Question

A 5.25mL sample of a 0.100M solution of an aromatic hydrocarbon dissolved in hexane is excited with a flash of light. The aromatic compound emits 16.2J of energy at an average wavelength of 379 nm.

-What percentage of the aromatic compound molecules emitted a photon?

Answer 1

molarity of hydrocarbon = 0.100M

volume of sample = 5.25mL = 5.25/1000L

moles of hydrocarbon = 0.1005.25/1000

                                = 5.2510^-4moles

no. of molecules of hydrocarbon = 5.2510^-4 6.022*10²³

                                               = 3.16110²⁰ molecules                                                  (1)

wavelength of light = 379nm = 37910^-9m

    using, E = hc/

                 = 6.62610^-34 2.99810⁸ / 37910^-9

                       = 5.24110^-19J

total amount of energy emitted = 16.2J

no. of photons emitted = 16.2 / 5.24110^-19

                                 = 3.09110¹⁹photons

percentage of molecules emitted a photon = no. of photons / no. of molecules 100

                                                              = 3.09110¹⁹ / 3.16110²⁰ 100

                                                              = 9.77%