Given:
lambda = 440 nm = 4.4*10^-7 m
Find energy of 1 photon first
use:
E = h*c/lambda
=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(4.4*10^-7 m)
= 4.518*10^-19 J
This is energy of 1 photon
Energy of 1 mol = energy of 1 photon * Avogadro's number
= 4.518*10^-19*6.022*10^23 J/mol
= 2.721*10^5 J
= 2.721*10^2 KJ
Answer: 272 KJ
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