Question

what is the minimum number of moles of 680 nm photons required to bring about the reduction of a mole of NADP⁺?

4

1

3

2

Answer 1

Light energy can be described as nhν, where n is a number of photons, h is Planck's constant, and ν is the frequency of the light.

E = nhν

n = 1 mole E= hν = hc/λ = (6.625 x 10^-34 J.s) x (3 x 10⁸ m) (10⁹ m) x 6.023 x 10²³ mol) / 680 nm/m = 176 kJ/mol

This process corresponds to a Gibbs Free Energy change of about +450 kJ/mol.

450 / 176 = 2.56 therefore minimum of 4 photons (+1 for equation)

Answer: 4

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