In a measurement of quantum efficiency of photosynthesis in green plants, it was found that 10 quanta of red light of wavelength 6850 Å were needed to release one molecule of `O_(2)`. The average energy storage in this process for 1 mol `O_(2)` evolved is 112 Kcal.
What is the energy conversion efficieny in this experiment?
Given: 1 cal =4.18 J, `N_(A)=6xx10^(23),h=6.63xx10^(-34)` J.s

Options

A. 23.5
B. 26.9
C. 66.34
D. 73.1

Answer

To solve the problem of calculating the energy conversion efficiency in the experiment measuring the quantum efficiency of photosynthesis, we will follow these steps:

### Step 1: Calculate the energy of one quantum of light
The energy \( E \) of a photon (quantum of light) can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 6.63 \times 10^{-34} \, \text{J.s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( \lambda = 6850 \, \text{Å} = 6850 \times 10^{-10} \, \text{m} \) (wavelength in meters)

Substituting the values:
\[ E = \frac{(6.63 \times 10^{-34} \, \text{J.s})(3 \times 10^8 \, \text{m/s})}{6850 \times 10^{-10} \, \text{m}} \]
Calculating this gives:
\[ E \approx 2.9 \times 10^{-19} \, \text{J} \]

### Step 2: Calculate the total energy for 10 quanta
Since 10 quanta of light are needed to release one molecule of \( O_2 \):
\[ \text{Total energy} = 10 \times E = 10 \times 2.9 \times 10^{-19} \, \text{J} = 2.9 \times 10^{-18} \, \text{J} \]

### Step 3: Convert the energy storage for 1 mole of \( O_2 \)
The average energy storage for 1 mole of \( O_2 \) evolved is given as 112 Kcal. We need to convert this into Joules:
\[ \text{Energy in Joules} = 112 \, \text{Kcal} \times 1000 \, \text{cal/Kcal} \times 4.18 \, \text{J/cal} = 468256 \, \text{J} \]

### Step 4: Calculate the energy per molecule of \( O_2 \)
To find the energy storage per molecule, we divide the total energy by Avogadro's number \( N_A = 6 \times 10^{23} \):
\[ \text{Energy per molecule} = \frac{468256 \, \text{J}}{6 \times 10^{23}} \approx 7.8 \times 10^{-19} \, \text{J} \]

### Step 5: Calculate the energy conversion efficiency
The energy conversion efficiency can be calculated using the formula:
\[ \text{Efficiency} = \left( \frac{\text{Energy stored per molecule}}{\text{Total energy for 10 quanta}} \right) \times 100 \]
Substituting the values:
\[ \text{Efficiency} = \left( \frac{7.8 \times 10^{-19} \, \text{J}}{2.9 \times 10^{-18} \, \text{J}} \right) \times 100 \approx 26.9\% \]

### Final Answer
The energy conversion efficiency in this experiment is approximately **26.9%**.

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