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6 Electronic Structure of Atoms Solutions to Exercises (b) Analyze/Plan. Use Equation 6.7 to calculate E, then = hc/AE. Solve. nᵢ = 3, = 2; = -2.18 10⁻¹⁸ 1 1 = -2.18 X 10⁻¹⁸ J (1/4-1/9) = hc/E = 6.626 10⁻³⁴ 2.998 X 10⁸ m/s = 6.56 10⁻⁷ m This is the red line at 656 nm. nᵢ = 4, = 2; = hc/E = 6.626 10⁻³⁴ J 2.998 10⁸ m/s = 4.86 10⁻⁷ m -2.18 10⁻¹⁸ J (1/4-1/16) This is the blue-green line at 486 nm. = 5, = 2; = hc/E = 6.626 10⁻³⁴ J 2.998 10⁸ m/s = 4.34 10⁻⁷ m -2.18 10⁻¹⁸ J This is the blue-violet line at 434 nm. Check. The calculated wavelengths correspond well to three lines in the H emission spectrum in Figure 6.11, so the results are sensible. 6.40 (a) Transitions with = 1 have larger E values and shorter wavelengths than those with = 2. These transitions will lie in the ultraviolet region. (b) nᵢ = 2, = 1; = hc/E = 6.626 10⁻³⁴ J 2.998 10⁸ m/s = 1.21 10⁻⁷ m -2.18 10⁻¹⁸ J (1/1-1/4) = 3, = 1; = hc/E = 6.626 X 10⁻³⁴ J-s 2.998 10⁸ m/s = 1.03 10⁻⁷ m -2.18 10⁻¹⁸ nᵢ = 4, nf = 1; = hc/E = 6.626 X 10⁻³⁴ 10⁻¹⁸ J S 2.998 10⁸ m/s = 0.972 10⁻⁷ m 2.18 J (1/1-1/16) 6.41 (a) 93.8 nm 1 10⁻⁹ m = 9.38 10⁻⁸ m; this line is in the ultraviolet region. 1 nm (b) Analyze/Plan. Only lines with nf = 1 have a large enough to lie in the ultraviolet region (see Solutions 6.39 and 6.40). Solve Equation 6.7 for recalling that is negative for emission. Solve. 148