Question

Assuming the wavelength for the fluorescence process is observed at 475 nm and the wavelength for the phosphorescence process occurs at 523 nm, calculate the energy released from 12.35 g of Calcium atoms (assume 1 electron is excited per calcium atom and every one of them relaxes via fluorescence or phosphorescence. NOTE: fluorescence occurs 2.45 times more frequently than phosphorescence)

Answer 1

Energy release

E= h c / Wl

h = Planck Constant = 6.626*10^-34 J s

c = speed of particle (i.e. light) = 3*10^8 m/s

E = energy per particle J/photon

WL = wavelength in meters

E(475 nm) = (6.626*10^-34)(3*10^8)/(475*10^-9) = 4.184*10^-19

E(523 nm) =  (6.626*10^-34)(3*10^8)/(523 *10^-9) = 3.800*10^-19

mol fo Ca = mass/MW = 12.35/40 = 0.30875

atoms of Ca = 0.30875*6.022*10^23 = 1.8592*10^23 atoms of Ca

then

Etotal = (Eparticle)(Particles) = (3.800*10^-20)*(1.8592*10^23) = 7064.96 J

a)

by fluorescence

Etotal = (Eparticle)(Particles) = (4.184*10^-19)*(1.8592*10^23) = 77788.928J

b)

by phosphorescence

Etotal = (Eparticle)(Particles) = (3.800*10^-19)*(1.8592*10^23) = 70649.6J