# Document 245865

```Problem Set 3
Introduction to Atomic Spectroscopy
1. Why does CaOH give a broad band spectrum while barium gives a line spectrum?
CaOH is a molecular species that has vibrational and rotational levels. Molecular
species are characterized by broad band spectra resulting from transitions among all
energy states. On the other hand, in barium atoms only transitions among electronic
energy states are possible since no rotational or vibrational levels exist; thus resulting
in line spectra.
2. What is meant by resonance fluorescence?
Resonance fluorescence is the emitted radiation from excited atoms at the same
wavelength as that of a resonance line. It can be obtained as a result of excitation by
an external source of radiation at the resonance wavelength.
3. At what conditions can a Stokes shift occur in atomic spectroscopy?
By Stokes shift we mean getting emission radiation at wavelengths longer than
excitation wavelengths.
Stoke's shift can occur in atomic spectroscopy when an electron is excited to a high
energy electronic level emits a photon and relaxes to a lower excited electronic level,
the emitted radiation (solid line) has lower energy, and thus longer wavelength.
4. What are natural line widths for atomic emission and absorption lines?
The natural line width of atomic absorption and emission lines is the line width
resulting from the uncertainty in the calculation of energy (and thus wavelength) due
to the uncertainty principle. It can be easily predicted that an uncertainty in
calculation of transition time would imply an uncertainty in energy:
∆E.∆t> h/4π
∆E = h/ 4π ∆t
Useful Point
∆t ranges from 10-7 to 10-9 s; the lifetime of the excited state (for emission lines).
Therefore an average of about 1x10-8 s can be assigned. However, the mean time for
an absorption transition is about 1x10-14 s, which reflects a much shorter time and thus
a larger uncertainty. It may be of interest to compare emission line width to absorption
line width for a specific transition, say for example at λ = 500 nm:
For the emission case we may write:
∆ν = 1/4π ∆t
∆ν = 1/(4π)1x10-8 = 108/4π s-1
ν = c/λ, differentiating this relation for finite values for ν and λ we get:
∆ν = -cλ−2 ∆λ
∆λ = ∆ν λ-2/c
∆λ = {108/4π s-1 (500 x 10-9 m)2}/2.998x108 m s-1
∆λ = 8.1x10-14 /4π m
∆λ = 8.1x10-4 /4π Ao
For the absorption case we may write:
∆ν = 1/ 4π ∆t
∆ν = 1/(4π )1x10-14 = 1014/4π s-1
ν = c/λ, differentiating this relation for finite values for ν and λ we get:
∆ν = -cλ−2 ∆λ
∆λ = ∆ν λ-2/c
∆λ = {1014 /4π s-1 (500 x 10-9 m)2}/2.998x108 m s-1
∆λ = 8.1x10-8 /4π m
∆λ = 8.1x102 /4π Ao
Therefore, since the absorption transition time is much shorter than emission time, the
uncertainty in calculation of transition time in absorption is much higher than
uncertainty in calculation of transition time for emission lines which is reflected in
parallel uncertainties in frequencies. An absorption line is supposed to be of larger
width than the emission line as a consequence of the uncertainty principle.
5. It is observed that in a hot flame the emission intensities 0f the sodium lines at
589.0 and 589.6 nm are greater in a sample solution that contains KCl than when this
compound is absent. Suggest an explanation.
Some of the sodium atoms are ionized at the high temperature of atomization. KCl
can be ionized easier than sodium and will thus be better ionized producing large
numbers of electrons. Ionized sodium can capture some of these electrons and can be
converted to the atoms again thus increasing the number of atoms that can be excited
to give emission. Therefore, signal increases in presence of potassium chloride.
6. The intensity of a line for atomic Cs is much lower in a natural gas flame, which
operates at 1800°C, than in a hydrogen-oxygen flame, whose temperature is 2700°C.
Explain.
A larger number of Cs atoms are excited at higher temperature, thus increasing the
signal.
7. Name a continuous type and a discrete type of atomizer that are used in atomic
spectrometry. How do the output signals from a spectrometer differ for each?
Flames is regarded as a continuous atomizers since samples are continuously
introduced through a nebulizer. An example of a noncontinuous (discrete) atomizer is
the electrothermal atomizer where a sample is introduced at one time. The signal from
a continuous atomizer is essentially constant throughout the time of sample
nebulization, The signal from a discrete atomizer will increase till a maximum is
reached followed by a decrease to zero after a very short time.
8. The Doppler effect is one of the sources of the line broadening in atomic
absorption spectroscopy. Atoms moving toward the light source encounter higher
frequency radiation than atoms moving away from the source. The difference in
wavelength, ∆λ, experienced by an atom moving at speed v (compared to one at rest)
is ∆λ/λ= v/c, where c is the velocity of light. Estimate the line width (in nanometers)
of the sodium D line (589.3 nm) when the absorbing atoms are at a temperature of (a)
2200 K and (b) 3000 K. The average speed of an atom is given by v = (8kT/πm)1/2,
where k is Boltzmann's constant, T is the absolute temperature, and m is its mass.
a. At 2200 K
The Boltzmann constant is K = 1.38x10-23 kg m s-2 K-1
Substitution in the relation ν = (8 KT/πm)1/2
ν = {(8 * 1.38x10-23 kg m s-2 K-1 * 2200 K)/ (3.14 * 23x10-3 kg/6.023x1023)}1/2
ν = 1423 m s-1
∆λ = νλο /c; Substitution in the Doppler shift equation we get:
∆λ = 1423 m s-1 * 5893 Ao/ 2.998x108 m s-1
∆λ = 0.028 Ao
b. The same procedure can be applied at 3000 K
ν = {(8 * 1.38x10-23 kg m s-2 K-1 * 3000 K)/ (3.14 * 23x10-3 kg/6.023x1023)}1/2
ν = 1662 m s-1
∆λ = νλο /c; Substitution in the Doppler shift equation we get:
∆λ = 1662 m s-1 * 5893 Ao/ 2.998x108 m s-1
∆λ = 0.033 Ao
Appreciate the Doppler shift increase at higher temperatures
9. For Na+ and Mg+ ions, compare the ratios of the number of ions in the 3p excited
state to the number in the ground state in
(a) a natural gas-air flame (2100 K).
(b) a hydrogen-oxygen flame (2900 K).
(c) an inductively coupled plasma source (6000 K).
Nj/No = Pj/Po exp –Ej/KT
Ej = hc/λ
Substitution in the above relation gives
Nj/No = Pj/Po exp
– hc/λKT
a.I. For sodium at 2100 K
Nj/No = 6/2 exp – {(6.6256x10-34 J s * 2.998x108 m s-1)/(5893 Ao * 10-10
m/Ao)(1.38x10-23 JK-1 * 2100 K)}
Nj/No = 2.6x10-5
a. II. For Mg+ we have the average resonance line at 2800 Ao
Nj/No = 6/2 exp – {(6.6256x10-34 J s * 2.998x108 m s-1)/(2800 Ao * 10-10
m/Ao)(1.38x10-23 JK-1 * 2100 K)}
Nj/No = 6.9x10-11
b.I. For Na at 2900 K
Nj/No = 6/2 exp – {(6.6256x10-34 J s * 2.998x108 m s-1)/(5893 Ao * 10-10
m/Ao)(1.38x10-23 JK-1 * 2900 K)}
Nj/No = 6.6x10-4
b.II. For Mg+ at 2900 K
Nj/No = 6/2 exp – {(6.6256x10-34 J s * 2.998x108 m s-1)/(2800 Ao * 10-10
m/Ao)(1.38x10-23 JK-1 * 2900 K)}
Nj/No = 6.1x10-8
c. I. For sodium at 6000 K
Nj/No = 6/2 exp – {(6.6256x10-34 J s * 2.998x108 m s-1)/(5893 Ao * 10-10
m/Ao)(1.38x10-23 JK-1 * 6000 K)}
Nj/No = 5.1x10-2
c. II. For Mg+ at 6000 K
Nj/No = 2/6 exp – {(6.6256x10-34 J s * 2.998x108 m s-1)/(2800 Ao * 10-10
m/Ao)(1.38x10-23 JK-1 * 6000 K)}
Nj/No = 5.7x10-4
10. In high-temperature sources, sodium atoms emit a doublet with an average
wavelength of 1139 nm. The transition responsible is from the 4s to 3p state.
Calculate the ratio of the number of excited atoms in the 4S state to the number in the
ground 3S state over the temperature range from an acetylene/oxygen , and for a
plasma source at 9000 oC
a. For acetylene/oxygen flame at 3000 oC
I. First calculate ratio between 4s to 3p
The excited state is the 4s (Pj = 2) and ground state is 3p (Po = 6)
Nj/No = Pj/Po exp
– hc/λKT
Substitution gives:
N4s/N3p = 2/6 exp – {(6.6256x10-34 J s * 2.998x108 m s-1)/(1139 nm * 10-9
m/nm)(1.38x10-23 JK-1 * 3273 K)}
N4s/N3p = 7x10-3
II. Calculate ratio from 3p to 3s (ground state)
Nj/No = 6/2 exp – {(6.6256x10-34 J s * 2.998x108 m s-1)/(589.3 nm * 10-9
m/nm)(1.38x10-23 JK-1 * 3273 K)}
N3p/N3s = 1.7x10-3
N4s/N3p * N3p/N3s = N4s/N3s = 7x10-3 * 1.7x10-3 = 1.2x10-5
b. For a plasma source at 9000 oC
I. First calculate ratio between 4sto 3p
The excited state is the 4s (Pj = 2) and ground state is 3p (Po = 6)
Nj/No = Pj/Po exp
– hc/λKT
Substitution gives:
N4s/N3p = 2/6 exp – {(6.6256x10-34 J s * 2.998x108 m s-1)/(1139 nm * 10-9
m/nm)(1.38x10-23 JK-1 * 9273 K)}
N4s/N3p = 8.53x10-2
II. Calculate ratio from 3p to 3s (ground state)
Nj/No = 6/2 exp – {(6.6256x10-34 J s * 2.998x108 m s-1)/(589.3 nm * 10-9
m/nm)(1.38x10-23 JK-1 * 9273 K)}
N3p/N3s = 0.215
N4s/N3p * N3p/N3s = N4s/N3s = 8.53x10-2 * 0.215 = 1.8x10-2
11. In the concentration range of 500 to 2000 ppm of U, there is a linear relationship
between absorbance at 351.5 nm and concentration. At lower concentrations the
relationship is nonlinear unless about 2000 ppm of an alkali metal salt is introduced
into the sample. Explain.
The extent of ionization will definitely be more pronounced when the concentration of
analyte decreases. When an alkali metal is added electrons are introduced in the
system, thus decreasing ionization of uranium.
12. Why is it not wise to make the instrument read zero for the absorbance of non
atomized matrix components?
Making the instrument read the absorbance of non atomized species as zero will limit
the sensitivity of the technique as well as its precision. The background absorption
should be removed in a way or another.
13. A high pressure xenon arc lamp produces a broad band spectrum from 200-1000
nm. Explain.
Xenon is expected to produce line spectrum as it is in the atomic state. However, at
high pressures it produces a broad band spectrum due to what is known as pressure
broadening resulting from collisions. Multiple collisions of xenon atoms make the
ground state assume different energies, and thus different emission wavelengths to the
extent that a continuum is produced.
14. Although a more homogeneous and smaller droplet size are obtainable using
ultrasonic nebulizers, pneumatic nebulizers are most often used. Explain why.
This is simply because pneumatic nebulization integration in instrumental designs is
more convenient.
15. Although the p orbital is degenerate, two lines are obtained for transition from the
3s to 3p in the sodium atom. Why?
This is because the energy of the p orbital will split to two energy levels depending on
the interaction of the magnetic field of the electron rotating around its axis and the
magnetic field resulting from rotation of the electron in the orbit. If both fields are
parallel, this is a high energy state (589.0 nm, repulsion) while a lower energy state
results when the two fields are opposite (589.6 nm, attraction).
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