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Atomic Emission Spectroscopy

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Atomic Emission Spectroscopy Chapter 9 - Atomic Emission Spectroscopy Consider Figure 9.1 and the discussion of Boltzmann distribution in Chapters 2 and 7. Before continuing in this chapter, speculate on why a flame is an excellent atomizer for use in atomic absorption but is a poor atomizer for use in atomic emission. Since the signal in AES is the result of relaxation from the excited state, the intrinsic LOD of AES is a function of the population density of the excited state. The flame in AAS is not nearly as hot as the sources used in AES. Use of a flame for AES would have a significant negative impact on the LOD of AES. Thinking in terms of quantum mechanic phenomena and instrumentation, list some similarities and differences between molecular absorption and molecular luminescence spectroscopic methods (see Figure 9.1). This is an open ended question so the instructor may need to adjust expectations accordingly. Because students were asked to discuss quantum mechanic phenomena, the students should not go into details outlining the similarities in instrumentation. Similarities Differences Both measure transitions between electronic UV-vis spectroscopy involves absorption of a quantum states photon while Luminescence involves the emission of a photon. Both techniques have relatively broad spectra as the result of superimposed vibrational and In UV-vis spectroscopy, the detector is measuring rotational states upon each electronic state power from an external source. In luminescence (vibronic). spectroscopy, the detector is measuring power as the excited state analyte molecules relax to the Both techniques involve photon energies that ground state. reside in the “UV-vis” portion of the EM spectrum. Use the Boltzmann distribution equation to calculate the percentage of magnesium atoms that are in the excited state (electronic transition from 3s to 3p, λ = 285.2 nm) under the conditions of (a) an air–acetylene flame at 2,955 K and (b) a plasma at 6,955 K. Assume that ge/gg = 3. Students are encouraged to review Example 9.1. The Boltzman distribution equation is ∆E Ne ge (− ) = e kT Ng gg … and −34 8 푚 ℎ푐 (6.626 푋 10 퐽푠)(2.99 푋 10 ⁄푠) ∆퐸 = = = 6.95 푋 10−19퐽 휆 285.2 푋 10−9푚 Therefore in the air-acetylene flame the population ratio would be, −19 ∆E 6.95 푋 10 퐽 Ne ge (− ) −[ −23 −1 ] = e kT = (3)푒 (1.38065 푥 10 퐽∙퐾 )(2,995퐾) = 1.506 푋 10−7 Ng gg And the percentage of magnesium atoms in the excited state at 2,995K would be 1.506 X 10-5 %. A more compact way of expressing this number would be to say that 0.1506 ppm of the magnesium atoms would be in the excited state at 2,995K. If we changed the atomizer to a plasma torch at 6,955K the population ratio would be −19 ∆E 6.95 푋 10 퐽 Ne ge (− ) −[ −23 −1 ] = e kT = (3)푒 (1.38065 푥 10 퐽∙퐾 )(6,955퐾) = 2.16 푋 10−3 Ng gg And the percentage of magnesium atoms in the excited state at 6,955K would be 0.216 %. A more compact way of expressing this number would be to say that 2160 ppm of the magnesium atoms would be in the excited state at 2,995K The Saha ionization equation allows us to estimate the ratio of gas phase atoms that will be ionized (ni) relative to the number of neutral atoms (n0 = ntotal – ni) under certain conditions. Assuming that the partial pressure of Mg atoms in the atomizer is 10-5 atm, ge/gg = 3, and the volume of the flame or plasma is 2.0 cm3, calculate the percentage of Mg atoms that will be ionized under the two atomizer conditions given in Problem 9.3. The first ionization energy of magnesium is 737.7 kJ/mol. 2 We can begin by calculating the Saha ratio (ni /n0) in general terms. We will first need to calculate the given ionization energy (737.7 kJ/mol) in terms of joules per atom: 737.7 푘퐽 1000 퐽 푚표푙 -18 퐸 = 푥 푥 = 1.2250 x 10 J/atom 푚표푙 푘퐽 6.022 푥 1023 푎푡표푚푠 1.225 x 10−18J 2 E −( ) n 2 −( i ) 3 2 −23 J i 21 k T 21 −3 (1.381 x 10 )(2955 K) = (2.41 x 10 )T3 ∙ e b = (2.41 x 10 m K2)(2955 K)3 ∙ e K n0 = 3.51103 x 1013 m-3 From the units, we see that this is a ratio calculated for a volume of 1 cubic meter (that is, per m3 or -3 times m ). If this were a direct ratio (that is, simply ni/no), it would not matter, but since the numerator of the ratio term we calculated is squared, we need to adjust to the volume of the flame: 2 13 3 ni 3.51103 x 10 m 3 7 ( ) = 3 x ( ) x 2.0 cm = 7.0221 x 10 n0 flame m 100 cm ni To get the percentage of ionized Mg atoms, we need x 100%. Since n0 is the number of unionized nt Mg atoms, we know that n0 = nt – ni, where nt is the total number of Mg atoms in the flame. We are given that the partial pressure of Mg in the flame is 10-5 atm, and we will treat this as an exact number to find nt – note that we need to distinguish between the nt as defined here (number of Mg atoms) vs. “n” as a common unit for moles, which we can designate as nt,moles −5 PV (10 atm)(0.0020 L) -11 nt,moles = = L∙atm = 8.24785 x 10 mol Mg atoms RT (0.08206 )(2955 K) mol∙K 23 −11 6.022 x 10 atoms 13 n = 8.24785 x 10 mol Mg atoms x = 4.96686 x 10 atoms total t mol We can derive a quadratic expression to allow us to extract ni: 2 2 ni 7 ni = 7.0221 x 10 so n0 = 7 n0 7.0221 x 10 2 ni also, n0 = nt - ni so = n − n 7.0221 x 107 t i 2 7 7 then ni = (7.0221 x 10 )nt − (7.0221 x 10 )ni 2 7 13 7 and ni = (7.0221 x 10 )(4.96686 x 10 ) − (7.0221 x 10 )ni 2 21 7 ni = (3.4878 x 10 ) − (7.0221 x 10 )ni 2 7 21 and finally ni + (7.0221 x 10 )ni − (3.4878 x 10 ) = 0 −푏 ± √−푏+4푎푐 10 We then use the quadratic equation ( ) to find ni = 5.9022 x 10 ions. 2푎 So percentage of ionized Mg atoms 10 푛푖 5.9022 푥 10 푥 100% = 13 푥 100% = 0.12% ionized at 2955 K 푛푡 4.96686 푥 10 We can follow the same procedure to find that 99.74% Mg atoms would be expected to be ionized at 6955 K. Instructors should follow this interesting result up with discussion introduced in Problem 9.5. The Saha equation does not take into account environment – it is based solely on the relationship between temperature and ionization energy. The inherently electron-rich, reducing environment of both the flame and the plasma prevents ionization at the level predicted by the Saha equation. In AES, one of the concerns of using too hot of a source is the generation of atomic ions instead of atomic atoms. The plasma source used in AES is much hotter than the atomizers used in AAS, yet we do not observe an overwhelming abundance of atomic ions in the sample spectrum. Speculate on why this is true (Hint: Think about the definition of and physical composition of plasma). A plasma is an energized atomic vapor resulting in a “soup” of free electrons and free cationic ions. The very high concentration of free electrons in the plasma result in the reduction of analyte cations generated during the atomization process. Find the cost per volume for a tank of high-purity argon in your area. What is the cost per minute of using (a) an ICP-AES and (b) a DCP-AES? The answer to this question will vary depending upon the price the student finds. This problem can also be a little tricky because the student will need to convert the volume of the gas from the volume of the pressurized cylinder to volume of argon at one atmosphere of pressure. A typical cylinder is 30 liters (often referred to as the water volume) and is purchased with a pressure typically near 2000 psi. Assuming constant temperature, if the new argon tank was pressurized to 2000 psi the pressure at one atmosphere (14.7 psi) would be found using P1V1=P2V2. Solving for V2 we obtain, 푃1푉1 2000푝푠 푋 30퐿 푉2 = = = 4081퐿 푃2 14.7푝푠 A typical ICP torch consumes 12 liters/min of argon so a typical argon tank will last approximately 4081퐿⁄ 340 푚푛 푡푎푛푘 = 12퐿 푡푎푛푘 ⁄푚푛 Assuming a price of $300/tank, an ICP torch costs $300⁄ 푡푎푛푘 = $0.88⁄ 340푚푛 푚푛 ⁄푡푎푛푘 A typical DCP torch consumes between 5 & 8 liters/min of argon. If we assume the high end of this range, an argon tank will last 4081퐿⁄ 510 푚푛 푡푎푛푘 = 8퐿 푡푎푛푘 ⁄푚푛 For a cost of $300⁄ 푡푎푛푘 = $0.59⁄ 510푚푛 푚푛 ⁄푡푎푛푘 You are the financial manager of an analytical laboratory that uses atomic emission to conduct environmental analysis. Assuming you run the instrument continuously for 6 hours on each workday, how much would you spend on argon each year (using your values from Problem 9.6)? Be sure to clearly indicate all assumptions you make to estimate the number of workdays each year.
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