Chapter 8 Problem Set
8-1. 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.
8-2. Resonance fluorescence is the emitted radiation from excited species 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.
8-3. 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.
En
Ex
Eo
8-4. The natural line width of atomic absorption and emission lines in 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
∆E = h/ ∆t
Useful Point
∆t ranges from 10-7 to 10-9 s; the lifetime of the excited state. Therefore an average of
about 1x10-8 s can be assigned. This mean is about 1x10-14 s for an absorption
transition. 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/ ∆t
∆ν = 1/1x10-8 = 108 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 s-1 (500 x 10-9 m)2}/2.998x108 m s-1
∆λ = 8.1x10-14 m
∆λ = 8.1x10-4 Ao
For the absorption case we may write:
∆ν = 1/ ∆t
∆ν = 1/1x10-14 = 1014 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 s-1 (500 x 10-9 m)2}/2.998x108 m s-1
∆λ = 8.1x10-8 m
∆λ = 8.1x102 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 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.
8-5. 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 thus increasing the number of atoms that can be excited to give
emission. Therefore, signal increases in presence of potassium chloride.
8-6. A larger number of Cs atoms are excited at higher temperature, thus increasing
the signal.
8-7. 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.
8-8. 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
8-9. Nj/No = Pj/Po exp –Ej/KT
Ej = hc/λ
Substitution in the above relation gives
– hc/λKT
Nj/No = Pj/Po exp
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
8-10. a. For acetylene/oxygen flame at 3000 oC
I. First calculate ratio between 4sto 3p
The excited state is the 4s (Pj = 2) and ground state is 3p (Po = 6)
– hc/λKT
Nj/No = Pj/Po exp
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)
– hc/λKT
Nj/No = Pj/Po exp
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
8-11. 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.