Atomic Emission and Molecular Absorption Spectra
v062513_6pm
Objective: The student will observe the atomic emission spectra of hydrogen using a spectroscope,
determine the identity of an unknown metal ion by conducting a flame test, and plot the absorption
spectrum of crystal violet.
In the eighteenth century, a Scottish physicist Thomas Melvill reported that
when he placed different substances in a flame, they emitted light of
different colors. For example, he noted that when table salt was passed
through a flame, the resulting emitted light was bright yellow which was a
result of sodium (Figure 1). This concept was something that had been
known for several centuries before Melvill, but what was important about his
Figure 1
discovery is that he found that when the emitted light was passed through a prism
and a small slit, there was a band of colors, or a spectrum, that was created that was unique to the
specific element in the flame. Figure 2 shows the atomic emission spectrum of sodium.
Figure 2
Throughout the nineteenth century, this idea of different elements producing a unique spectrum was
utilized as a way of identifying what elements were in substances and was even used to discover new
elements. J.J. Balmer, a teacher from Switzerland, made an important discovery in 1885. Balmer was
working with hydrogen and its atomic spectrum and he knew that hydrogen emitted light at 4 different
wavelengths. Through trial-and-error, he discovered that there was a pattern that the emissions followed
that could be represented by the
1
1
1
Balmer-Rydberg equation,
= R
2
λ
nf
ni2
(1)
where nf = 2 for lines appearing in the visible light region, ni is a whole number greater than 2, R is the
Rydberg constant (1.097 x 10-2 nm-1) , and λ is the wavelength of the emission line in nanometers.
Note that the Rydberg constant necessarily has units of nm-1 because n’s are unitless, wavelength in this
model is expressed in nm, and its reciprocal is used.
Over the next thirty years, the picture of the atom would begin to take
shape. Work by J.J. Thompson, E. Rutherford and other scientists showed
that the atom was made of posi tively charged protons and neutral neutrons
found within the nucleus, and that the nucleus was surro unded by a
negatively charged electron cloud. There were several models that tried to
describe atomic structure, but one of the most important models was
proposed by Niels Bohr in 1913. Observing the atomic spectrum for
hydrogen and the work done by Rydberg and Balmer, Bohr suggested that
the hydrogen atom was made up of a nucleus with an electron “orbiting“
Spectroscopy 1
Figure 3
that nucleus. He also indicated that the electron could only occupy specific energy levels and that it
would not be found between energy levels.
The Bohr model of the atom can be seen in Figure 3.
Bohr proposed that the electron in the hydrogen atom would exist in a
stable, lowest energy or “ground state” orbital shown on Figure 4 as n =
1. However, when the atom is irradiated by light, the electron absorbs
that energy (where the energy of the incoming light matches the
difference in the energy level between the orbitals) which results in the
electron being excited to a higher energy orbital. The electron would
then relax to a lower, more stable energy level. For a more in depth
discussion of atomic spectroscopy and the Bohr model, see Chapter 7
Section 3 of Tro 2nd Edition.
The energy difference between the energy levels can be determined
using the equation:
ΔE = hν (2)
Rearranging the speed of light equation (c = νλ) to solve for frequency,
the speed of light divided by the wavelength can be substituted in
equation 2 for the frequency to give the equation:
ΔE = hc!
!!!!!!!!!!!!!!!!!λ#
Figure 4
(3)
In equations 2 and 3, h is Planck’s constant (6.626 x 10-34 J s), ν stands for the frequency, c is the speed
of light (2.998 x 108 m/s), and λ is the wavelength of the light in meters.
Bohr found that using his model he could accurately predict the wavelengths of the lines on the
hydrogen emission spectrum that Balmer had observed and described using the Balmer-Rydberg
equation. He also determined that if an electron fell from an orbital higher than n=2 to the level n=2, the
energy emission would be observed in the visible region of the spectrum. The problem with Bohr’s
model of the atom and the Rydberg-Balmer equation was that they could not accurately predict the
wavelengths of spectral lines for atoms with more than one electron due to the fact that this model and
equation do not take into account electron-electron repulsions that exist in elements with more than one
electron. Though the Bohr model has its fallacies, the idea of different energy states is still applicable to
all atoms.
- Molecular Absorbance
We’ve seen that atoms are capable of absorbing
and emitting light at specific wavelengths, but
absorption of light also occurs at the molecular level.
It’s this light absorption by molecules that allows us to
sense many of the colors we see on an everyday basis.
For example, the reason we perceive grass as being
green is due to the molecules of chlorophyll within the
grass. As seen in Figure 5, chlorophyll absorbs light in
the red and violet regions of the visible electromagnetic
spectrum. The primary wavelength of light that is not
Figure 5. Absorption spectrum of chlorophyll
Spectroscopy 2
absorbed is around 500 nm which corresponds with the green region of the spectrum. This particular
wavelength of light is reflected back to the eye and we see that the grass is green.
An effective qualitative method of determining the color of a
substance that will be observed is by looking at an artist’s color
wheel (Figure 6). The color absorbed and the color observed tend
to be complementary. For chlorophyll the color absorbed is red
and the color observed is green. As you can see from the color
wheel these two colors are complementary.
So how do molecules absorb light? Similar to atomic absorption,
molecules have internal energy levels that are responsible for the
specific wavelengths of light absorbed. When molecules form,
molecular orbitals are formed in which the electrons are found.
Figure 6. Color wheel relating color with
Similar to atomic orbitals, molecular orbitals have different
wavelength of light.
energy levels in which electrons can be found. Molecular orbitals
and molecular orbital theory is discussed in chapter 10 Section 8 of Tro 2nd Edition.
The best way to discuss light absorption by molecules is to look at the structure of a couple of molecules
that absorb light. The molecules in Figure 7 are beta-carotene, the molecule responsible for the orange
color of carrots, and the second structure is crystal violet.
Figure 7. Stucture of beta-carotene (left) and crystal violet (right).
What do these two molecules have in common? The answer is that both molecules have conjugated
bond systems. A conjugated bond system means that the molecules have alternating single (as illustrated
by a single line) and double (as illustrated by two parallel lines, =) bonds. The presence of the
conjugated bond system results in a delocalization of electrons.
Just as in atoms, high-energy states exist for molecules. When electrons in the molecule are exposed to
energy that matches the difference in energy between two molecular orbital levels, the electrons will
transition to a higher energy molecular orbital, leading to a molecular excited state. In the case of
molecules with conjugated bonds, the energy difference in the molecular orbital levels is less due to the
delocalization of the electrons. As a result, the wavelength of energy falls within the visible region of the
electromagnetic spectrum. This allows the molecule to absorb visible light and produces the colors that
we see.
Spectroscopy 3
Procedure:
All measurements should be recorded using the correct number of significant figures appropriate to the
equipment being used. All calculated values should be recorded to the correct number of significant
digits. Carefully read Chapter 1 Section 7 of Tro 2nd Edition to re-familiarize yourself with significant
digits and being consistent with them in mathematical operations.
IF THE SPEC 20 IS NOT ALREADY ON, TURN IT ON NOW TO ALLOW THE
INSTRUMENT TO WARM UP BEFORE USE.
Flame Test of an Unknown Salt
1. Choose an unknown salt and record its # on your data sheet.
2. Make a loop at the end of a paperclip.
3. Set up and light the Bunsen burner.
4. Scoop a small amount of the unknown salt with the loop end of the paper clip and move the tip of
the paper clip into the flame of the Bunsen burner.
5. Record the color of the flame on the data sheet.
Atomic Spectrum of Hydrogen
1. Plug in and turn on the power supply with the lamp in place.
2. Set up the spectroscope by moving the slit toward the center of the discharge tube while being
careful not to touch the tube.
3. Record the wavelength of the hydrogen emission lines on the data sheet.
4. Turn off the power supply and allow the discharge tube to cool.
Absorption Spectrum of Crystal Violet – Spec 20 Calibration
1. With no cuvette in the sample compartment, set
transmittance
Absorbance/Transmittance
the wavelength of the Spec 20 to 440 nm using
absorbance
595
0.0
Mode Selector
concentration
the wavelength selector knob. Be sure that the
factor
wavelength range selector on the front of the
instrument is set to the correct wavelength
range.
2. Set the Spec 20 to show transmittance using the
Wavelength
absorbance/transmittance selector.
selector knob
3. Turn the zero adjust knob until the transmittance
reading shows a value of 0.
4. Fill a cuvette approximately 2/3 full with
ON/
distilled water, wipe the outside of the cuvette
OFF
with a KimWipe, then place the cuvette into the
100% T Adjust
sample compartment and close the lid.
blank-­‐filled cuvette
0% T Adjust
5. Turn the 100% adjust knob until the Spec 20
empty cavity
Wavelength shows a transmittance reading of 100.
Filter Range lever
6. Remove the cuvette containing the distilled
water and set it aside. You will need the cuvette containing the distilled water when you begin taking
readings of the crystal violet.
Absorption Spectrum of Crystal Violet
Spectroscopy 4
1. Change the absorbance/transmittance knob to read absorbance.
2. Fill a second cuvette approximately 2/3 full of crystal violet solution, wipe the outside of the cuvette
with a KimWipe, then place the cuvette into the calibrated instrument and close the lid.
3. Record the absorbance value on the table provided on the data sheet.
4. Remove the crystal violet cuvette and set it aside.
5. Change the Spec 20 wavelength to the next wavelength shown on the data sheet table using the
wavelength selector.
6. Change the absorbance/transmittance knob to read transmittance.
7. Turn the zero adjust knob to set the transmittance equal to 0.
8. Place the cuvette containing the distilled water from the calibration step into the instrument. Using
the 100% adjust knob set the transmittance equal to 100. Remove the cuvette containing the distilled
water and set it aside.
9. Change the absorbance/transmittance knob to read absorbance.
10. Place the cuvette containing the crystal violet solution into the instrument and record the absorbance
on the data sheet.
11. Repeat steps 4-10 for all wavelengths shown on the data sheet.
Spectroscopy 5
Plotting Data in Excel
The exact instructions to use will depend on your version of Excel. The following set of
instructions and pictures are similar for all versions of Excel.
1. First, enter the values of your independent variable (x-axis values) into column A, and the
values of the dependent variables (y-axis values) into column B. For the crystal violet plot,
the wavelength values are the independent variables and the absorbance values are the
dependent variables.
2. Using the mouse, highlight the values in columns A and B.
Spectroscopy 6
3. Select “Insert” tab, select the “Scatter” tab and then select the plot type.
0.6 0.5 0.4 0.3 Series1 0.2 0.1 0 0 100 200 300 400 500 600 700 -­‐0.1 4. Click on layout under chart tools
a. Chart Title – Wavelength vs Absorbance for (compound name)
5. Format the axis.
a. Axis titles
i. X-axis – Wavelength
ii. Y-axis – Absorbance
b. Right click on the X-axis, click on format axis, and change the minimum and
maximums to include your data without much excess. Make sure to fix the minimum
and maximum. Do the same for the Y-axis.
Spectroscopy 7
Wavelength vs Absorbance 0.6 Absorbance 0.5 0.4 0.3 0.2 0.1 0 350 400 450 500 Wavelength Spectroscopy 8
550 600 Pre-Laboratory Assignment
1. An electron in a lithium atom moves from the 2p orbital to the 2s orbital with a ΔE of
2.96 x 10 -19 J. When the transition occurs, energy equal to ΔE is released in the form of a photon.
What is the wavelength of the light that is emitted?
2. Does the emission line determined question1 fall in the visible region of the electromagnetic
spectrum? If so, what color is the light that is emitted?
3. Which of the following molecules, based on its structure, would most likely absorb light? Circle all
that apply and explain your reasoning.
Spectroscopy 9
Data Sheet
Flame Test
Unknown # __________
Color _____________
Atomic Spectrum - Hydrogen
λ1 = _____________ nm
color = _____________
λ 2 = _____________ nm
color = _____________
λ 3 = _____________ nm
color = _____________
λ 4 = ______________ nm
color = _____________
Molecular Absorbance Spectrum – Crystal Violet
Wavelength (nm)
Absorbance
Wavelength (nm)
440
570
460
580
480
590
500
600*
adjust filter lever
520
610
540
620
560
640
Spectroscopy 10
Absorbance
Results
1. Using the chart provided, what is the identity of the metal in the unknown from the flame test ?
Ion
K
Na
Li
Cu
Sr
Color
Lilac
Yellow
Red
Green
Red/Orange
Unknown? _______________________________
2. What is the approximate wavelength of the light emitted from the unknown? From the wavelength,
determine ΔE for the transition.
Color
Red
Red/Orange
Orange
Orange/Yellow
Yellow
Yellow/Green
Green
Green/Blue
Blue
Blue/Violet
Violet
Wavelength(nm)
701
622
609
597
587
577
535
492
474
455
423
Wavelength? _____________________________
ΔE (J/photon)? ____________________________
ΔE (kJ/mol)? ____________________________
Spectroscopy 11
3. Calculate the energy of each photon associated with each of the wavelengths of the hydrogen lines.
Lower energy Purple line
Higher energy Purple line
Teal-blue line
Red line
Spectroscopy 12
4. Using the Rydberg equation, calculate the energy level (ninitial) from which the electron fell to
produce the emission. Remember: when visible light is produce the electron falls to the 2nd energy
level. See p 301 of Tro 2nd ed.
Lower energy Purple line
Higher energy Purple line
Teal-blue line
Red line
Spectroscopy 13
5. Indicate the electron transition (with arrows) and associated energy for each of the four
wavelengths of light observed for hydrogen on the diagram below.
6. Plot the absorbance vs. wavelength data of crystal violet in Excel. At what wavelength does
the maximum absorbance of light occur?
7. Does the absorbance obtained explain the color of crystal violet? Explain.
Spectroscopy 14
Post-Laboratory
1. The Balmer-Rydberg equation fails to accurately predict the quantized energy levels for
atoms with more than one electron.
a. What physical force is absent in the hydrogen atom, but present in all other atoms, that
allows the Rydberg equation to work for hydrogen?
b. Can multi-electron atoms still give line spectra? Explain.
2. Use a resource to identify two elements that were discovered because of a unique, previously
unobserved, line in their emission spectrum using a Bunsen flame.
3. The line spectrum of a certain
system is shown in the figure at
right. These lines result from
transitions among the four levels
shown.
n"="4"
n"="3"
A"
B"
C"
D"
E"
n"="2"
a. Which of the photons is the
most energetic?
!"""""shorter"wavelengths"
observed"line"spectrum"of"system"
b. Add an arrow to the level
diagram representing the
transition associated with the
longest wavelength photon.
n"="1"
c. Give the levels associated with
each of the following photons:
energy"levels"of"system"
Photon B: _______
Photon C: _______
Spectroscopy 15
F"
4. Phenolphthalein is an indicator that is colorless in the presence of an acid and pink in the
presence of a base as you discovered during the acid / base titration lab. Given knowledge
obtained in this laboratory, explain why phenolphthalein behaves as it does in acid/base
solutions. (Note: the CO2-, called a carboxyl group, contains a charge which indicates extra
electrons on the oxygen.)
O"
O"
O"
O"
O"
O"H"
phenolphthalein"
colorless"in#acid#
phenolphthalein"
pink"in#base#
O"H"
O"
Spectroscopy 16