Question

During photosynthesis, chlorophyll absorbs light of wavelength $440\text{nm}$ and emits light of wavelength $670\text{nm}$. What is the energy (in kJ/mol) available for photosynthesis from the absorption-emission of a mole of photons?

• A. 85.20 $kJmo{l}^{-1}$
• B. 105.20 $kJmo{l}^{-1}$
• C. 98.40 $kJmo{l}^{-1}$
• D. 93.30 $kJmo{l}^{-1}$

Answer 1

The correct option is D 93.30 $kJmo{l}^{-1}$

$\Delta E={\left[\frac{nhc}{\lambda }\right]}_{\text{absorbed}}-{\left[\frac{nhc}{\lambda }\right]}_{\text{emitted}}$

where, $\Delta E$ = Energy available
n = number of photons
$\lambda$ = wavelength of the radiation
h = Planck's constant
c = speed of light
$=nhc[\frac{1}{{\lambda }_{\text{absorbed}}}-\frac{1}{{\lambda }_{\text{emitted}}}]$

$=6.023\times {10}^{23}\times 6.626\times {10}^{-34}\times 3\times {10}^8[\frac{1}{440\times {10}^{-9}}-\frac{1}{670\times {10}^{-9}}]$

$=0.1197[2.272\times {10}^6-0.492\times {10}^6]$
$=0.0933\times {10}^6\text{J/mol}=93.3\text{kJ/mol}$