Question

If the n=2 level in an atom has an energy of -5.220 EV and the n=2 to n=1 transition results in the emission of light with a wavelength of 522.89 nm, what is the ionization energy (in EV) of an electron in the n=1 level?

Answer #1

1 ev = 1.60218e-19

5.22 ev = (5.22)*1.60218e-19 = 8.3633796*10^-19 J

R = 8.3633796*10^-19

for

WL = 522.89 nm

find ionization energy of electron 1

dE = energy of transition, i.e. energy of 522.89

E = hc/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

E = ( 6.626*10^-34 )(3*10^8)/(522.89*10^-9) = 3.80156*10^-19 J

so..

dE =3.80156*10^-19 J

dE = Efinal - Einitial

3.80156*10^-19 = 8.3633796*10^-19 - E1

-E1 = (3.80156*10^-19) - (8.3633796*10^-19) =- 4.56181*10^-19 J

E1 = 4.56181*10^-19 J

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