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1.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
= h
ν
.
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 10
8
m s
–1
) / (1.89 x 10
–6
m) = 1.59 x 10
14
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 10
14
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:
ε
n
= (–21.8 x 10
–19
J) / n
2
;
where
n = 1,2,3,4,.
............
Thus,
ε
1
= –21.8 x 10
–19
J;
ε
2
= –5.45 x 10
–19
J
ε
3
= –2.42 x 10
–19
J;
ε
4
= –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:
n
initial
=
4
→
n
final
=
3
2. A series of spectral lines in emission is characterized by
a common value
of
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 Spring '09
 pH

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