Light and Spectra
difference in energy between lower and higher electron
energy levels. When the electrons return to lower energy
levels, energy is emitted in the form of a photon. Since
atoms have a number of energy levels available, there are
a number of different energies that can be absorbed and
released. For example, the visible portion of the spectrum
for mercury contains the following colors:
INTRODUCTION
Light and color have intrigued humans since antiquity. In
this experiment, you will consider several aspects of light
including:
a. The visible spectrum of colors (red to violet)
b. Bright line spectra as emitted by an excited gas or
solid
c. The relationship between color, wavelength, fre‑
quency and energy.
BACKGROUND AND THEORY
Light is a form of energy called electromagnetic radiation.
It has wavelength and frequency. Wavelength, λ (lambda),
is the distance between adjacent wave crests. Visible
light has wavelengths in the 400 nm to 700 nm range
(1 nm = 10‑9 m). Frequency, ν (nu), tells how many waves
pass by a point in a second. Violet light, with a wavelength
of 400 nm, has a frequency of 7.5 x 1014 per second, many
more waves than are shown in the diagram.
λ
λ
COLOR
violet
blue
green
λ , nm
405
436
492
COLOR
yellow
orange
red
λ , nm
579
623
689
The energies of the emitted wavelengths can be correlated
with the energies of electron transitions from higher to lower
energy levels. During an energy high → energy low electronic
transition, the greater the difference between the energy
levels, the larger is the energy released, and the shorter is
the wavelength of the photon emitted. In the energy level
diagram of a generic atom shown below, the left-hand arrow
represents an electron releasing energy as it drops from the
6th to the 2nd energy level.
n=6
n=5
ν = 2 waves
ν = 3 waves
per time line
per time line
The energy of electromagnetic radiation varies directly with
the frequency, and inversely with the wavelength. Thus,
violet light is of higher energy than red light, which has
wavelengths in the 650 nm range. White light, such as that
from an ordinary incandescent light bulb, is a mixture of
wavelengths in the visible range. When white light strikes
a prism or diffraction grating, the light is dispersed into a
continuous spectrum of visible colors. A non-continuous
spectrum occurs when an electric current passes through
a gaseous element in a gas discharge tube or when metal
ions are put into a flame. This type of spectrum, called a
bright line spectrum, is not continuous but instead contains
only certain colors at particular wavelengths. A bright line
spectrum is due to transitions of electrons between energy
levels in atoms or ions. When an atom or ion absorbs energy,
such as from a flame or electrical source, it absorbs only
certain discrete amounts of energy. These amounts are the
n=4
n=3
n=2
The energy is released in the form of a photon. This photon
will have more energy (and have a shorter wavelength) than
the photon given off in the transition represented by the
right-hand arrow. In terms of energy,
(n 6 → n 2 ) > (n 3 → n 2 )
The larger energy would be in the violet-blue portion of the
spectrum. The smaller energy would be in the red-orange
portion of the spectrum.
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Frequency and Energy
Continuous Spectrum
The relationship between wavelength and frequency is:
λν = C
C, the speed of light, equals 3.00 x 108 m/s. The violet line
with λ = 405 nm from the mercury spectrum would have a
frequency calculated as follows:
3.00 x 108 m ⋅ s− 1
C
ν
=
= 7. 41x1014 s− 1
ν=
405x10− 9 m
λ or
Find the tall, clear light bulb connected to a variable power
transformer. Turn the transformer on with the dial set to high.
Hold the diffraction grating as instructed above to observe
the filament in the light bulb. A full spectrum is visible.
Now t urn off the room lights, and observe the spectrum
as the rheostat is turned down, then back up. Notice how
the blues and greens diminish in intensity more than the
reds diminish as the temperature of the filament decreases,
then reappear as the temperature increases. The theory
available 100 years ago predicted that as the temperature
increased, the spectrum would completely shift over into
the ultraviolet, with no reds or yellows or greens remaining.
This inadequate theory had physicists mumbling about the
“ultraviolet catastrophe”. Theoretical physicists looked to
develop theories consistent with experimental data. Max
Planck developed a new theory which has held up for close
to a century. The equation: E = hν is part of the theory.
Notice that λ is written as 405 x 10-9 m, rather than 405
nm.
To find the energy of a photon, the Planck relationship is
used:
E = hν
where h, the Plank constant, = 6.63 x 10-34 J·s.
To find the energy of the violet line,
E = ( 6. 63x10− 34 J ⋅ s) ( 7. 41x1014 s −1 ) = 4. 91x10 −19 J
This is the energy emitted when an electron in one particu‑
lar energy level in a mercury atom drops down to a lower
particular energy level. 4.91 x 10-19 J might seem extraor‑
dinarily small, but in terms of a mole of atoms, there would
be 296 kJ·mol-1. (Can you verify this?).
Line spectra
The conditions used in this part of the experiment, namely,
high voltage through a low pressure sample of an element,
produce a characteristic line spectrum for each element.
Each element is labeled and set-up properly. Plug in the
high voltage source if necessary.
This is a good time to make a general observation about
quantities. You will hear people say, and perhaps you say,
“that seems like an awfully small amount”. The response
to this is, “compared to what?”. A mercury atom, one of the
really heavy atoms, weighs about 3 x 10-22g. Compared to
your weight, a mercury atom weighs “nothing”. Compared
to a neon atom, the mercury atom weighs 10 times as much.
A thousand percent heavier! The moral of the story is:
Consider everything in perspective.
Hydrogen Spectrum: With the lights still out, observe the
hydrogen spectrum. The closer you get, the clearer it appears.
Observe the lines and draw them on the chart below.
red
yellow
green
blue
violet
Mercury Spectrum: Observe the lines and draw them on
the chart below. Just as H and Hg differ, so each atom has
a unique spectrum.
EXPERIMENT
Using the diffraction grating:
Diffraction gratings split a light source into its component
wavelengths. To use the grating, hold it so the “MADE IN
U.S.A.” is on the bottom facing you. The right-side card‑
board frame should be between your dominant eye and the
light source, blocking the light source from your eye. The
plastic should be to the left of your eye. You will see the
spectrum by glancing into the plastic.
red
yellow
green
blue
violet
Neon Spectrum: Simply observe. Notice the number of
lines in the red and yellow portion of the spectrum.
The unique spectrum given by each element provides a simple
means of analysis for elements. After exciting a substance
with a spark or in a flame, the position of the spectral lines
produced gives the identity of the elements in the substance,
and the intensity of the lines gives the quantity of the ele‑
ments. Of course, instruments that do this have data bases
built in that allow for the experimental data to be compared
with previously compiled data.
Light
source
Julie
Enriquez
2
Also, notice that the lines have different widths. This is
typical of spectral lines, and spectroscopists rank lines
according to their appearance and abundance. Hence the
designations s, p, d, and f for electron orbitals (from sharp,
principal, diffuse and fundamental.
CALCULATIONS
Hg Spectrum. The one you observed in the lab is a low
resolution spectrum. The spectral charts posted on the walls
show higher resolution spectra of a number of substances.
Look at a chart and sketch the spectral lines of mercury
roughly to scale in the strip below: (numbers are λ in nm)
Hydrogen Spectrum.
The hydrogen atom is the simplest atom and a detailed
study of its spectrum leads to valuable insights. Shown in
Figure 1 below are the energy levels in the hydrogen atom
as calculated from the Schrödinger equation. The numbers
on the left give the energy of each level in joules. The
number on the right gives the principal quantum number
of each level. The names along the top are the names of the
scientists who first characterized the spectrum corresponding
to these transitions. Shown in Figure 2 is a black and white
representation of the experimentally observed spectrum of
the hydrogen atom. IR is infrared, VIS is visible and UV is
ultraviolet. The wavelength of each line in nanometers is
listed below the line. Figure 2, which is experimental, can
be calculated from Figure 1, which is theoretical.
750 700 650 600 550 500 450 400
Notice that the spectrum has three lines shown in the yellow
region. Record the actual wavelengths from the chart, then
calculate the frequency and energy of each of these three
lines in the mercury spectrum:
λ
ν
E
_______
_______
_______
_______
_______
_______
_______
_______
_______
Figure 1: Energy Levels and Some Possible Transitions for Hydrogen Atoms
Lyman
Ultraviolet
Balmer
Visible
Paschen
Infrared
n=5
-0.872
n=4
-1.36
n=3
7→3 6→3 5→3 4→3
-5.45
n=2
6→2 5→2 4→2 3→2
Each arrow represents an energy
release, ∆E, as an electron goes
from a higher energy level to a
lower one. The energy is given
off as photons. The photons are
in the part of the spectrum
indicated above the arrows.
λ in nanometers
3
656
955
1282
1876
1005
Figure 2: H Atom Spectrum
IR
Far IR
Visible
UV
122
103
97.3
n=1
5→1 4→1 3→1 2→1
486
434
410
397
389
-21.8
1094
Energy/10-19 Joules
-2.42
4052
n = ∞ (13 or 14)
n=8
n=7
n=6
n values
0
-0.341
-0.445
-0.606
To give consistency to the appearance of the energies calcu‑
lated, all energies will be shown with 10-19 as the exponent.
Then the energy of the 97.3 nm line is 20.4 x 10‑19 J. This is
very close to the absolute value of the ∆E for the n 5 → n1
transition previously calculated, which was 20.9 x 10-19 J.
Upon close inspection of the left and right y-axis coordi‑
nates of Figure 1, it is seen that for the n 4 → n 1 transition,
represented by the second from the left vertical arrow:
Hydrogen Spectrum Calculations
According to spectral theory, each line in the spectrum
comes as an electron drops from a higher to a lower energy
level. In Figure 1 on page 57, the various vertical arrows
show a number of such possible transitions. The energy of
a transition, for example, the one indicated by the left hand
vertical arrow in Figure 1, can be calculated by subtracting
the energy of the initial state from the energy of the final
state (found on the left of the diagram):
∆E = E final − Einitial
= ( −21. 8x10
− 19
J ) − ( −0.872 x10
− 19
∆E = ( −21. 8x10− 19 J ) − ( −1. 36x10 −19 J)
= 20. 4 x10 − 19 J
This is exactly the same value as the Planck calculation for
the λ = 97.3 nm given just above. So we can conclude that
the spectral line at 97.3 nm is caused by an electron drop‑
ping from n = 4 to n = 1 in the hydrogen atom.
J)
= ( −20. 9x10− 19 J)
(The negative value indicates energy released.)
This ∆E is for an n 5 → n1 transition (see the right side of
Figure 1 for n values).
Each line in the hydrogen spectrum shown in Figure 2 has
an energy which can be calculated using the Planck relation‑
ship. The far right hand line has λ = 97.3 nm, or 97.3 x 10-9
m. To calculate the energy, first find the ν of the line:
ν=
Then
Your understanding of the calculations and discussion on
this page are crucial for the completion of the calculation
table below. If you have difficulty with the calculations,
reread this section. If the difficulties continue, see the
instructor.
The calculations just performed are summarized in the first
line of the table below. Continuing with the wavelengths
from Figure 2, complete the first three columns in the table.
By analyzing the energy levels in Figure 1, find which
transition is responsible for each spectral line, and indicate
the transition in the last column.
3.00 x108 m ⋅ s− 1
= 3. 08x1015 s− 1
97.3x10 −9 m
(
E = ( 6. 63x10− 34 J ⋅ s) 3. 08x1015 s −1
)
= 2.04 x10− 18 J
CALCULATION TABLE
λ in nm
1
97.3
2
103
3
122
4
389
5
397
6
410
7
434
8
486
9
656
10
955
11
1005
12
1094
13
1282
14
1876
15
4052
ν in seconds ‑1
∆E in Joules
n high → n low
3.08 x 10 15
20.4 x 10 ‑19
4 →1
4
Name_________________________________________ Grade___________ Date ___________
The Rydberg equation is a famous relationship that first showed the order of the hydrogen spectrum. To calculate the fre‑
quency, the equation has the form:
 1
1
ν = RH 2 − 2 
 n1 n2 
In this equation, n1 is the same as nlow in the table, and n2 is the same as nhigh in the table. Use the values for ν, nlow and
nhigh from the first entry in the calculation table on the previous page, and solve this equation for RH (3 significant digits).
Use the values of ν, nlow and nhigh from the 5th entry and again calculate RH.
RH (from 1)
=
________________s-1
RH (from 5)
=
________________s-1 Now use this value of RH (averaged if they differ) and the values of n1 and n2 in the 10th entry to calculate ν for the 10th
data.
ν (calculated)
=
________________s-1
Compare this calculated value to the value of the observed ν in the second column of the 10th entry.
ν (experimental) =
________________s-1
Does the calculated value agree with the observed value?______ If not, repeat your calculations.
QUESTION
When the Sun’s spectrum is carefully observed, a great number of spectral lines are apparent. If the lines are viewed in
groups, various elements are identified to cause specific groups of lines. For example, each of the lines seen in the mercury
spectrum occurs in the Sun’s spectrum.
Helium is named after Helios, the Greek name for the Sun, because evidence for its existence was observed in 1868 by
examining the sun’s spectrum. Since helium was not found on earth until 1895, how do you suppose observers knew they
had found a new element in 1868?
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