Question

During photosynthesis, chlorophyll absorbs light of wavelength $440nm$ and emits light of wavelength $670nm$. What is the energy available for photosynthesis from the absorption-emission of a mole of photons?

Answer 1

$\Delta E={\left[\frac{Nhc}{\lambda }\right]}_{absorbed}\cdot -{\left[\frac{Nhc}{\lambda }\right]}_{evolved}$
$=Nhc[\frac{1}{{\lambda }_{absorbed}}-\frac{1}{{\lambda }_{evolved}}]$
$=6.023\times {10}^{23}\times 6.626\times {10}^{-34}\times 3\times {10}^8[\frac{1}{440\times {10}^{-9}}-\frac{1}{670\times {10}^{-9}}]$
$=0.1197[2.272\times {10}^6-1.492\times {10}^6]$
$=0.0933\times {10}^{6}J/mol=93.3kJ/mol$.