An electron moves from an excited state back to the ground state releasing 4.56 x 10-19 J of energy. a) Find the frequency of the photon of light that would be released. b) Find the
wavelength in nm.

Explanation:

The first step in this problem is to convert the wavelength, λ, to a frequency, ν, and

then calculate the energy of the photon using ε = hν. In such problems you must take

care to use a consistent set of units. We will use S.I. units. Thus,

λ = 18900 Å = 18900 x 10–10 m = 1.89 x 10–6 m

The frequency associated with this wavelength is given by:

ν = c/λ = (3.00 x 108 m s–1 ) / (1.89 x 10–6 m) = 1.59 x 1014 s–1

The energy ε associated with a photon of this frequency is:

ε = h ν = (6.626 x 10–34 J s) (1.59 x 1014 s–1) = 1.05 x 10–19 J

Note -- the ENERGY of this photon corresponds to a DIFFERENCE between

energy levels of the H atom

The energy levels for the H atom are given by:

εn = (–21.8 x 10–19 J) / n2 ; where n = 1,2,3,4,

Thus, ε1 = –21.8 x 10–19 J; ε2 = –5.45 x 10–19 J

ε3 = –2.42 x 10–19 J; ε4 = –1.36 x 10–19 J

We now seek an ENERGY LEVEL DIFFERENCE which matches the energy of the

photon.

We see that the electronic transition which gives rise to an EMITTED photon with

energy ε = 1.05 x 10–19 J is:

ninitial = 4 → nfinal = 3

2. A series of spectral lines in emission is characterized by a common value of nfinal -- i.e.

the quantum number of the level at which the transition terminates. Since the 18900 Å

line arises from a transition in which ninitial = 4 and nfinal = 3, the series to which it

belongs is characterized by nfinal = 3. The series limit for this series of emission lines is

the wavelength that would correspond to the transition ninitial = ∞ to nfinal = 3

From the energy level expression given in problem1 we see that:

ε∞ = (–21.8 x 10–19 J) (1/ ∞2) = 0

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