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
.