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Chapter 7 The Quantum-Mechanical Model of the Atom 143 7.97 For the orbital in the Excel® spreadsheet, call the columns as follows: column A as r and column as Make the values for r column A as follows: 0-200. In column B, put the equation for the wave function written as follows: = (POWER(1/3.1415,1/2))*(1/POWER(53,3/2))*(EXP(-A2/53)).Go to make chart, choose xy scatter. e.g., sample values 1s r (pm) Y (1s) 0.0016 0.0014 0 7.000146224 0.0012 1 7.000143491 2 7.000140809 1s wavefunction 0.001 0.0008 3 7.000138177 0.0006 4 7.000135594 0.0004 0.0002 5 7.00013306 0 6 7.000130573 0 50 100 150 200 250 r (pm) For the 2s orbital in the same Excel spreadsheet, call the columns as follows: column A as r and column C as Use the same values for r in column A: 0-200. In column C, put the equation for the wave function written as follows: = Go to make chart; choose xy scatter. e.g., sample values 2s 0.0006 r (pm) (2s) 0.0005 0 7. 0000516979 0.0004 1 7. 000050253 2 7. 0000488441 2s wavefunction 0.0003 3 7. 0000474702 0.0002 4 7. 0000461307 0.0001 5 7. 0000448247 0 50 100 150 200 250 6 7. 0000435513 -0.0001 r (pm) Note: The plot for the 2s orbital extends below the x-axis. The x-intercept represents the radial node of the orbital. 7.99 Given: threshold frequency = 2.25 X 10¹⁴ s⁻¹; = 5.00 X 10⁻⁷ m Find: of electron Conceptual Plan: and then and then KE Solution: = (6.626 X 10⁻³⁴ (2.25 X 10¹⁴ = 1.491 10⁻¹⁹ J; E = (6.626 X 10⁻³⁴ 5.00 10⁻⁷ m X = 3.976 X 10⁻¹⁹ J X KE = 3.976 X 10⁻¹⁹ J - 1.491 X 10⁻¹⁹ J = 2.485 X 10⁻¹⁹ J; KE = 1/2mv²; substituting and rearranging gives 2.485 v² = = 5.455 X 10¹¹ 1 X 10⁻³¹ kg) v = 7.39 X 10⁵ m/s Check: The units of the answer (m/s) are correct. The magnitude of the answer is reasonable for the speed of an electron. 7.101 Given: t = 5.0 fs; = 722 nm Find: and Conceptual Plan: t and then Copyright © 2017 Pearson Education, Inc.