Instrumental Analysis
CHEM 790
Problem Set 4
Problems 6-1, 6-2, 6-8, 6-12, 6-13, 6-14, 6-15, 6-18, 6-19
6-1.
Coherent radiation has a narrow range of frequencies and a phase relationship that
is constant with time.
The dispersion of a transparent substance is the variation of the refractive index of
the material with respect to wavelength.
Anomalous dispersion is frequency ranges in which there are sharp changes in the
refractive index of the medium.
The work function is the minimum energy required to remove an electron from a
material, usually metal.
The ejection of electrons from the surface of a metal upon illumination is the
photoelectric effect
The ground state of a molecule is the lowest energy state of the molecule.
Electronic excitation is the process by which a photon is absorbed and promotes
and electron into an excited state.
Blackbody radiation is due to the incandescence of solids upon heating. It is a
continuum radiation source.
Fluorescence is the emission of a photon upon return of an excited state species to
the ground state with the same term state multiplicity.
Phosphoresence is the emission of a photon upon return of an excited state species
to a ground state of a different term symbol multiplicity. The change is multiplicity gives
phosphorescence a longer lifetime.
Resonance fluorescence is fluorescence where the emitted photon is of exactly the
same energy as the excitation photon, i.e. there is no Stokes shift.
A photon is the subatomic particle of radiation is is chargeless and massless, but
has a spin angular momentum quantum number of 1.
Absorptivity is the proportionality constant that relates the absorption of radiation
to the concentration of the absorbing species and the pathlength. Chemists usually refer to
the molar absorptivity of a molecule in units of M-1 cm-1 (per Molar per centimeter).
Wavenumbers are equivalent to 1/wavelength and are usually used by
spectroscopists since they are linear with respect to energy. Wavelength is not linear
with respect to energy.
Relaxation is the process by which a species in the excited state returns to the
ground state. It may involve the emission of radiation (fluorescence, phosphorescence),
or heat (vibrational relaxation).
Stokes shift is the energy difference between the absorbed photon and the emitted
photon. It corresponds to the energy of vibrational levels.
6-2
An X-ray photon at 6.24Å has a wavelength of 6.24 x10-10 meters. The
frequency of this radiation is ν = c/λ or = 3.00 x 108 m/s / 6.24 x10-10 meters or 4.80 x
1017 Hz. The energy of this photon is E = hν or (6.6207 x 10-34 J s * 4.80 x 1017 Hz) or
3.18 x 10-16 joules which is 1986 eV since 1 J = 6.2415 x 1018 eV.
6-8
255 kJ/mol are required to break the bonds in AgI. 255,000 J / mol is 4.23 x 10-19
J/molecule. A photon with and energy of 4.23 x 10-19 J would be able to break the AgI
bond. Since ν=E/h, the frequency needs to be at least 4.23 x 10-19 J / 6.6207 x 10-34 J s
or 6.39 x 1014 Hz. This is a wavelength (λ = c/ν) of 3.00 x 108 m/s / 6.39 x 1014 Hz
4.69 x 10-7 meters, or 469 nm.
6-12
The reflective loss of is given by
Ir / Io = (n2 – n1)2 / (n2 + n1)2
or
Ir = Io * (n2 – n1)2 / (n2 + n1)2
Since the refractive index of air and quartz are 1.00 and 1.55, respectively,
Ir
Ir
Ir
Ir
= Io
= Io
= Io
= Io
* (1.55 –1.00)2 / (1.55 + 1.00)2
* (0.55)2 / (2.55)2
* 0.302 / 6.50
* 0.0464 , thus 95.4% is transmitted
Or 4.6% is lost to reflection at the first air/quartz interface, or 95.4% is
transmitted into the quartz window.
Of the 95.4% of the light that enters the front quartz window, 4.64% of this light
is reflected when it exits the window (this is 95.4 x 4.64 or 4.39% of the original total
light). 95.4% - 4.39%, or 91.0 % of the total light crosses the center of the cuvette and
reflects off the front face of the rear wall of the cuvette. Again 4.64% of the light is
reflected, or 91.0 x 4.64 or 4.22% of the total light. The 91.0% - 4.22% or 86.7% of the
light leaving the back wall of the cuvette loses another 4.64% of the light remaining (86.7
x 4.64%, or 4.02% of the original total light), so 86.7% – 4.0% gives 82.7% of the
original total light intensity leaves the back wall of the cuvette. Thus 100% - 82.7% or
17.3 % of the light is lost to reflection.
6-13
The energy must be concentrated into photons to provide the energy to eject
electrons from the surface, dilution of the energy in a wavefront over the full surface
would not allow for electron ejection. Also, a wave model of radiation does not describe
the observed quantized nature of the photoelectric effect.
6-14
Abs = -log T or
T = 10-Abs or
% T = 100% x 10-Abs
so 0.278 Abs = 52.7% T
1.499 Abs = 3.16% T
0.039 Abs = 91.4% T
6-15
Abs = -log T
So 29.9 % T = 0.524 Abs
86.1 % T = 0.065 Abs
2.97 % T = 1.53 Abs
6-18
The concentration is 3.78 x 10-3 M and the transmittance in a 2 cm cell is 0.212
which is an Abs of 0.673. Thus the molar absorptivity is ε = A/bc or = 0.673 /(3.78 x 10-
3
M)(2cm) or 89.0 M-1cm-1. Increasing the transmittance by a factor of 3 from 0.212 to
0.636 decreases the Abs to 0.196, note this is not 3x less! The concentration needed to
achieve 0.196 Abs in a 1 cm cell is, c = A/εb = 0.196/ (89.0 M-1cm-1)(1 cm) = 2.20 x
10-3 M.
6-19
A transmittance of 9.53% is a Abs of 1.02. Using c = A/εb , c = 1.02 / (3.03 x
103 M-1cm-1)(2.5 cm), or a concentration of 1.34 x 10-4 M.
Problems 7-1, 7-2, 7-6, 7-8, 7-11, 7-12, 7-13, 7-18
7-1
Gratings disperse radiation linearly across the focal plane, whereas prisms
disperse in a non-linear fashion (short wavelengths disperse to a greater degree than
longer wavelengths. Thus, fixed slit widths can be used with grating monochromators
and prisms require variable slits to maintain a fixed effective bandwidth.
7-2
Quantitative analyses require greater sensitivity and using larger slit widths and
more light intensity (radiant power) allows for lower detection limits. Qualitative
analyses need higher resolution and thus require smaller slit widths.
7-6
Spontaneous emission is the random process by which an excited state relaxes to
the ground state with the emission of a photon. It generates incoherent monochromatic
radiation. Stimulated emission is the relaxation of an excited state to the ground state
using a photon of the same energy. The emitted photon is of the same energy and phase
as the photon which stimulated the emission which leads to coherent monochromatic
light.
7-8
The effective bandwidth of a filter is the full width at half-height of the light
transmitted through the filter.
7-11
The dispersion of 400 nm to 800 nm radiation is wider in a glass prism
monochromator than in a fused silica prism monochromator.
7-12
Using equation 7-6, nλ = d(sin i + sin r), d = nλ / (sin i + sin r). So for the values
given, d = (1) (400 nm) / (sin 45° + sin 5°) = 400 nm / (0.7071 + 0.0871) = 504 nm /
blaze or 1985 blazes / mm.
7-13
The resolution of a monochromator given by R = λ / Δλ = nN. For this example, n
= 1 (first order), N = 15 mm x 84 lines/mm = 1260, so R = 1260. If λ = 1200 cm-1, Δλ
= λ / R or 1200 cm-1/ 1260 or 0.952 cm-1. The lines at 1200 cm-1 must be 0.951 cm-1
apart to be resolved.
7-18
The resolving power of a monochromator is R = λ / Δλ = nN which for this
example, nN is (1)(3 cm x 1500 lines/mm x 10 mm/cm) = 45,000. The first order
reciprocal linear dispersion, D-1, is equal to dcosr/nf which for this example is ((1 mm x
106 nm/mm) / 1500 lines) (≈ 1) / 1 (1600 mm), or 0.416 nm/mm. x 10-4 mm/m. For the
second order reciprocal linear dispersion, n = 2, and D-1 is 0.21 nm/mm.
Problems 8-4, 8-5, 8-9
8-4
The natural line widths of atomic emission and absorption are determined by line
broadening due to the uncertainty principle and are typically 10-5 nm FWHH.
8-5
In the presence of KCl, ionization of sodium is suppressed because of the high
concentration of electrons from the ionization of potassium. In the absence of KCl, some
of the sodium atoms are ionized, which leads to a lower emission intensity for atomic Na.
8-9
The population of the excited state can be determined using the Boltzmann
Distribution,
Nj/N0 = (gj/g0)exp(-Ej/kT)
For Na, there are two distinct 3p excited states at 5890Å and 5896Å with energies of Ej =
hc/λ or (6.63x10-34 J•s* 3.00 x108 m/s)/(5890 x 10-10 m) = 3.37 x 10-19 J and 3.32 x 1019 J. There are two equal energy states (degenerate states) for one electron in the 3s
orbital (two electron spins, ± ½, in one orbital or 2 x 1 = 2) so g0 = 2, and gj = 6 since
there are six degenerate states for a 3p orbital (2 electron spins in three orbitals, 2 x 3 =
6).
For a natural gas – air flame, the ions in the excited state with a wavelength of 5890Å,
Nj/N0 = (gj/g0)exp(-Ej/kT)
Nj/N0 = (6/2)exp(-3.37 x 10-19 J /1.38065x10-23 J/K •1800K)
= 3 exp(-13.38)
= 4.62 x 10-6 of the ions are in this excited state
for the ions in the other excited state (5896Å)
Nj/N0 = (gj/g0)exp(-Ej/kT)
Nj/N0 = (6/2)exp(-3.32 x 10-19 J /1.38065x10-23 J/K •1800K)
= 3 exp(-13.35)
= 4.73 x 10-6 of the ions are in this excited state
for the hydrogen – oxygen flame
Nj/N0 = (6/2)exp(-3.37 x 10-19 J /1.38065x10-23 J/K •2950K)
= 3 exp(-8.27)
= 7.68 x 10-4 of the ions are in the 5890Å excited state
Nj/N0 = (6/2)exp(-3.32 x 10-19 J /1.38065x10-23 J/K •2950K)
= 3 exp(-8.15)
= 8.65 x 10-4 of the ions are in the 5896Å excited state
for the ICP torch
Nj/N0 = (6/2)exp(-3.37 x 10-19 J /1.38065x10-23 J/K •7250K)
= 3 exp(-3.36)
= 0.103 or 10.3% of the ions are in the 5890Å excited state
Nj/N0 = (6/2)exp(-3.32 x 10-19 J /1.38065x10-23 J/K •7250K)
= 3 exp(-3.31)
= 0.109 or 10.9% of the ions are in the 5896Å excited state
For Mg+, the wavelengths are 2803Å and 2796Å, or 7.09 x 10-19 J and 7.11 x 10-19 J, the
statistical weights are the same, so the equations are
Nj/N0 = (gj/g0)exp(-Ej/kT)
Nj/N0 = (6/2)exp(-7.09 x 10-19 J /1.38065x10-23 J/K •1800K)
= 3 exp(-0.285)
= 1.22 x 10-12 of the ions are in the 2803Å excited state
for the ions in the other excited state (5896Å)
Nj/N0 = (gj/g0)exp(-Ej/kT)
Nj/N0 = (6/2)exp(-7.11 x 10-19 J /1.38065x10-23 J/K •1800K)
= 3 exp(-0.286)
= 1.12 x 10-12 of the ions are in the 2796Å excited state
At 2950K, these become
8.26 x 10-8 of the ions are in the 2803Å excited state
7.86 x 10-8 of the ions are in the 2796Å excited state
At 7250K there are
2.51 x 10-3 of the ions are in the 2803Å excited state
2.46 x 10-3 of the ions are in the 2796Å excited state
Problems 9-1, 9-3, 9-11, 9-12, 9-16, 9-20
9-1
A releasing agent is a substance which reacts with an interfering substance to
remove its interference with the analyte in an AA experiment.
A protective agent binds to the analyte to produce a volatile species which aids in
the atomization of the analyte.
An ionization suppressor is a species added to the AA sample which prevents
ionization of the analyte.
Pressure broadening is a broadening of the line widths in AA due to the collision
of the analyte with other atoms/ions in the heated medium.
Atomic absorption measurements typically employ hollow-cathode lamps which
are high voltage lamps with a tungsten anode and a metal cathode made from the analyte
of interest in an low pressure of Ne or Ar.
Sputtering is the process by which ionized gas in a hollow cathodes lamp collides
with and releases atoms from the cathode. These atoms are responsible for the radiation
emitted by the hollow cathode lamp.
Self-absorption is the absorption of radiation by the sample where the radiation
comes from an emission by the sample and not from the lamp.
Spectral interference is due to absorption/emission of an interfering species that
lies within the resolution of the monochromator of the analyte of interest.
Chemical interference is due to chemical processes during atomization that effect
the absorption characteristics of the analyte.
When the interfering species is known, it is possible to add a large excess of it to
the samples and blank to buffer the effects in the sample. This is known as a radiation
buffer.
Doppler broadening is broadening of atomic absorption/emission lines due to the
movement of the analyte towards/away from the detector.
9-3
An electrothermal ionizer is more sensitive than a flame atomizer because and all
of the sample is atomized, it is atomized in a short time period and remains in the optical
path longer. Flame atomizes lose sample during the atomization process (down the
burner drain) and the atoms fly through the optical path in a millisecond.
Gratings disperse radiation linearly across the focal plane, whereas prisms
disperse in a non-linear fashion (short wavelengths disperse to a greater degree than
longer wavelengths. Thus, fixed slit widths can be used with grating monochromators
and prisms require variable slits to maintain a fixed effective bandwidth.
9-11
The resolution of a monochromator is given by R = λ / Δλ = nN. For this
example, n = 1 (first order), λ = 500 nm, Δλ = 0.002 nm, and N = X mm x 2400
lines/mm, or
500 nm / 0.002 nm = 1 x X mm x 2400 lines/mm
or
104 mm = X
9-12
Using Figure 9-3, I measure the following temperatures, at 1400K at 2 cm, 1750K
at 3 cm, 1863K at 4 cm, and 1830K at 5 cm. Using the Boltzmann equation, I can
compare the # of K in the samples at various temperatures relative to each other (set 2
cm, 1400K = 1.0). The energy of the excited state is 766.5 nm above the ground state, or
Ej = hc/λ or (6.63x10-34 J•s* 3.00 x108 m/s)/(766.5 x 10-9 m) = 2.29 x 10-19 J. Since N0
and (gj/g0) are the same for all samples, they fall out of the equation upon division.
N3cm/N2cm= exp(-Ej/kT) / exp(-Ej/kT)
N3cm/N2cm= exp[(-2.29 x 10-19 J / 1.38065x10-23 J/K •1750K )] /
exp[(-2.29 x 10-19 J /1.38065x10-23 J/K •1400K)]
N3cm/N2cm= exp (-9.48) / exp (-11.84)
= 10.6 (the signal is 10x larger at 3 cm than at 2 cm)
N4cm/N2cm= exp(-Ej/kT) / exp(-Ej/kT)
N4cm/N2cm= exp[(-2.29 x 10-19 J / 1.38065x10-23 J/K •1863K )] /
exp[(-2.29 x 10-19 J /1.38065x10-23 J/K •1400K)]
N4cm/N2cm= exp (-9.03) / exp (-11.84)
= 16.6 (the signal is 16x larger at 4 cm than at 2 cm)
N5cm/N2cm= exp(-Ej/kT) / exp(-Ej/kT)
N5cm/N2cm= exp[(-2.29 x 10-19 J / 1.38065x10-23 J/K •1830K )] /
exp[(-2.29 x 10-19 J /1.38065x10-23 J/K •1400K)]
N5cm/N2cm= exp (-9.19) / exp (-11.84)
= 14.1 (the signal is 14x larger at 5 cm than at 2 cm)
9-16
The peak labeled atomize is the unknown sample peak and has a peak height of 9
mm. The standards have peak heights of 0.8 mm (0.05 µg/mL), 16 mm (0.1
µg/mL).and 30 mm (0.2 µg/mL). These standards yield a linear standarization
curve of
Intensity = 1 mm + 145.71 (mm per µg/mL)
So the sample with an intentsity of 9 mm, has a concentration of
9 mm = 1mm + 145.71 (mm per µg/mL)
or 0.055 µg/mL
9-20
The 5 µL aliquots of standards give absorbance values of 0.396 (0.250 ppm Pb)
and 0.559 (0.450 ppm Pb) after treatment. These provide a standarization curve of :
Intensity = 0.142 + 1.015 [Pb in ppm]
The blood sample gave a signal of 0.444 after treatment, so it appears to have a
[Pb in ppm] value of
0.444 = 0.142 + 1.015 [Pb in ppm]
[Pb in ppm] = 0.297 ppm
Problems 10-1, 10-2, 10-10, 10-11
10-1
An internal standard is an analyte added to the sample in a fixed amount. The
signal from the internal standard is used to correct the data for calibration curves.
10-2
The plasma torch allows excitation of multiple elements due to its high
temperature. This makes it easy to detect multiple elements under the same conditions
which is difficult with a flame emission spectrometer.
10-10
Internal standards are typically used in plasma emission spectrometry to allow for
correction for instrumental drift.
10-11
Arbritary Units
The plot is shown at right,
the regression line equation is given.
Data 1
80
The standard deviation of the slope
is
70
60
sm = √[s2n / (nΣxi2 – (Σxi)2)]
the std. dev. of the intercept is
sb = √[s2Σxi2/ (nΣxi2 – (Σxi)2)]
The std. dev. of the signal is
0.83187 = s and n=5.
Sum of the square of the variances is
0.032703, the sum of the variances
squared is 0.0061247.
Arbritary Units
50
40
30
20
y = 3.18 + 9.2x R= 0.99959
10
0
0
1
2
3
4
µgrams
5
6
7
8
so
sm = √[s2n / (nΣxi2 – (Σxi)2)] = √[0.831872 * 5 / (5*0.032703)/0.0061247)] = √3.46/26.7
= 0.36
For sb = √[s2Σxi2/ (nΣxi2 – (Σxi)2)] = √[0.8318720.0327032 /(5*0.032703)/0.0061247)]=
= √0.00074 / 26.8 = 5.25 x 10-3
The average blank signal is 4.93, the averages for samples A, B, and C are 28.6, 40.7, and
72.6. Using the linear fit, these correspond to 2.23µg, 3.54µg, and 7.00µg.
The abs. std dev of sample A is 0.0349, the rel std dev is 0.016.
The abs. std dev of sample B is 0.0618, the rel std dev is 0.0206.
The abs. std dev of sample C is 0.0538, the rel std dev is 0.027.