Atomic Spectra – Lab Report Example

The Atomic Spectrum as Unique Element Signature and Evidence for Discrete Energy s in the Atom Introduction: Prior to the 1900s, scientistshave studied the discontinuous spectra produced from gaseous elements, which were in contrast to the continuous spectra produced from white light or sunlight. These mostly empirical studies were later explained in the early 1900s by scientists who were studying the internal structure of the atom.
In 1913, Bohr proposed his atomic model of fixed orbits with discrete energy states. His model was able to provide an explanation for the earlier works of Johann Balmer, Johannes Rydberg and Walter Ritz on the discontinuous line spectra of gases. The model suggested that electrons can only travel within the orbits or make a jump from one orbit to another. This is done by absorbing or emitting energy that is equal to the energy difference between the two orbits. The discrete energy emitted produces the distinct lines found in an atomic emission spectrum.
This relation between atomic structure, atomic spectrum and energy transfer is now being utilized in various disciplines. The composition of distant stars are being determined by studying the spectrum of the light emitted by the stars. Analysis in the environmental sciences make use of atomic spectra to detect the presence and/or quantity of certain target elements. In our daily lives, this principle can be seen in the differently-colored lasers, neon lights and street lamps,and even fireworks.
The purpose of this experiment is to demonstrate atomic emission and its discrete nature; to characterize emission spectra in terms of color, wavelength and equivalent photon energy; and to identify elements based on their respective emission spectra.
To demonstrate atomic emission and its discrete nature, several atomic/ionic samples were brought to excited states in discharge tubes and/or open flames. These emissions were observed directly and through a which generated discontinuous line spectra for the samples.. The colors of the line spectra were used to estimate the corresponding wavelengths. Using the Planck-Einstein equation, the wavelengths were converted to the corresponding energy of the photon emitted by the atoms. Identification of unknown samples were made through observation and characterization of the flame color and spectral lines.
Worksheet

1. General
a. (2 pts.) Why do elements and ions exhibit line emission spectra? (not continuous spectra)
Based on the atomic structure (in the Bohr Model, as well as in Quantum Atomic Theory), the electrons of an atom can only assume discrete energy states. When thermal or electrical energy is applied, atoms become excited and their electrons assume higher energy states. As the electrons return to their normal or ground states, they release energy that is also in discrete quantities. Because this energy (in the form of photons) has discrete values, the corresponding photon wavelengths will be discrete. The photons, in turn, produce a discontinuous line spectra instead of a continuous one.

b. (1 pt.) Why do different elements have different line spectra?
Different elements have distinct sets of atomic orbital energies. This means that electrons of two
elements, A and B, occupying their respective 2s or 3p orbitals will have different energy states. Because the set of atomic orbital energies is unique for each element, the set of energy emissions and the corresponding line spectrum will also be unique for each element.

c. (2 pts.) Would an element emit the same line spectrum if a different excitation source
was used (electricity vs. flame?) Explain.
As discussed in the previous answers, the line spectrum is unique for each element. Ideally, any excitation source would produce the same line spectrum.
However, because different sources will probably have different energies, this will affect the excitation of the atom. In a lower energy source set-up, there will be less energy to send electrons to higher energy states. The resulting spectrum will not exhibit spectral lines
representative of higher energy emissions (smaller wavelengths), and will thus be incomplete relative to the higher energy source set-up.

d. (1 pt.) Do elements also exhibit line absorption spectra?
When an external radiation is incident on an element, certain wavelengths of the external radiation are absorbed by the atoms, and are equivalent to the change in energy of the electrons in the excited state. Through the transmitted radiation, elements will exhibit line absorption spectra, depicted as dark lines, superimposed on the continuous spectrum. The dark lines indicate the absence of the absorbed wavelengths in the transmitted radiation.
2. Discharge tubes:
a. (1 pt.) Give the wavelength of the lines present in one of the discharge tubes.
The continuous visible spectrum shows the corresponding wavelength for different colors of light. From this, the wavelength of the lines produced by the oxygen discharge tube can be determined.

Color
Wavelength (nm)
purple
400-450
blue
450-520
green
520-560
orange
580-600
red
600-630
yellow
560-580

b. (2 pts.) Give the energy (J/photon) and (kJ/mol photons) for each spectral line.
The corresponding photon energy of each spectral line can be computed from the Planck- Einstein Equation. The energy of a photon is expressed as:
E=hc/λ
where
E = photon energy, (J/photon)
h = Plancks constant, (6.62606896×10-34 J·s)
c = speed of light, (299,792,458 m/s)
λ = wavelength of spectral line, (m)
(J/photon) can be multiplied by the Avogadros number, 6.022×1023 , and divided by 1000 to convert the unit to (kJ / mol photons).

Sample calculation for purple spectral line (using average wavelength):
Energy of photon = hc/λ
= (6.62606896×10-34 J·s × 299,792,458 m/s )
425 nm × (1×10-9 m / nm)
= 4.67 x10-19 J / photon

= 4.67 x10-19 J / photon × 6.022×1023 photons/mol photons
1000 J / 1 kJ
= 281 kJ / mol photons

Energy
Color
Average wavelength (nm)
(x10-19 J/photon)
(kJ/mol photons)
purple
425
4.67
281
blue
485
4.10
247
green
540
3.68
222
orange
590
3.37
203
red
615
3.23
195
yellow
570
3.48
210

c. (1 pt.) Which color has the most energy and which has the least?
The table above shows that the purple spectral line has the highest energy, whereas the red spectral line has the lowest energy.

3. Flame ionization:
a. (2 pts.) What was the identity of your unknown salt? Explain how you arrived at this
conclusion.
Unknown Salt # 114 produced a red-orange glow in the flame, and spectral lines of red, orange and green. These properties are also present in the the known Calcium Chloride sample. Thus, Unknown Salt #114 is most likely Calcium Chloride.


b. (3 pts.) Were there any ambiguities in determining the identity of the unknown? List
sources of error in the experimental method. Which source of error was most
problematic?
There are ambiguities in identifying the unknown when the resulting flame color for the different salts are very similar or indistinguishable.
Errors in the experimental method may be due to the subjectivity of the human eye in color perception, the presence of impurities from previously tested solutions, and fluctuations in the energy of the excitation source.
Among the errors, the subjectivity of the human eye in perceiving and interpreting colors was the most problematic. This is because some students may be able to perceive more shades of a single color than other students. Some may even possess a wider vocabulary of colors. When the recorded but erroneous color observation is used to measure wavelengths and compute for energy, the original error is also passed on.
References:
1. Brown, T.E. Chemistry: The Central Science, 9th ed; Prentice Hall: New Jersey, 2004.
2. Monk P.M.S. Physical Chemistry;John Wiley and Sons, West Sussex, England, 2004.
3. Softley, T.P. Atomic spectra: Volume 19 of Oxford chemistry primer; Oxford Science Publications, England, 1994.
4. Tipler, P. Physics for Scientists and Engineers: Electricity, Magnetism, Light, and Elementary Modern Physics (5th ed.); W. H. Freeman: New York, 2004.