Question

# Assuming the wavelength for the fluorescence process is observed at 475 nm and the wavelength for...

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)

#### Homework Answers

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

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