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6 Electronic Structure of Atoms Solutions to Exercises Einstein postulated that light itself is quantized, that the minimum energy of a photon. (a quantum of light) is directly proportional to its frequency, E = hv. If a photon that strikes a metal surface has less than the threshold energy, no electron is emitted from the surface. If the photon has energy equal to or greater than the threshold energy, an electron is emitted and any excess energy becomes the kinetic energy of the electron. 6.23 Analyze/Plan. These questions deal with the relationships between energy, wavelength, and frequency. Use the relationships E = hv = to calculate the desired quantities. Pay attention to units. Solve. (a) E = hv = 6.626 10⁻³⁴ J 6.75 10¹² = 4.47 10⁻²¹ J (b) = 6.626 10⁻³⁴ J m/s 1 1 10⁻⁹ nm m = 6.17 J (c) hc/AE 1s 10⁸ m = 6.92 10⁻⁸ m = nm 6.24 Analyze/Plan. These questions deal with the relationships between energy, wavelength, and frequency. Use the relationships E = hv = to calculate the desired quantities. Pay attention to units. Solve. (a) = = 2.998 S 10⁸ m 1 nm 1 1 10⁻⁹ nm m = 10¹⁴ = 4.61 10¹⁴ (b) = = 6.626 10⁻³⁴ J - 4.6123 10¹⁴ s⁻¹ = 3.06 10⁻¹⁹ J (c) The energy gap between the ground and excited states is the energy of a single 650 nm photon emitted when one electron relaxes from the excited to the ground state. = 3.06 10⁻¹⁹ J 6.25 Analyze/Plan. Use E = pay close attention to units. Solve. (a) E = = 6.626 10⁻³⁴ J 2.998 1s 10⁸ m 3.3 1 µm 1 1 10⁻⁶ µm m = 6.0 10⁻²⁰ J E = = 6.626 10⁻³⁴ J 2.998 1s 10⁸ m 0.154 1 nm 1 1 10⁻⁹ nm m = 1.29 10⁻¹⁵ J Check. (6.6 3/3.3) ≈ 6 (6.6 3/0.15) ≈ 120 10⁻¹⁷ ≈ 1.2 The results are reasonable. We expect the longer wavelength 3.3 µm radiation to have the lower energy. (b) The 3.3 µm photon is in the infrared and the 0.154 nm (1.54 m) photon is in the X-ray region; the X-ray photon has the greater energy. 6.26 E = hv AM : 6.626 10⁻³⁴ J 1010 10³ = 6.69 10⁻²⁸ J 144