EXPERIMENT 5 – Optical Spectroscopy
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Observe emission spectra from a variety of sources using a spectrometer. Design and
implement a procedure capable of determining the composition of a solution that contains two or
more ionic salts. Construct a partial energy-level diagram for hydrogen.
I
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Optical spectroscopy involves the measurement and analysis of electromagnetic radiation
(a.k.a. light). Many of the properties of light are conveniently described by means of a classical
wave model. Within this model, light waves are characterized by such variables as frequency and
wavelength. The frequency (
ν
), which describes the number of wave crests passing a given point
per second for the light wave, is inversely proportional to the wavelength (
λ
), the distance
between successive wave crests. The light we can see, visible light, corresponds to a very small
portion of the electromagnetic spectrum, from about 400 nm (violet) to 800 nm (red). The light
wave frequency and wavelength are related to one another by the equation
c
=
λν
(
1
)
where
c
is the speed of the light wave. The speed of light in a vacuum is 3.00 × 10
8
m/s.
For phenomena where the classical wave model of light proves insufficient, a particle
model is invoked. In the particle model, light is composed of a stream of discrete particles called
photons.
The energy (
E
) of each photon is directly proportional to its frequency and inversely
proportional to its wavelength,
λ
ν
hc
h
E
=
=
(
2
)
where the proportionality constant
h
is Planck’s constant (6.6261×10
–34
J
⋅
s). Therefore,
electromagnetic radiation can be described by either the wave or particle model; the model
applied is the one that accurately describes the phenomenon being investigated.
In spectroscopy, light is used as a means of probing matter. One means of probing matter
with light uses the phenomenon of absorption. When an atom, molecule, or ion absorbs a photon,
its energy increases. The energy change of the atom must be equivalent to the energy of the
photon. Thus the absorbed wavelengths of light reveal the differences between energy levels:
E
3
= –8.0 kJ/mol
Δ
E
3,1
= E
3
– E
1
=
31.5 kJ/mol
E
3,2
1
= hcN
A
/
E
3,1
=
3800 nm
E
2
= –17.0 kJ/mol
E
2,1
= E
2
– E
1
=
22.5 kJ/mol
E
3,1
2
= hcN
A
/
E
2,1
=
5320 nm
E
2,1
E
3,2
= E
3
– E
2
=
9.0 kJ/mol
E
1
= –39.5 kJ/mol
3
= hcN
A
/
E
3,2
=
13301 nm
λ
1
λ
2
λ
3
Figure 5.1: For an atom with only three energy states as shown above, there are
only three wavelengths of light that can be absorbed, each illustrated by an arrow.
The values of the three absorbed wavelengths of light, as calculated from the
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