Question

In a measurement of quantum efficiency of photosynthesis in green plants, it was found that 10 quanta of red light of wavelength $6850\stackrel{\circ }{A}$ were needed to release one molecule of $O_2$. The average energy storage in this process is 112 kcal/mol of $O_2$ evolved. The energy conversion efficiency in this experiment is:

• A. 23.75%
• B. 26.90%
• C. 66.23%
• D. 73.19%

Answer 1

The correct option is B 26.90%

According to Planck’s quantum theory $E=\frac{hc}{\lambda }$
where E = energy of 1 quanta of the radiation
h = Planck's constant $=6.626\times {10}^{-34}Js$
c = speed of light =$3\times {10}^{8}m/s$
$\lambda$ = wavelength of the radiation $=6850\stackrel{\circ }{A}=6850\times {10}^{-10}m$
So, $E=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{6850\times {10}^{-10}}=2.9\times {10}^{-19}J$

Energy of 10 quanta $=10\times 2.9\times {10}^{-19}=29\times {10}^{-19}J$

Given: average energy stored $=112kcal/mol$ of $O_2$
We know 1 calorie = 4.18 J
Therefore, energy of 1 molecule $=\frac{112\times 4.18\times {10}^3}{6\times {10}^{23}}=7.80\times {10}^{-19}J$

% efficiency $=\frac{7.80\times {10}^{-19}}{29\times {10}^{-19}}\times 100=26.90\%$