Question

A student is studying the photodissociation of I_{2 to} I when a sample I₂ is irradiated with a power of 279 mW at 590 nm for 41 seconds. Assuming complete absorption of the incident radiation , what is the quantum yield of the reaction

Answer 1

Quantum yield () = No. of molecules of I₂ dissociated in a given time / No. of photons absorbed at the same time

= No. of moles of I₂ dissociated in a given time / No. of Einsteins of radiation absorbed at the same time

The power of radiation = 279 mW = 279*10^-3 W = 279*10^-3 J/s

Now, the amount of energy transferred = 279*10^-3 J/s * 41 s = 11.439 J

1 Einstien = 11.97 J mol^-1 / 590*10^-7 cm = 202881 J

Therefore, the no. of Einsteins of radiation absorbed = 11.439/202881 = 5.638*10^-5