Just for fun!
Not that you would, BUT if you ever have to take a breathalyzer test…why should you think: ORANGE?!
Breathalyzers use an orange solution of dichromate ions that react with alcohol and turn into green chromate. A visible spectrophotometer is used to measure the absorption of these two colors to determine how ‘orange’ our ‘green’ the solution has become and whether you have any alcohol in your breath.
How Stuff Works: How Breathalyzers Work
Did you know spectroscopy has shown us that the universe is expanding?
You might have heard of Doppler radar to forecast the weather, but how does a “Doppler shift” tell us that the universe is expanding? http://www.britannica.com/eb/article- This website gives some more applications and uses for spectroscopy in the UV and visible spectrum:
Wikipedia: Ultraviolet-Visible Spectroscopy


Jokes
An electron sitting in a prison asked a second electron cellmate, "What are you in for?"
To which the latter replied, "For attempting a forbidden transition."

Q: Why does hamburger have lower energy than steak?
A: Because it's in the ground state.

Q: What happens when electrons lose their energy?
A: They get Bohr'd.


Smiley face image taken from Microsoft Office Clip Art

Spectroscopy of Atoms¹

Goals

• To understand the relationship between the color emitted by atoms and their electronic structure.
• To understand the relationship between the absorption spectra of atoms and their electronic structure.
• To use color emitted by metal atoms in a flame to identify the metal in an unknown salt. 
• To use the color of a solution to predict the nature of its visible absorption spectrum.

Background

I. Light, Absorption, Emission and Transmission

• Absorption: The energy of an incident photon is used to promote an electron in the substance to a higher energy level. The photon ceases to exist.
• Emission: Energy is released by photon production in a material. Physical processes that result in emission include fluorescence, phosphorescence, and black body radiation. 
• Transmission: Photons pass through a material without being blocked.

The radiant energy emitted by the sun (or other stars) contains all possible wavelengths of electromagnetic radiation ("light"). We can see due to reactions that occur when this light is absorbed by our eyes. The portion of electromagnetic radiation to which the retina of the human eye responds is called the visible light region of the electromagnetic spectrum: 

The fact that the radiation emitted by the sun contains a mixture of radiation wavelengths may be demonstrated by passing sunlight through a prism.  A prism transmits light only after bending the light's path; the degree to which light's path is bent is related to the wavelength of the light.  When sunlight (or other "white" light, containing all possible visible wavelengths) is passed through a prism, each component color of the white light is bent to a different extent by the prism, resulting in a beam of white light being spread out into a complete rainbow of colors.  Such a rainbow pattern is called a continuous spectrum.

Figure 1:  The spectrum of white light.²  White light enters the prism from the right and exits to the left. When white light is incident on a prism, the component wavelengths are spread out into a continuous rainbow spectrum.

It was discovered that the use of a narrow slit in a spectroscope between the prism and the source of white light sharpened and improved the quality of spectra produced by a beam of white light.

II. Emission

Fluorescence occurs when an excited electron drops to a lower energy level. A photon is released, whose energy corresponds to the difference between the two energy levels.

For example, neon gas will glow with a bright red color when excited with a sufficiently high electrical voltage; this is a property used in neon signs.  When energy is applied to these atoms, electrons within the atoms move from their normal states (ground states) to states of higher potential energy (excited states). See Figure 2.  Later, the atoms “relax,” giving off energy in the form of light (Figure 2). When returning to the ground state, the electrons will emit energy (ΔE = hν). This emission is called fluorescence. For the simple atom shown in Figure 2, six emission lines would be observed, corresponding to the difference in energy between the initial and final states of the downward arrows.

Figure 2: A diagram showing transitions of electrons between energy levels in an atom. The upward arrows indicate excitation of ground-state electrons. The downward arrows indicate emission. The frequency of light absorbed or emitted corresponds to the energy difference between initial and final states (ΔE = hν).

If the light emitted by atoms of a particular element is passed through a prism and is viewed with a spectroscope, only certain sharp bright-colored lines are seen in the resulting spectrum. See Figure 3. The positions of these colored lines occur in the same location that they would in a broad white-light spectrum.  However, a line spectrum only shows a few wavelengths because only a few wavelengths are emitted by fluorescence from an atomic gas. Unlike the broad spectrum emitted by a source such as the sun, pure atomic gases emit a limited number of wavelengths because there are a limited number of energy transitions available to them due to the quantized nature of their energy levels. The experimental demonstration of bright line spectra implies a regular, fixed electronic structure for the atom and led to an enormous amount of research to discover exactly what that structure is.

Figure 3:  The line spectrum of lithium gas.

Black body radiation occurs when objects are heated. Any warm object will emit photons, and its temperature determines the emission spectrum. An emission spectrum due to black body radiation has one very broad peak, unlike fluorescence, indicating that black body radiation does not involve electrons changing energy levels (if it did, the emission spectrum would be a line spectrum). At low temperatures, a black body radiation spectrum is a broad low-energy peak, and higher energy photons are added to the emission spectrum as the object gets hotter. Mammals emit infrared light because they are warm-blooded. A warmer object will emit radiation at higher energies. For example, a fireplace poker will begin emitting visible light when it gets very hot (“red hot”).  As it gets even warmer, it will emit even higher frequencies, to include the entire visible range (“white hot”).  

III. Absorbance and Beer's Law

When light passes through a substance, some of it may be absorbed. The transmitted light can be measured by a spectrophotometer (see below). The amount of light that gets through the sample is given in units of % Transmittance (%T) or absorbance (A), where A= -log (T). Absorbance is related to the concentration of the absorbing chemical compounds. The mathematical relationship between concentration and absorbance is called Beer's Law:

A = εcl

Where A = absorbance, ε = extinction coefficient (molar absorptivity), c = concentration, and l = path length. This means that absorbance is directly proportional to concentration. In other words, if you double the concentration of the absorbing species, the absorbance will double.

IV. Absorption Spectra in Solution

One of the most obvious properties of any substance is its color.  The green color of leaves is due to chlorophyll absorption, the orange of a carrot or tomato arises from carotenes, and the red of blood results from hemoglobin.  For solid objects, the color is characteristic of the spectrum (in the visible region) of light scattered by the substance when white light (or sunlight) shines on it (light that is not absorbed is scattered).  For solutions or transparent objects, the color is characteristic of the spectrum (in the visible region) of light transmitted by the substance when white light (or sunlight) shines through it (light that is not absorbed is transmitted).  The quantitative measure of color that we use is the absorption spectrum, a plot of absorbance vs. wavelength (or frequency), indicating the absorption of radiation by the substance.  The absorption spectrum is complementary to the transmission of light.   Chlorophyll is green because it absorbs strongly in the blue (435 nm) and red (660 nm) regions of the spectrum.   Thus, the "transmission window" is left around 550 nm, which corresponds to green light.

Figure 4:  The absorbance spectrum of chlorophyll.

When light is sent through a solution containing a colored substance, some of the light will be absorbed.  The absorption spectrum, like an emission spectrum, has characteristic energies at which the atom (or ion or molecule) absorbs light.  This is due to the fact that atoms, ions, and molecules all have specific ground and excited states.  Thus, only these discrete states can be occupied by the electrons of the substance. Upon absorption of light, electrons in the substance are promoted to the excited states.  The absorption spectrum can help us determine the amount of colored substance we have in a solution.  Later in the semester we will make use of Beer’s Law, which states that absorbance of a solution is proportional to the concentration of absorbing atoms/molecules in the solution.

Figure 5:  Energy levels of a substance in solution. 

Absorbance spectra of atoms in solution look very different from absorbance spectra of atomic gases (compare Figures 3 and 4).  Atomic gases absorb only discrete frequencies.  Absorption peaks in solution appear broader because solvent molecules interact with each atom in solution, making their levels slightly different.  Each absorber absorbs slightly different wavelengths of radiation, resulting in a broad absorption peak.

V. Instruments for measuring emission and absorption spectra

Spectroscope

In order to examine the emission spectra of a number of elements, you will use a simple spectroscope of the sort depicted in Figures 6 and 7.  The spectroscope includes four major features:  a slit for admitting a narrow, collimated beam of light; a prism or diffraction grating that spreads the incident light into its component wavelengths; a telescope for viewing the spectrum; and an illuminated reference scale against which the spectrum may be viewed.  In our experiments, we will not use the reference scale extensively. You may use it to get a general idea of the correlation between wavelength and observed color.

Figure 6: How a spectroscope works.

Figure 7: Wavelength scale.

Spectrophotometer

In order to measure an absorption spectrum, a spectrophotometer, also known as a spectrometer, is used.  A simple version of this instrument is shown in Figure 8.  The light from a lamp is split into its component wavelengths by a prism or diffraction grating.  This selected light is sent through our solution, and the detector then measures the amount of light that reached it.  A computer compares how much light reached the detector when no sample was in place (100% transmission, T) with how much light reached the detector when a sample was in place (< 100% transmission) and converts this to an absorbance reading according to the relationship A = -log T.  For example, if only 90% of the light is transmitted when the sample is in place, the absorbance at that wavelength is A = -log T (or -log (0.90) = 0.046 = A).  A plot of Absorbance vs. Wavelength is then generated.  In our case, we will measure visible light absorbance, but one can also design spectrometers to measure ultraviolet (UV) light absorption or infrared (IR) light absorption.

Figure 8: Components of a simple spectrophotometer.  The spectrum is recorded as Absorbance vs. Wavelength.

Synopsis

There are three major portions to this lab.  In Part I, you will observe fluorescence from various metal salts heated in a flame. You will then be given one unknown which you will identify based on the color of the flame upon its heating. 

In Part II, you will determine emission wavelengths more quantitatively, using spectroscopes.  You will be able to identify line spectra of mercury, hydrogen, and other atoms by observing gas discharge lamps, in which atoms are electrically excited and emit light depending on their atomic energy levels.  You will be able to use the information you obtain to identify the gas in overhead fluorescent lights.

In Part III, you choose a colored solution and guess its absorption spectrum.  Then you will use a spectrophotometer to measure the absorption spectrum.  Part III is different not only because you will measure absorption rather than emission, but also because you will be studying atoms in solution rather than gaseous atoms.

Preparation

Reading Assignment:

• Description of Experiment—see below.
• Atomic Theory—Zumdahl, Sec. 12.1-12.5, pp. 510-530; note the visible spectrum relating wavelengths and colors, p. 512.

Questions:

Fill out the prelab worksheet that can be found at the end of the Experiment section.

Print:

Print and bring to lab Workshop 3 in pdf or Word format. Please print double-sided. If you think it would be beneficial to you, feel free to attempt the workshop before you come to lab.

Experiment

To print instructions, select the portion that you with to print, choose File/Print, and choose "selection" to prevent printing the entire document.

Safety

• Wear safety glasses.
• Place all waste solutions in the waste containers provided. 
• Clean up after yourselves:
  • Volumetric flasks get rinsed and go back where they came from
  • Volumetric pipettes go tip-up in cylinder in front of room
  • Plastic cuvettes go in the trash
  • Test tubes and disposable pipettes go in the glass waste boxes

Materials and Equipment

• 0.5 M solutions of NaCl, KCl, LiCl , BaCl₂, CaCl₂
• Unknown salt solution for flame test
• Dilute HCl
• Test tubes
• Nichrome wire loops
• Bunsen burners
• Elemental lamps, incandescent lamp, half-coated fluorescent lamp
• Spectroscopes
• Colored solutions
• Plastic cells for absorbance measurements
• Diode array spectrometers

Instructions

Students will work in pairs and may do these four parts in any order. 

Part 1: Emission of light from atoms in flames

Obtain a test tube holder, five test tubes and a nichrome wire loop.  Fill each test tube with a small amount of a known salt solution and label them. Obtain a 50 mL beaker and put approximately 30-mL of dilute HCl into it.  Light a Bunsen burner and adjust the flame so that you have a clear inner blue flame – this is the hottest part of the flame. Clean the wire by immersing it into the HCl solution and then holding it in the hottest portion of the flame. Impurities left on the wire from previous experiments will impart a color to the flame (compare the color to the flame without the wire in it). The salt that burns bright orange is the most difficult to clean off the wire, so make sure that the flame does not burn bright orange.  If the impurities remain after the wite tip glows hot for 10 seconds, start again with another rinse.  After you are satisfied that the wire is clean, dip the wire into one of the sample solutions and then into the flame.  Some of the solute metal ions capture free electrons from the flame and become electronically neutral metal atoms in excited states. These metal atoms lose energy and give off light characteristic of the element.  Write down the colors for each salt you burn in the flame. Remember to clean the wire between each measurement.

Obtain another test tube and fill it with a small amount of an unknown solution.  Be sure to note the letter of your unknown, e.g.Unknown E.  This unknown will contain the same salt as one of the five known salt solutions measured above.  Conduct the same test as above.

Part 1 Practice questions (you do not need to turn these in):

1. What is the identity of the metal in your ‘unknown’ salt solution?  Be sure to note the letter of your unknown and justify your answer.
2. Why do different elements exhibit different colors in a flame?  To help justify your answer, please state why one salt solution would emit a yellow flame and another salt solution would emit a green flame.
3. An electron emits orange light when transitioning between one pair of energy levels and green light when transitioning between a different pair of energy levels. Which pair of energy levels is further apart?

Part 2:  Using a spectroscope to measure atomic emission line spectra

The simple spectroscope you will use has the same design as the one shown in Figure 6. It allows you to separate the wavelengths of light into a spectrum.

Hold the eyepiece of the spectroscope up to your eye.  Look through the slit at the small end at a bright light source (like a lamp).  You will see the light through a vertical slit on the left and the spectrum of the light source projected onto a wavelength scale on the right.  Adjust the position of the spectroscope until the light source, as seen through the slit, is as bright as possible.  If the wavelength scale is too dim to read, try to use a white background for the back of the small window of the scale.

When the light source contains discrete spectral lines, one or more vertical colored lines will appear on the wavelength scale and you will be able to read each line's wavelength.  The large numbers on the scale are hundreds of nanometers. See Figure 7 for how to read the scale.

With your spectroscope, look at the following lamps (found in the hoods): Hydrogen, Mercury, Helium, and Sodium, and the half-coated fluorescent light.  Note the wavelengths, colors, and intensities of the lines generated by each light source. Note the colors that the lamps appear to the unaided eye. Observe an incandescent light bulb with your spectroscope and note how it is different from the elemental lamps observed earlier.

Part 2 Practice questions (you do not need to turn these in):

1. Why are the spectra produced by the Hydrogen and Mercury lamps different?
2. What is the difference between what you see with the spectroscope when viewing the elemental lamps and what you see when you look at the incandescent light bulb with your spectroscope?  Are they producing light in the same way? Why or why not?
   [Note: www.howstuffworks.com may be helpful in answering these questions.]
3. Which gas is present (from the 4 you observed in lab) in the fluorescent lamps above you?  Please justify your answer.
4. If you were to observe your heated unknown salt solution through the spectroscope, could you conclusively identify it now?  Why might this be more conclusive?
5. An unknown solution from lab produced an orange flame when combusted.  How would a spectroscope help you decide if it was CaCl₂ or NaCl?

Part 3: Obtain an absorption spectrum of a colored ion in solution

Obtain one of the colored solutions from your Instructor. Based on the color of your solution, what wavelengths of visible light would you expect to be absorbed by the solution?  Answer this in your lab notebook before taking a spectrum.  Remember that the colors you see are not absorbed by your sample.

Take two square plastic cuvettes. Fill each ~3/4 full, with 1) deionized water, and 2) colored solution. Take care not to touch the smooth sides of the cuvette, where the light will travel through.

Take absorption spectra (400-750 nm) of your two colored solutions using the Vernier/Ocean Optics spectrometers found in the back of the lab. Take care to place the ridged sides of the cuvette out of the path of light. Use deionized water as your background. Directions are given in the Appendix.  Mark the wavelength of maximum absorbance and print the spectrum or save it electronically. Be sure the spectrum makes it into your lab notebook.

Part 3 Practice questions:

1. What did you predict the absorption spectrum of your colored solution to look like?
2. How closely does your measured solution match your prediction?
3. What color(s) were absorbed and transmitted by your colored solution?
4. How does the chemistry in Part 3 differ from the chemistry seen in Part 2? (Hint: There are two different ways)
5. If you had a more concentrated solution, thus a darker colored solution, would you expect the absorbance reading to increase or decrease? Use Beer's Law (see "Background" above) to explain your answer.
6. What would you expect to happen to the peak wavelength if you diluted your sample by a factor of two?
7. According to Beer's Law, what (quantitatively) would you epxect to happen to the peak absorbance if you diluted your sample by exactly a factor of two?

Part 4: Absorption spectrum of a colored filter

Take a look at the following absorption spectra of colored filters. 

Click here for absorption spectra

Part 4 Practice question:

For each spectrum, predict what color the filter would appear based on the absorption spectrum given to you.  [To aid in justifying your answer, note the dominant peaks and dips by wavelength and indicate what color corresponds to each wavelength. Remember that the filter will appear to have the color of the wavelengths that are not absorbed.]

Workshop 3: Limiting Reagents and Stoichiometry

Click here to download Workshop 3: Uncertainty in pdf or Word format

Prelab & Results

Click here for Prelab worksheet: pdf or Word format

Click here for Results worksheet: pdf or Word format

References

¹Part of this lab is a modified version of "Atomic Spectroscopy" In Experimental Chemistry by James Hall, Houghton Mifflin Co., Boston 1997.

²Wikipedia, Prism (optics) 2/13/08.

Created By: Adilia James '07 and Sarah Coutlee '07
Maintained By: Nick Doe
Date Created: July 3, 2006
Last Modified: June 15, 2009
Expiration Date: July 3, 2007