Question

The photochemical formation of molecule Z uses a radiation of 300 nm. The sample is  irradiated with 100W of this radiation. The quantum yield is 0.2. If the sample absorbs only  50% of the radiation falling on it, then find the amount of Z formed in one second

Answer 1

Photon energy is E = hc/l = 6.626*10^-34*3*10⁸/3*10^-7 = 6.626*10^-19 J

Power is energy per second, hence, number of photons per second is:

N(photon) = 100/6.626*10^-19 = 1.509*10²⁰

The amount of Z formed per second is:

N(Z) = N(photon)*0.5(absorption)*0.2(quantum yield) = 1.509*10¹⁹ molecules per second, or,

1.509*10¹⁹/6.02*10²³ = 2.507*10^-5 mole per second.