Answer:
9.86 × 10²⁴ photons
Explanation:
Let's consider the photosynthesis global reaction.
6 CO₂(g) + 6 H₂O(l) → C₆H₁₂O₆(s) + 6 O₂(g)
We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.
ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)
where,
ni are the moles of reactants and products
ΔG°f(p) are the standard free Gibbs energies of formation of reactants and products
For the balanced equation for 1.00 mol of glucose, ΔG° is:
ΔG° = [1 mol × ΔG°f(C₆H₁₂O₆(s)) + 6 mol × ΔG°f(O₂(g))] - [6 mol × ΔG°f(CO₂(g)) + 6 mol × ΔG°f(H₂O(l))]
ΔG° = [1 mol × (-910.6 kJ/mol) + 6 mol × 0 kJ/mol] - [6 mol × (-394.4 kJ/mol) + 6 mol × (-237.1 kJ/mol)]
ΔG° = 2878 kJ
The energy of each photon (E) can be calculated using the Planck-Einstein's equation.
E = h . c. λ⁻¹
where,
h is the Planck's constant
c is the speed of light
λ is the wavelength
E = (6.626 × 10⁻³⁴ J.s) × (3.00 × 10⁸ m/s) × (680 × 10⁻⁹ m)⁻¹ = 2.92 × 10⁻¹⁹ J
The minimum number of photons required is:
[tex]2878\times 10^{3} J.\frac{1photon}{2.92\times 10^{-19}J} =9.86 \times 10^{24}photon[/tex]
