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Chapter 7

In your own words, explain why the features in molecular absorption

spectroscopy are broader than the features seen in AAS.

This is an open-ended questions: Instructors will need to adjust their expectations accordingly. However, students should discuss the nature of vibronic vs purely electronic transitions. In their discussion, students should mention the fact that in a molecule each electronic transition can initiate and terminate in one of many superimposed vibrational and/or rotational quantum states. In an atom the electronic states are discrete.

Use Example 7.1 as a guide and demonstrate the wavelength dependence of lifetime broadening by calculating the minimum theoretical bandwidth at 200 nm, 400 nm, and 700 nm, assuming a lifetime of 10 ns for each transition. Compare your answers to Example 7.1.

From Example 7.1 we see that 𝛥𝛦 =ℏ

2𝛥𝑡 and for a life time of 10 ns

𝛥𝛦 =ℏ

2𝛥𝑡=

1.0546 𝑥 10−34𝐽 ∙ 𝑠

2(1 𝑥 10−8𝑠)= 5.3 𝑥 10−27𝐽

and the bandwidth (𝛥𝜆) = 𝛥𝛦𝜆2

ℎ𝑐

At 200 nm, =

𝛥𝜆 = 𝛥𝛦𝜆2

ℎ𝑐= 5.3 𝑥 10−27𝐽

(200 𝑥 10−9𝑚)2

(6.626 𝑥 10−34𝐽 ∙ 𝑠)(2.99 𝑥 108 𝑚/𝑠)= 1.07 𝑋 1015𝑠 = 1.07𝑓𝑠

At 400 nm


ℎ𝑐= 5.3 𝑥 10−27𝐽

(400 𝑥 10−9𝑚)2

(6.626 𝑥 10−34𝐽 ∙ 𝑠)(2.99 𝑥 108 𝑚/𝑠)4.28 𝑋 1015𝑠 = 4.28𝑓𝑠

At 700 nm


ℎ𝑐= 5.3 𝑥 10−27𝐽

(700 𝑥 10−9𝑚)2

(6.626 𝑥 10−34𝐽 ∙ 𝑠)(2.99 𝑥 108 𝑚/𝑠)1.311 𝑋 1014𝑠 = 13.11𝑓𝑠

Use Example 7.1 as a guide and demonstrate the lifetime dependence of

lifetime broadening by calculating the theoretical minimum bandwidth for three different

transitions, all centered at 500 nm transition with lifetimes of 10 μs, 10 ns, and 10 fs,

respectively.

For 10μs: Δ𝐸 = ℎ

2Δ𝑡=

1.0546 𝑥 10−34 𝐽𝑠

2(10 𝑥 10−6 𝑠) = 5.273 x 10-30 J

∆𝜆 = Δ𝐸𝜆2

ℎ𝑐= (5.273 𝑥 10−30𝐽)

(500 𝑥 10−9𝑚)2

(6.63 𝑥 10−34𝐽𝑠)(2.998 𝑥 108𝑚

𝑠) = 6.63 x 10-18 m = 6.6 x 10-9 nm

For 10 ns: See Example 7.1

For 10 fs: : Δ𝐸 = ℎ

2Δ𝑡=

1.0546 𝑥 10−34 𝐽𝑠

2(10 𝑥 10−15 𝑠) = 5.273 x 10-21 J

∆𝜆 = Δ𝐸𝜆2

ℎ𝑐= (5.273 𝑥 10−21𝐽)

(500 𝑥 10−9𝑚)2

(6.63 𝑥 10−34𝐽𝑠)(2.998 𝑥 108𝑚

𝑠) = 6.63 x 10-9 m = 6.6 nm

Above we stated that the promotion of the sodium 3s electron into the 3p orbital can occur parallel or antiparallel to the angular momentum of the 3p orbital ( J = S + L or J = S – L). Derive the term symbols for the two possible excited states.

S =2s +1 = 2(½) +1 = 2 L = P = 1 J = S ± L = 3 or 1 Possible term symbols are

𝑃32 𝑜𝑟 𝑃1

2

The 4s 4p transition in potassium experiences a splitting of the 4p orbital

in a manner similar to that seen for sodium. The 4s 4𝑝12⁄ transition occurs at a frequency

of 3.89 × 1014 Hz and the 4s 4𝑝32⁄ transition occurs at a frequency of 3.91 × 1014 Hz.

Calculate the energy difference in joules and the wavelength difference in nanometers for these two transitions.

E = h6.626 X 10-34 Js)(3.91 X 1014s-1 – 3.89 X 1014s-1) = 1.33 X 10-21 J

= c/ = 2.99 X 108 m/s)/(3.91 X 1014s-1 – 3.89 X 1014s-1) = 0.015 m

Doppler broadening is a significant source of line broadening in atomic

spectroscopy. Using a maximum velocity of 2000 𝑚

𝑠for an analyte atom in a flame, use

Equation 7.3 to determine the line broadening (Δλ) associated with a transition centered at 500 nm. Report your answer in nanometers.

E and frequency are directly proportional, so we can use the difference in frequency to get the difference in energy. Δν = (5.99588916-5.99580916)x1014 Hz = 8.00 x 109 Hz ΔE = h(Δν) = (6.626 x 10-34 Js)(8.00 x 109 s-1) = 5.3 x 10-24 J For Δλ, we should to calculate the independent wavelengths then get the difference. 4p1/2 : c=λν λ = (2.998 x 108 m/s)/(5.99588916 x 1014 s-1) = 5.0000924 x 10-7 m = 500.00924 nm 4p3/2 : c=λν λ = (2.998 x 108 m/s)/( 5.99580916 x 1014 s-1) = 5.0001591 x 10-7 m = 500.01591 nm Δλ = 0.0067 nm

Show the wavelength dependence of Doppler broadening by repeating

Exercise 7.6 for transitions centered at 200 nm, 420 nm, and 680 nm.

For the 200 nm transition:

𝐸 = ℎ(𝑐+2000𝑚/𝑠)

200 𝑥10−9 𝑚 = 9.9324 x 10-19 J. Convert this back to wavelength(𝜆 =

ℎ𝑐

𝐸): 199.9987 nm

𝐸 = ℎ(𝑐−2000𝑚/𝑠)


ℎ𝑐

𝐸): 200.0013 nm

so Δλ = 0.0026 nm For the 420 nm transition:

𝐸 = ℎ(𝑐+2000𝑚/𝑠)


ℎ𝑐

𝐸): 419.9983 nm

𝐸 = ℎ(𝑐−2000𝑚/𝑠)


ℎ𝑐

𝐸): 420.0072 nm

so Δλ = 0.0089 nm For the 680 nm transition:

𝐸 = ℎ(𝑐+2000𝑚/𝑠)


ℎ𝑐

𝐸): 679.9955 nm

𝐸 = ℎ(𝑐−2000𝑚/𝑠)


ℎ𝑐

𝐸): 680.0045 nm

so Δλ = 0.0091 nm

Exercise 7.8: Doppler broadening of atomic transitions is temperature dependent and the

effect on the bandwidth ( compared to an atom with no velocity is ∆𝜆

𝜆=

𝜈

𝑐 , where is the

velocity of an atom and c is the velocity of light. The relationship between an atom’s velocity

and temperature is given by the equation 𝜈 = √8𝑘𝑇

𝜋𝑚, where k = the Boltzmann constant, T is

temperature (Kelvin) and m is the mass of the atom in kilograms. Calculate the effect of Doppler broadening on the bandwidth of the calcium 422.7 nm line in an 1800°C acetylene flame.

1800°C (2073.15K) acetylene flame,

𝜈 = √8𝑘𝑇

𝜋𝑚=

√8 (1.38064853 𝑋 10−23 𝑚2𝐾𝑔

𝑠2𝐾) (2073.17𝐾)

(3.142)(6.655 𝑋 10−26𝐾𝑔)= 1046.46 𝑚𝑒𝑡𝑒𝑟𝑠/𝑠𝑒𝑐𝑜𝑛𝑑

𝑚2𝐾𝑔

𝑠2

∆𝜆 = 𝜆 (𝜈

𝑐) = 422.7 𝑋 10−9𝑚 (

1046.46 𝑚𝑠⁄

2.99 𝑋 108 𝑚𝑠⁄

) = 1.4794 𝑋 10−12𝑚 ≈ 1.5 𝑓𝑚

List and explain the factors contributing to the electronic bandwidths seen in atomic

transitions.

This is an open-ended questions: Instructors will need to adjust their expectations accordingly however discussion should be centered on the following topics,

1. Lifetime broadening

2. Magnetic field broadening

3. Pressure broadening

4. Doppler broadening

Why is it desirable to use an element specific source instead of a continuum source in AAS?

This is an open-ended questions: Instructors will need to adjust their expectations accordingly however discussion should be centered on the following topics,

1. Spectral bandwidth of continuum sources vs. element specific sources and the effect on sensitivity of the technique.

2. Doppler Broadening in a source vs. in the flame.

Explain how the use of the HCL improves the analytical selectivity of AAS.

This is an open-ended questions: Instructors will need to adjust their expectations accordingly however this question is designed specifically to reinforce the idea that the lower pressures and temperatures in the HCL relative to the temperatures in the flame allow for absorptions by the analyte to occur within the narrower bandwidth of an HCL emission.

Explain how the construction of the HCL extends the life of the lamp.

This is an open-ended questions: Instructors will need to adjust their expectations accordingly, however students should mention the fact that the cathode in the lamp is shaped like a cup or a bowel as a means of trapping any sublimed atoms from the cathode in a relatively small volume of space. The trapped atoms can then redeposit on the cathode surface and therefore extend the life of the lamp.

Discuss the considerations made in choosing a nebulization technique.

This is an open-ended questions: Instructors will need to adjust their expectations accordingly however reasonable answers should include the advantages and disadvantages of the pneumatic and Babington nebulizers. Specifically, the discussion should mention that the pneumatic nebulizer is the most common nebulizer for flame AAS however the Babington nebulizer is often preferred for samples that are relatively viscous or samples that have a high ionic strength.

Explain why the nebulization step is fundamentally responsible for determining the lower

limit for quantitative analysis in AAS.

The reproducibility of the nebulization step ultimately controls the concentration of analyte in the

optical path of the spectrometer. Statistically, the 95% confidence limit of a measurement is related to the standard deviation (s) of replicate measurements by the equation

C.L. = ∓𝑡 𝑠

√𝑁

See Chapter 22 for a review of statistical tools for data analysis.

Problem 7.15: Using Example 7.2 as a guide, calculate the temperature needed to create gaseous

sodium atoms with 50% of the atoms in the excited state. Is 3,400°C a realistic temperature?

We need to solve equation 7.5 for T

Ne

Ng=

ge

gge(−

∆EkT

)

From Example 7.2 we now that E = 3.36 X 10-19J and ge

gg= 1. The Boltzman constant has a value of

1.38064853 𝑋 10−23𝐽𝐾−1 Make appropriate substitutions

0.5 = 1𝑒−(

3.36 𝑋 10−19𝐽(1.381 𝑋 10−23𝐽𝐾−1)(𝑇)

)

Take the natural log (ln) of both sides of the equation to obtain

−0.693 = −3.36 𝑋 10−19𝐽

(1.381 𝑋 10−23𝐽𝐾−1)(𝑇)

Solve for (T)

𝑇 =3.36 𝑋 10−19𝐽

(1.381 𝑋 10−23𝐽𝐾−1)(0.693)= 35,101.05 𝐾 = 34,828 ℃

The second half of the question asked if 3,400°C a realistic temperature. As we saw in Example 7.2, a flame temperature of 3,400°C is only capable of an excited state population ratio of 0.13%.

Using your own words, write a short paragraph that describes the relationships between

maximum sensitivity, flame temperature, and flame height.

This is an open-ended questions: Instructors will need to adjust their expectations accordingly. This is a relatively complex question. Sensitivity is improved by a hotter flame only if the increased temperature results in an increase in the concentration of atomic analyte atoms. Sensitivity can be diminished if the increased flame temperature results in excited state analyte. However as we saw in Problem 7.15, to produce a significant concentration of excited state analyte atoms, the flame temperature would have to be extremely hot. For most circumstances, a hotter flame improves the sensitivity. The maximum concentration of analyte atoms is also a function of how long the sample exists within the flame. Since flames are dynamic, adjustment of the flame height is used to find the region of the flame that contains the highest concentration of analyte atoms.

Under what analytical parameters is FAAS a more suitable technique than GFAAS? Under what analytical parameters is GFAAS a more suitable technique than FAAS?

This is an open-ended questions: Instructors will need to adjust their expectations accordingly. Implicit in the student’s answer will be a thoughtful definition of the word “suitable”. For instance if low detection limits are of paramount concern, the GFAAS is the more suitable technique. On the other hand if throughput and reproducibility are of paramount concern, then FAAS is the more suitable technique.

Assume you own a water analysis laboratory and one of the services you provide is the determination of magnesium and calcium in well water. Your overhead (e.g., utilities, rent, payroll, chemicals, supplies) is approximately $1,200/week. Using your autosampler tray, you can process 400 samples/day in flame mode and 150 samples/day using furnace mode. Operating at full capacity:

(a) What is your “breakeven” price per measurement if you ran all of your samples in flame mode at full capacity?

(b) What is your “breakeven” price per measurement if you ran all of your samples in furnace

mode at full capacity?

Some assumptions will have to be made to complete this question so the instructor may need to adjust their expectations accordingly. If we assume a 5 day work week the lab can process

𝐹𝑙𝑎𝑚𝑒 𝑚𝑜𝑑𝑒 = (5𝑑𝑎𝑦𝑠

𝑊𝑜𝑟𝑘𝑊𝑒𝑒𝑘⁄ ) (400

𝑠𝑎𝑚𝑝𝑙𝑒𝑠𝑑𝑎𝑦⁄ ) = 2000

𝑠𝑎𝑚𝑝𝑙𝑒𝑠𝑊𝑜𝑟𝑘𝑊𝑒𝑒𝑘

⁄

𝐹𝑢𝑟𝑛𝑎𝑐𝑒 𝑚𝑜𝑑𝑒 = (5𝑑𝑎𝑦𝑠

𝑊𝑜𝑟𝑘𝑊𝑒𝑒𝑘⁄ ) (150

𝑠𝑎𝑚𝑝𝑙𝑒𝑠𝑑𝑎𝑦⁄ ) = 750

𝑠𝑎𝑚𝑝𝑙𝑒𝑠𝑊𝑜𝑟𝑘𝑊𝑒𝑒𝑘

⁄

a) To break even in flame mode the lab would have to charge

𝐵𝑟𝑒𝑎𝑘 𝐸𝑣𝑒𝑛 =

$1,200𝑤𝑒𝑒𝑘⁄

2000 𝑆𝑎𝑚𝑝𝑙𝑒𝑠𝑤𝑒𝑒𝑘

⁄= $0.60/𝑠𝑎𝑚𝑝𝑙𝑒

b) To break even in furnace mode the lab would have to charge

𝐵𝑟𝑒𝑎𝑘 𝐸𝑣𝑒𝑛 =

$1,200𝑤𝑒𝑒𝑘⁄

750= $1.60/𝑠𝑎𝑚𝑝𝑙𝑒

Using Beer’s law, explain why GFAAS has lower detection limits than FAAS.

Beer’s law is A = bc where the variables “b” and “c” are path length and concentration respectively.

Both of these parameters can be manipulated by the analyst. Although the path length “b” is typically

shorter in GFAAS relative to FAAS, the concentration “c” of analyte in the optical path is often several

orders of magnitude greater in GFAAS relative to FAAS.

Problem 7.20 Outline the stoichiometry showing that the addition of 0.1ml of a 1000 ppm standard to 100 grams of soil represents a 1 ppm spike. Recall that the definition of ppm for an aqueous solution is

ppm = 𝑚𝑔

𝐿

and the definition of ppm for a solid matrix is

ppm = (𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑎𝑛𝑎𝑙𝑦𝑡𝑒

𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑚𝑎𝑡𝑟𝑖𝑥) 𝑥 106.

The total mass of analyte contained in 0.1ml (0.0001 L) of a 1000 ppm standard is

1000𝑚𝑔

𝐿 𝑥 0.0001 𝐿 = 0.1 𝑚𝑔

Adding 0.1 mg (1 X 10-4 g) to 100 grams of soil represents

(1.0 𝑋 10−4𝑔(𝑎𝑛𝑎𝑙𝑦𝑡𝑒)

100 𝑔(𝑠𝑜𝑖𝑙)) 𝑥 106 = 1.0 𝑝𝑝𝑚

Problem 7.21: Imagine you have performed a standard addition for the analysis of barium in soil and obtained the following data. Using a spreadsheet, plot your data and determine the original concentration of barium in the soil sample.

Students will need to make an XY scatter plot with absorbance on the y-axis and added Ba on the x-axis. Students will then need to fit a trend line (least squares regression) and extrapolate the data backward to the x-intercept. The negative of the x-intercept will be the original concentration in the sample matrix. The students graph should resemble this one. Solving for the negative of the x-intercept yields an original soil concentration of 3.14 ppm.

Show how you would make a series of four standards ranging from 1 to 100 ppm given the following constraints. Your stock solution is 1000 ppm. The available pipettes have volumes of 2 mL, 3 mL, and 5 mL. The available volumetric flasks have volumes of 10 mL, 25 mL, 50 mL, and 100 mL. See the “Be Flexible” sidebar.

Example 7.4 demonstrates how a students might go about this exercise.

This is an open-ended questions and students will find various means to satisfy the experimental parameters: Instructors will need to adjust their expectations accordingly.

Students will be expected to make use of the equation

C1V1 = C2V2

Below are some sample calculations. The lowest concentration standard an analyst can make from the available equipment would be to dilute 2 ml of stock to 100 ml. This would result in a 20 ppm standard. 20ppm standard:

𝑉1 =𝐶2𝑉2

𝐶1=

(20𝑝𝑝𝑚)(100𝑚𝐿)

1000𝑝𝑝𝑚= 2.00𝑚𝐿 𝑜𝑓 𝑠𝑡𝑜𝑐𝑘 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

In order to make a 20ppm standard from a 1000ppm stock, you would take 2.00mL of stock solution and dilute it to 100mL. If the analyst needs a standard that is lower than 20ppm, the analyst will need to make a dilution from the 20 ppm standard. Propagation of error will be a factor. At this point, the analyst can repeat the above process by decrease the volume of the volumetric flask from 100ml to 50ml and then to 25ml to produce a 40ppm and 80ppm standard respectively. 40ppm standard:

𝑉1 =𝐶2𝑉2

𝐶1=

(40𝑝𝑝𝑚)(50𝑚𝐿)


In order to make a 40ppm standard from a 1000ppm stock, one would take 2.00mL of stock solution and dilute it to 50mL.

80ppm standard:

𝑉1 =𝐶2𝑉2

𝐶1=

(80𝑝𝑝𝑚)(25𝑚𝐿)


In order to make an 80ppm standard from a 1000ppm stock, one would take 2.00mL of stock solution and dilute it to 25mL.

100ppm standard:

𝑉1 =𝐶2𝑉2

𝐶1=

(100𝑝𝑝𝑚)(50𝑚𝐿)


In order to make a 100ppm standard from a 1000ppm stock, one would take 5.00mL of stock solution and dilute it to 50mL.

Table 7.4 contains the data from an AAS analysis of

calcium found in powdered milk. The data table contains the absorption values for four standards ranging from 5 to 90 ppm, and it also contains absorption values (in replicates of five) for three different brands of powdered milk (M1–M3). The procedure for preparing the powdered milk samples for AAS analysis involved digesting 0.05 grams of powdered milk in 50 mL of acid followed by a final dilution with water to a volume of 100 mL.

(a) Use a spreadsheet to construct a calibration curve using data from your four standards. Construct a trend line (least squares regression line) and determine the equation for the slope of your line.

(b) Using the equation for the trend line found in part (a), determine the mean ppm calcium for each powdered milk sample and report the 90% confidence limit

(c) Using your data from part (b) and your knowledge of the experimental procedure, calculate the mass percentage of calcium in each powdered (dry) milk sample.

a) The calibration curve should resemble the one shown here.

b) This is best done with a spreadsheet. Program your spreadsheet to solve the equation from your regression line for “x” and input the absorbance values for each trial as “y”. Then use your spreadsheet to calculate the average and the 90% C.L. See Chapter 22 for a review of data reduction statistical tools and spreasheets. Milk Sample 1: 56.5 + 1.38 ppm

Milk Sample 2: 41.7 + 1.38 ppm Milk Sample 3: 28.9 + 1.38 ppm

c) Milk Sample 1: 56.5 ppm = 56.6 mg/L x 0.1 L = 5.66 mg Ca in the sample.

% Ca = (0.00566 g)/(0.05 g) x 100 = 11.3 % Ca

Milk Sample 2: 41.7 ppm = 41.7 mg/L x 0.1 L = 4.17 mg Ca in the sample.

% Ca = (0.00417 g)/(0.05 g) x 100 = 8.34 % Ca

Milk Sample 3: 28.9 ppm = 28.9 mg/L x 0.1 L = 2.89 mg Ca in the sample.

% Ca = (0.00289 g)/(0.05 g) x 100 = 5.78 % Ca

Explain why the analytical selectivity of FAAS is not affected by Doppler broadening and collision energy transfer events in the flame.

This is an open-ended questions: Instructors will need to adjust their expectations accordingly. However, the answer should include a discussion of the HCL source and the fact that Doppler broadening in the source the source is less than in the flame. Therefore the effects of Doppler broadening are limited by the source and not the flame.

What is the typical relative error from sample to sample in

(a) FAAS

(b) GFAAS

In your own words, explain how the background noise from flame emissions is attenuated in AAS.

This is an open-ended questions: Instructors will need to adjust their expectations accordingly. However the student’s answer should include a discussion of the use of a chopper to add a high frequency component to the source radiation and the use of a high pass filter to attenuate the low frequency emissions from the flame. Students should be referred to Chapter 5 if they require review of signal processing strategies.

Explain how the use of an HCL in AAS reduces the need for a monochromator of high spectral purity. In other words, how does the HCL allow us to get away with using a cheaper monochromator?

This is an open-ended questions: Instructors will need to adjust their expectations accordingly. This Exercise is essentially the inverse of Exercise 7.3. The answer is essentially the same as Exercise 7.3. The relatively narrow bandwidth of the HCL emission restricts absorption of analyte within the flame to those emission bands.

What is the purpose of using a D2 lamp collinear with the optical path in AAS?

This is an open-ended questions: Instructors will need to adjust their expectations accordingly. The D2 lamp is part of the “background correction” of the instrument. Students should mention the fact that source radiation from the HCL and radiation from a continuum source (D2 lamp) are alternately sent through the analyte in the flame by use of a chopper. The chopper modulates the HCL and the deuterium lamp radiation at 180 degrees out of phase. Therefore, the flame is alternately exposed to the D2 source and the HCL source. During exposure to the D2 source, a background absorption reading is made and during exposure to the HCL, an analyte plus background absorption reading is made. A lock-in amplifier is used to subtract the two readings.

The 3s 3p transition in Na is at about 589 nm. The same Mg+ transition is at about 280 nm. Use the Boltzmann equation (Equation 7.5) to calculate the fraction of atoms or ions in the excited state for each analyte at 2,000°C.

Ne

Ng=

ge

gge

(− ∆E

kT) Eq. 7.5

Sodium: E for the 589 transition in sodium was determined in Example 7.2 to be 3.36 X 10-19J. Substitution into Equation 7.5 yields

Ne

Ng= 1𝑒

−(3.36 𝑋 10−19𝐽

(1.381 𝑋 10−23𝐽𝐾−1)(2273.15𝐾))

= 0.000022471 ≈ 22.5 𝑝𝑝𝑚

Mg+: we need to first convert 280nm into joules.

𝐸 = ℎ𝑐

𝜆=

(6.6261 𝑋 10−34𝐽𝑠)(2.99 𝑋 108 𝑚𝑠⁄ )

(280 𝑋 10−9𝑚)= 7.076 𝑋 10−19𝐽

Substitution into Equation 7.5 yields

Ne

Ng= 1𝑒

−(7.076 𝑋 10−19𝐽

(1.381 𝑋 10−23𝐽𝐾−1)(2273.15𝐾))

= 1.6245 𝑋 10−10 ≈ 0.16245 𝑝𝑝𝑏

Demonstrate the lifetime dependence of Heisenberg broadening by calculating the theoretical minimum bandwidth for three different transitions, with lifetimes of 10 ms, 10 ns, and 10 fs respectively for transitions centered at

(a) 200 nm (b) 400 nm (c) 700 nm (d) 900 nm

Students are encouraged to review Example 7.1 for assistance with this problem.

The relevant equations are

𝛥𝛦 =ℏ

2𝛥𝑡𝛥𝜆 = 𝛥𝛦

𝜆2

ℎ𝑐

Programing these equations into a spreadsheet, the following correlation table between lifetimes and wavelength was obtained.

Transition Lifetime (s) E (joules) (200nm) (400nm) (700nm) (900nm)

10 ms 5.23 X 10-37 1.06 X 10-6 fs 4.22 X 10-6 fs 1.29 X 10-5 fs 2.14 X 10-5 fs

10 ns 5.23 X 10-43 1.06 fs 4.22 fs 12.9 fs 21.4 fs

10 fs 5.23 X 10-49 1.06 ns 4.22 ns 12.9ns 21.4 ns

Determine the ground state and first excited state term symbols for each of the following elements.

(a) Ca (b) Mg (c) Ti (d) K

a) The ground state electron configuration for calcium is Ca:[Ar]4s2, therefore, s=0 and the total spin S = 2s+1 = 1. The H.O.M.O. is an s-orbital therefore L = 0 and J = s + L = 0. So the ground state term symbol is 1S0. The first excited state of calcium would create the following electron configuration

Ca:[Ar]4s13d1. s = 0 and S = 2s+1 =1. The H.O.M.O. is now a d-orbital therefore L = 2 and J = s + L = 2. The first excited state term symbol would therefore be 1D2.

b) The ground state electron configuration for magnesium is Mg:[Ne]3s2, therefore, s=0 and the total spin S = 2s+1 = 1. The H.O.M.O. is an s-orbital therefore L = 0 and J = s + L = 0. So the ground state term symbol is 1S0. The first excited state of magnesium would create the following electron configuration

Mg:[Ne]3s13p1. s = 0 and S = 2s+1 =1. The H.O.M.O. is now a p-orbital therefore L = 1 and J = s + L = 1. The first excited state term symbol would therefore be 1P1.

c) The ground state electron configuration for titanium is Ti: [Ar]4s23d2, therefore, s=1 and the total spin S = 2s+1 = 3. The H.O.M.O. is an d-orbital therefore L = 2 and J = s + L = 3. So the ground state term symbol is 3P3. The first excited state of titanium would create the following electron configuration

Ti:[Ar]4s23d14p1. s = 1 and S = 2s+1 =3. The H.O.M.O. is now a p-orbital therefore L = 1 and J = s + L = 2. The first excited state term symbol would therefore be 3P2.

d) The ground state electron configuration for potassium is K:[Ar]4s1, therefore, s=½ and the total spin S = 2s+1 = 2. The H.O.M.O. is an s-orbital therefore L = 0 and J = s + L = 2. So the ground state term symbol is 2S2. The first excited state of potassium would create the following electron configuration

K:[Ar]4s03p1. s = ½ and S = 2s+1 =2. The H.O.M.O. is now a d-orbital therefore L = 2 and J = s + L = 4. The first excited state term symbol would therefore be 2D4.

Use Equation 7.2 to determine the pressure broadening of sodium analyte atoms in a flame AAS experiment. Assume a pressure of 1 atm, a flame temperature of 2,300°C, and an average mass of perturbing particles in the flame of 15 g/mol. Hint: You will need to use the ideal gas law to estimate Np.

The relevant equation is

Δ𝜈𝑝 = 𝜎𝑁𝑝

√[2𝜋𝑅𝑇 (1

𝑀1+

1𝑀2

)]

𝜋

This exercise requires the student to draw on previous knowledge. First, the student needs to

make an educated estimate of (the cross sectional area of sodium) and second the student needs to find Np (the number density of perturbing particles per unit volume (m3 to maintain SI units). A close approximation of the cross sectional area can be calculated from the atomic radius. The atomic radius of sodium is ~1.8Å. Therefore the cross sectional area is approximately A =

r2 = 1.131 X 10-19 m2. Np can estimated from the ideal gas law however the ideal gas law will give you Np as a function of liters and we then need to convert liters to cm3. Rearranging the ideal gas law we obtain

𝑛

𝑉=

𝑃

𝑅𝑇=

1 𝑎𝑡𝑚

(0.082057𝐿 𝑎𝑡𝑚𝐾 𝑚𝑜𝑙

) (2573.15 𝐾)= 4.73 𝑋 10−3

𝑚𝑜𝑙𝑒𝑠

𝐿

𝑁𝑃 = (4.73 𝑋 10−3 𝑚𝑜𝑙𝑒𝑠𝐿

) 1𝐿

1000 𝑐𝑚3 𝑥 (

100 𝑐𝑚

1 𝑚)

3

𝑥 6.022 𝑥 1023𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠

𝑚𝑜𝑙= 2.852 𝑥 1024 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠

𝑚3

We need to convert the molar masses given into SI unites, and into units compatible with NP (particles, rather than moles).

Particle mass of Na = 23𝑔

𝑚𝑜𝑙 𝑥

1 𝑘𝑔

1000 𝑔 𝑥

1 𝑚𝑜𝑙

6.022 𝑥 1023 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 = 3.819 x 10-26 kg/particle

Average particle mass = 15𝑔

𝑚𝑜𝑙 𝑥

1 𝑘𝑔

1000 𝑔 𝑥

1 𝑚𝑜𝑙


When we substitute these values into Eq. 7.2, we use the energy form of R (8.314 J/mol-K) because we are calculating a value, frequency, that is fundamentally related to energy. However, since we are working at the particle level, we need to use an R-value that relates to

particles rather than moles; also, recall that a Joule is equivalent to a 𝑘𝑔∙𝑚2

𝑠2:

R = 8.314 𝐽

𝑚𝑜𝑙 𝑥

1 𝑚𝑜𝑙

6.022 𝑥 1023 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑥

𝑘𝑔∙𝑚2

𝑠2

𝐽⁄ = 1.38 𝑥 10−23

𝑘𝑔∙𝑚2

𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 ∙ 𝑠2

Substitution into equation 7.2 yields

Δ𝜈𝑝 = (1.131 X 10−19 m2) (2.85 𝑋 1024𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠

𝑚3)

√[2(3.14) (1.38 𝑥 10−23 𝑘𝑔 ∙ 𝑚2

𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 ∙ 𝑠2) 2573 𝐾 (1

3.819 𝑥 10−26 𝑘𝑔𝑝𝑎𝑟𝑡

+1

2.491𝑥10−26 𝑘𝑔𝑝𝑎𝑟𝑡

)]

3.14

Δ𝜈𝑝 = 4.57 𝑋 108𝑠−1

To put this in more familiar units, we can convert to Δλ.

𝑐 = 𝜆𝜈 or 𝜈 = 𝑐𝜆−1

We can differentiate to get 𝑑𝜈 = −𝑐𝜆−2𝑑𝜆 or |Δ𝜆| = Δ𝜈𝜆2

𝑐

Figure 7.5 shows us that the primary absorption and emission line for sodium is centered around

λ = 589.3 nm.

|Δ𝜆| = Δ𝜈𝜆2

𝑐=

(4.57 𝑥 108 𝑠−1)(589.3 𝑛𝑚)2

2.998 𝑥 1017𝑛𝑚𝑠⁄

= 5.3 x 10-4 nm

Assuming an atomization temperature of 1,400°C in a graphite furnace AAS experiment,

determine the line broadening () associated with the d-lines in a sodium atom. Report your answer in nanometers. Hint: See Exercise 7.11



√[2𝜋𝑅𝑇 (1

𝑀1+

1𝑀2

)]

𝜋




𝑛

𝑉=

𝑃

𝑅𝑇=

1 𝑎𝑡𝑚


) (1673.15 𝐾)= 7.28 𝑋 10−3


𝐿

𝑁𝑃 = (7.28 𝑋 10−3 𝑚𝑜𝑙𝑒𝑠𝐿

) 1𝐿

1000 𝑐𝑚3 𝑥 (100 𝑐𝑚

1 𝑚)

3



𝑚3



𝑚𝑜𝑙 𝑥

1 𝑘𝑔

1000 𝑔 𝑥

1 𝑚𝑜𝑙



𝑚𝑜𝑙 𝑥

1 𝑘𝑔

1000 𝑔 𝑥

1 𝑚𝑜𝑙




𝑠2 :

R = 8.314 𝐽

𝑚𝑜𝑙 𝑥

1 𝑚𝑜𝑙


𝑘𝑔∙𝑚2

𝑠2

𝐽⁄ = 1.38 𝑥 10−23

𝑘𝑔∙𝑚2




𝑚3 )

√[2(3.14) (1.38 𝑥 10−23 𝑘𝑔 ∙ 𝑚2

𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 ∙ 𝑠2) 1673.15 𝐾 (1

3.819 𝑥 10−26 𝑘𝑔𝑝𝑎𝑟𝑡

+1

2.491𝑥10−26 𝑘𝑔𝑝𝑎𝑟𝑡

)]

3.14

Δ𝜈𝑝 = 4.89 𝑋 108𝑠−1




𝑐

Figure 7.5 shows us that the primary absorption and emission line for sodium is centered around

λ = 589.3 nm.


𝑐=

(4.89 𝑥 108 𝑠−1)(589.3 𝑛𝑚)2

2.998 𝑥 1017𝑛𝑚𝑠⁄

= 5.7 x 10-4 nm

Repeat Exercise 7.12 for a flame AAS experiment assuming a flame temperature of 2,000°C. Compare the resolution of the two techniques and comment on the effect of atomization temperature on resolution as a result of Doppler broadening.



√[2𝜋𝑅𝑇 (1

𝑀1+

1𝑀2

)]

𝜋




𝑛

𝑉=

𝑃

𝑅𝑇=

1 𝑎𝑡𝑚


) (2273.15 𝐾)= 5.36 𝑋 10−3


𝐿

𝑁𝑃 = (5.36 𝑋 10−3 𝑚𝑜𝑙𝑒𝑠𝐿

) 1𝐿

1000 𝑐𝑚3 𝑥 (

100 𝑐𝑚

1 𝑚)

3



𝑚3



𝑚𝑜𝑙 𝑥

1 𝑘𝑔

1000 𝑔 𝑥

1 𝑚𝑜𝑙



𝑚𝑜𝑙 𝑥

1 𝑘𝑔

1000 𝑔 𝑥

1 𝑚𝑜𝑙




𝑠2 :

R = 8.314 𝐽

𝑚𝑜𝑙 𝑥

1 𝑚𝑜𝑙


𝑘𝑔∙𝑚2

𝑠2

𝐽⁄ = 1.38 𝑥 10−23

𝑘𝑔∙𝑚2




𝑚3)

√[2(3.14) (1.38 𝑥 10−23 𝑘𝑔 ∙ 𝑚2

𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 ∙ 𝑠2) 2273.15 𝐾 (1

3.819 𝑥 10−26 𝑘𝑔𝑝𝑎𝑟𝑡

+1

2.491𝑥10−26 𝑘𝑔𝑝𝑎𝑟𝑡

)]

3.14

Δ𝜈𝑝 = 3.28 𝑋 108𝑠−1




𝑐

Figure 7.5 shows us that the primary absorption and emission line for sodium is centered

around λ = 589.3 nm.


𝑐=

(3.28 𝑥 108 𝑠−1)(589.3 𝑛𝑚)2

2.998 𝑥 1017𝑛𝑚𝑠⁄

= 3.8 x 10-4 nm

What we learn from comparing Exercise 7.12 & 7.13 is that as the temperature increased from 1400 to 2000oC, the intrinsic line width decreases by

(3.8 𝑋 10−4) − (5.7 x 10−4)

5.7 x 10−4 𝑿 𝟏𝟎𝟎% = −𝟑𝟑%

This seems counter-intuitive until we consider the boundaries we placed on the situation. In

each case, we assumed the pressure remained constant at 1 atm. With constant pressure but

non-constant volume, as temperature increases, the particle number density decreases,

yielding fewer collisions (the fundamental basis of pressure broadening). In fact, we can see

that broadening in a constant-pressure system is related to:

Δ𝜈 ≈ (𝑇)−1√𝑇

In this case, the reciprocal function has a greater effect than the root function, so as

temperature increases, broadening decreases.

In an enclosed system, such as a gas lamp, the situation would be reversed. As temperature

increases, the pressure in the system increases proportionally, yielding more collisions, and

thus greater broadening.

Explain why the use of element specific sources such as the HCL or EDL negate the concerns exposed by Exercise 7.13.

The HCL and EDL operate at cooler temperatures and at lower pressures. Therefore the linewidth of the emission is narrow compared to the line widths for absorption created in the flame. Therefore the resolution of the absorption event is controlled by the HCL or EDL lamp and not by the flame temperature.

Discuss the steps involved in converting the sample matrix into analyte atoms.

This is an open-ended questions: Instructors will need to adjust their expectations accordingly.

If the sample matrix is liquid, then the sample can often be filtered and aspirated directly. If the sample matrix is a solid, it is typically digested in acid, partially neutralized, filtered and then aspirated into the instrument. Students should be expected to expand on all of the steps involved.

Discuss the benefits and disadvantages of Smith-Hieftje background correction. The main advantage of Smith-Hieftje background correction is the fact that one can subtract the absorption reading during high current operation from the absorption reading during normal operation, allowing the analyst to correct for absorption due to closely spaced line spectra from interfering species. A notable (and costly) disadvantage of using Smith-Hieftje background correction is the fact that the high currents needed significantly shorten the lifetime of the hollow cathode lamps.

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