{of}{\text{CO}}_{\text{2}}\text{×Energy of mole of photon at 680nmλ}}$

$\begin{array}{c}\text{Efficiency}\text{in}\text{\%=}\frac{\text{mole}{\text{CO}}_{\text{2}}}{\text{8}\text{mole}\text{photons}}\text{×}\frac{\text{mole}\text{photons}}{\text{42}\text{kcal}}\text{×}\frac{\text{112}\text{kcal}}{\text{mole}{\text{CO}}_{\text{2}}}\text{×100}\\ \text{=33\%}\end{array}$

Conclusion

The efficiency of conversion of light energy into chemical energy after photon capture is 33%.