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analyte ions are separated from electrons and negative ions by a suitable field and are accelerated into the mass spectrometer itself. 11-6. Lasers are used for sampling in ICPMS by exposing a solid sample to an intense, pulsed laser beam, which rapidly vaporizes (ablates) the sample. The resulting vapor is carried into the ICP torch where atomization and ionization occurs. The resulting gaseous mixture then enters the mass spectrometer for ion analysis. 11-7. Isobaric interferences are encountered when the isotopes of two elements have the same mass. A second type of spectroscopic interference occurs from molecular species that have the same mass as that of an analyte ion. A third type of interference is from matrix species that combine with the analyte and reduce the analyte signal as a result. 11-8. An internal standard is often used in preparing calibration curves for ICPMS in order to compensate for random error from instrument drift and noise, torch instabilities, and some matrix effects. The internal standard chosen should be an element that is normally absent from the sample, but one that has an atomic mass and ionization potential close to that of the analyte. Principles of Instrumental Analysis, 6th ed. Chapter 11 3 11-9. In an isotope dilution experiment, a known weight of the analyte that has been prepared from a non-naturally occurring isotope of one of the elements in the analyte is added to the sample to be analyzed and mixed thoroughly until homogenization is assured. A known weight of the sample is then taken and a fraction of the highly purified analyte is isolated and weighed. The amount of enriched analyte is then determined by a mass spectrometric measurement. As shown in Section 32 D-1, these data then permit the calculation of the percentage of the analyte in the original sample. 11-10. In a glow-discharge mass spectrometric analysis, the sample is made part of the cathode of a glow-discharge atomizer, such as that shown in Figure 8-12. Here, the solid sample is bombarded by a stream of argon ions that have been accelerated through a potential of 5 to 15 kV. Analyte atoms are sputtered from the cathode surface and converted by collisions with electrons and argon ions into positive ions. These are then passes into the mass spectrometer for analysis. 11-11. In secondary-ion mass spectrometry, the sample is bombarded with a stream of 5- to 20- eV positive ions, such as Ar+ or Cs+. The ion beam is formed in an ion gun. The impact of the beam on the surface of the sample results in secondary ion formation containing analyte ions, which then pass into the mass spectrometer for analysis. Skoog/Holler/Crouch Chapter 12 Principles of Instrumental Analysis, 6th ed. Instructor’s Manual 1 CHAPTER 12 12-1. λ0 = 12398/V Equation 12-2 λ0 = 12398/(90 × 103) = 0.138 Å 12-2. Equation 12-2 gives the minimum voltage needed to produce an emission line of wavelength λ. Thus, solving for Vmin Vmin = 12398/λ (a) For U, Kβ1 = 0.111 Å and Lβ1 = 0.720 Å Table 12-1 For U Kβ1 Vmin =12398/0.111 = 1.12 × 105 V or 112 kV For U L β1 Vmin =12398/0.720 = 17.2 kV Similarly, λ for Kβ1, Å Vmin for Kβ1, kV λ for Lβ1, Å Vmin for Lβ1, kV (a) For U 0.111 112 0.720 17.2 (b) For K 3.454 3.59 No Lβ1 line ---- (c) For Rb 0.829 15.0 7.075 1.75 (d) For W 0.184 67.4 1.282 9.67 12-3. Here we assume that a linear relationship exists between the atomic number Z for the elements in question and the square root of the frequency ν of their Kα lines (see Figure 12-3). This linear relationship can then be used to determine the frequencies and wavelengths of the other elements as shown in the spreadsheet. Principles of Instrumental Analysis, 6th ed. Chapter 12 2 Principles of Instrumental Analysis, 6th ed. Chapter 12 2 12-4. Proceeding as in Pb. 12-3, we construct the following spreadsheet: 12-4. Proceeding as in Pb. 12-3, we construct the following spreadsheet: Principles of Instrumental Analysis, 6th ed. Chapter 12 3 12-5. ln P0/P = μMρx Equation 12-3 ln (1/0.478) = 49.2 cm2/g × 8.9 g/cm3 × x x = 0.738/(49.2 × 8.9) = 1.69 × 10–3 cm 12-6. μM = WKμK + WIμI + WHμH + WOμO Equation 12-4 (a) M 11.00 39.1 126.9 89.00 16.0 = 16.7 + 39.2 + 0 + 1.50 100.00 166.0 166.0 100.00 18.0 μ ⎛ ⎞ ⎛× × ×⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎞⎟⎠ = 4.92 cm2/g 12-7. ln P0/P = (1/0.965) = μMρx = 2.74 × 2.70x x = 0.0356/(2.74 × 2.70) = 4.82 × 10–3 cm 12-8. ln P0/P = ln(1/0.273) = μMρx = μM × 0.794 × 1.50 μM = 1.298/(0.794 × 1.50) = 1.09 (a) μM = 1.09 = 2 × 39.2 2I W = 0.0139 or 1.39% I 2I W 2 (b) μM = 1.09 = 2 × 39.2 + 2I W ETOH 24 160.70 + 0.00 + 1.50 46 46 W ⎛ ⎞× ×⎜ ⎟⎝ ⎠ 1.09 = 78.4 + 0.887W 2I W ETOH + W 2I W ETOH = 1 Solving the two equations yields = 2.62 × 10 2I W –3 or 0.262% Principles of Instrumental Analysis, 6th ed. Chapter 12 4 12-9. nλ = 2d sin θ Equation 12-6 (a) For topaz, d = 1.356 Å For Fe Kα at 1.76 Å 1 × 1.76 = 2 × 1.346 sin θ θ = 40.46° and 2θ = 80.9° The following data were obtained in the same way. (a) Topaz (d = 1.356 Å) (b) LiF (d = 2.014 Å) (c) NaCl (d = 2.820 Å) Fe (1.76 Å) 80.9 51.8 36.4 Se (0.992 Å) 42.9 28.5 20.3 Ag (0.497 Å) 21.1 14.2 10.1 12.10. Using the same method used in Solution 12-9, we find (a) 2θ = 135° (b) 99.5° 12-11. Vmin = 12398/λ Equation 12-2 (a) Vmin = 12398/3.064 = 4046 V or 4.046 kV Likewise for the others, (b) 1.323 kV (c) 20.94 kV (d) 25.00 kV 12-12. Following the least-squares procedure outlined in Appendix 1, Section a1D, the spreadsheet shown was constructed. The unknown was found to contain 0.149 ± 0.002 %Mn Principles of Instrumental Analysis, 6th ed. Chapter 12 5 Skoog/Holler/Crouch Chapter 13 Principles of Instrumental Analysis, 6th ed. Instructor’s Manual 1 CHAPTER 13 PROBLEM SOLUTIONS 13-1. (a) %T = 10–A × 100% = 10–0.038 × 100% = 91.6% Proceeding in the same way, we obtain (b) 11.0% (c) 39.9% (d) 57.4% (e) 36.7% (f) 20.3% 13-2. (a) A = – log T = – log (15.8/100) = 0.801 Proceeding in the same way, we obtain (b) 0.308 (c) 0.405 (d) 0.623 (e) 1.07 (f) 1.27 13-3 (a) %T = 10–0.038/2 × 100% = 95.7% Similarly, (b) 33.2% (c) 63.2% (d) 75.8% (e) 60.6% (f) 45.1% 13-4. (a) A = – log (2 × 15.8/100) = 0.500 Similarly, (b) 0.007 (c) 0.103 (d) 0.322 (e) 0.770 (f) 0.968 13-5. 4 6.23 mg KMnO = c 3 410 g KMnO L − × 4mg KMnO 4 4 1 mol KMnO 158.034 g KMnO × 5= 3.942 10 M−× A = – log T = – log 0.195 = 0.710 ε = A/(bc) = 0.710/(1.00 × 3.942 × 10–5) = 1.80 × 104 L mol–1 cm–1 13-6. 5.24 mg A = c 100 mL 310 g A−× mg A 1000 mL× 1 mol A L 335 g A × 4= 1.564 10 M−× A = – log T = – log 0.552 = 0.258 ε = A/(bc) = 0.258/(1.50 × 1.564 × 10–4) = 1.10 × 103 L mol–1 cm–1 Principles of Instrumental Analysis, 6th ed. Chapter 13 2 13-7. (a) A = εbc = 9.32 × 103 L mol–1 cm–1 × 1.00 cm × 3.79 × 10–5 M = 0.353 (b) %T = 10–A × 100% = 10–0.353 × 100% = 44.4% (c) c = A/εb = 0.353/(9.32 × 103 × 2.50) = 1.52 × 10–5 M 13-8. (a) A = εbc = 7.00 × 103 L mol–1 cm–1 × 1.00 cm × 3.49 × 10–5 M = 0.244 (b) A = εbc = 7.00 × 103 L mol–1 cm–1 × 2.50 cm × 3.49 × 10–5/2 M = 0.305 (c) For (a), %T = 10–0.244 × 100% = 57.0%; For (b) %T = 10–0.305 × 100% = 49.5% (d) A = – log T = – log (0.570/2) = 0.545 13-9. ε = 7.00 × 103 L mol–1 cm–1 Fe mg Fe = 7.9c 3 g Fe 10 L mg