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6 Electronic Structure of Atoms Solutions to Exercises nᵢ = 6.626 9.38 X X 10⁻³⁴ 10⁻⁸ J-s m 2.18 2.998 X 10⁻¹⁸ 10⁸ m/s J. = 6 (n values must be integers) nᵢ = 6, = 1 Check. From Solution 6.40, we know that nᵢ > 4 for A = 93.8 nm. The calculated result is close to 6, so the answer is reasonable. 6.42 (a) 2626 nm 1 10⁻⁹ m = 2.626 X 10⁻⁶ m; this line is in the infrared. 1 nm (b) Absorption lines with nᵢ = 1 are in the ultraviolet and with nᵢ = 2 are in the visible. Thus, nᵢ ≥ 3, but we do not know the exact value of nᵢ. Calculate the longest wavelength with nᵢ = 3 = 4). If this is less than 2626 nm, nᵢ > 3. A = hc/E = 6.626 X 10⁻³⁴ 2.998 X 10⁸ m/s = 1.875 X 10⁻⁶ m -2.18 X 10⁻¹⁸ J (1/16-1/9) This wavelength is shorter than 2.626 10⁻⁶ m, so nᵢ > 3; try = 4 and solve for nf as in Solution 6.41. Note that is positive because we are dealing with absorption. = 6 6.43 ; Change mass to kg and velocity to m/s in each case. Solve. (a) 50 1 hr km X 1000 1 km m X 60 1 min hr X 1 60 min = 13.89 = 14 m/s A = 6.626 X 10⁻³⁴ kg X 85 1 kg 13.89 1s m = 5.6 10⁻³⁷ m (b) 10.0 1000 = 0.0100 kg A = 6.626 10⁻³⁴ kg X 0.0100 1 kg 250 1s m = 2.65 X 10⁻³⁴ m (c) We need to calculate the mass of a single Li atom in kg. 6.94 1 mol Li Li 1000 1kg X 6.022 10²³ Li atoms = 1.152 10⁻²⁶ = 1.15 10⁻²⁶ kg 1 mol A = 6.626 X 1.152 1 10⁻²⁶ kg X 2.5 X 1s 10⁵ m = 2.3 10⁻¹³ m (d) Calculate the mass of a single O₃ molecule in kg. 48.00 g O₃ 1kg 1 mol 1 mol O₃ X 1000 6.022 O₃ molecules = 7.971 10⁻²⁶ = 7.97 10⁻²⁶ kg 149