A modern compact fluorescent lamp contains 1.4 mg of mercury. if each mercury atom in the lamp were to emit asingle photon of wavelength 254 nm, how many joules of energy would be emitted?

First calculate the energy per photon for Î" of 254 nm (254 * 10^-9 m) : E = h*c/Î" where h is the Planck constant h = 6.626x10^-34 Js c = speed of light = 2.998 x 10^8 m s-1 Î" = wavelength m E = energy in J photon-1 hence E = [6.626*10^-34*2.998*10^8] / [254 * 10^-9] = 7.821*10^-19 J mol of mercury = 0.0014/200.59 = 6.979*10^-6 mol number of atoms (x Avogadro Const.) = (6.979*10^-6) * (6.022*10^23) Joules produced = (6.979*10^-6) * (6.022*10^23) * (7.821*10^-19) = 3.29 J