Question

A 7.00 mL ampule of a 0.130 M solution of naphthalene in hexane is excited with a flash of light. The naphthalene emits 13.0 J of energy at an average wavelength of 349 nm. What percentage of the naphthalene moleules emitted a photon?

Answer 1

mol of naphthalene = M* V= 0.130*7*10^-3 = 0.00091 mol

1 mol = 6.022*10^23 molecules

0.00091 mol --> (0.00091)(6.022*10^23) molecules = 5.48*10^20

theoretical yield:

For the wavelength:

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= (6.626*10^-34)(3*10^8)/(349*10^-9)

E = 5.695*10^-19 J/photon

Actual energy

E per molecule = E/molescules = 13/ ( 5.48*10^20) = 2.3722*10^-20 J/molecule

ratio

photons / molecules = ( 2.3722*10^-20) / (5.695*10^-19) = 0.0416 molecule per photon