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6 Electronic Structure of Atoms Solutions to Exercises FM : 6.626 X 10⁻³⁴ J 98.3 X 10⁶ = 6.51 10⁻²⁶ J The FM photon has about 100 times more energy than the AM photon. 6.27 Analyze/Plan. Use E = to calculate J/photon; Avogadro's number to calculate J/mol; photon/J [the result from part (a)] to calculate photons in 1.00 mJ. Pay attention to units. Solve. (a) = = 6.626 325 10⁻⁹ 10⁻³⁴ m X 2.998 S 10⁸ m = 6.1122 10⁻¹⁹ = 6.11 X 10⁻¹⁹ J/photon (b) 6.1122 X 10⁻¹⁹ J X 6.022 X 10²³ photons = 3.68 10⁵ J/mol = 368 kJ/mol 1 photon 1 mol (c) 6.1122 1 photon X 10⁻¹⁹ J 1.00 mJ X 1 1 mJ J = 1.64 10¹⁵ photons Check. Powers of 10 (orders of magnitude) and units are correct. (d) If the energy of one 325 nm photon breaks exactly one bond, one mol of photons break 1 mol of bonds. The average bond energy in kJ/mol is the energy of 1 mol of photons (from part b) 368 kJ/mol. 941 10³ J 1 mol 6.28 X = 1.563 X 10⁻¹⁸ = 1.56 10⁻¹⁸ J/photon mol N₂ 6.022 10²³ photons = hc/E = 6.626 1.563 X 10⁻³⁴ 10⁻¹⁸ J J X 2.998 1s m = 1.27 10⁻⁷ m = 127 nm According to Figure 6.4, this is ultraviolet radiation. 6.29 Analyze/Plan. E = gives J/photon. Use this result with J/s (given) to calculate photons/s. Solve. (a) The ~1 10⁻⁶ m radiation is infrared but very near the visible edge. (b) Ephoton = = 6.626 987 10⁻³⁴ 10⁻⁹ m 2.998 1s 10⁸ m = 2.0126 10⁻¹⁹ = 2.01 10⁻¹⁹ J/photon 0.52 32 J X 2.0126 1 photon 10⁻¹⁹ J = 8.1 10¹⁶ photons/s Check. (7 3/1000) ≈ 21 X ≈ 2.1 J/photon (0.5/30/2) = 0.008 10¹⁹ = 8 10¹⁶ photons/s Units are correct; powers of 10 are reasonable. 6.30 (a) 3.55 mm = 3.55 m; the radiation is microwave. (b) = = 6.626 3.55 X 10⁻³⁴ 10⁻³ m 2.998 1s 10⁸ m = 5.5957 10⁻²³ = 5.60 10⁻²³ J/photon 145