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6 Electronic Structure of Atoms Solutions to Exercises 6.17 Analyze/Plan. These questions involve relationships between wavelength, frequency, and the speed of light. Manipulate the equation = to obtain the desired quantities, paying attention to units. Solve. (a) = 2.998 S 10⁸ m X 10 1 µm 1 10⁻⁶ m = 3.0 X 10¹³ (b) = c/v; 2.998 X S m X 5.50 1s 10¹⁴ = 5.45 10⁻⁷ m (545 nm) (c) The radiation in (b) is in the visible region and is "visible" to humans. The 10 µm (1 10⁻⁵ m) radiation in (a) is in the infrared region and is not visible. (d) 50.0 X 1 1s 10⁶ 2.998 S 10⁸ m = 1.50 10⁴ m Check. Confirm that powers of 10 make sense and units are correct. 6.18 (a) = 2.998 m X 1 10⁻⁵ m = 6.0 X 10¹² (b) = c/v; 2.998 S 10⁸ m X 1s 10⁸ = 1.2 m (c) Neither of the radiations in (a) and (b) can be observed by an X-ray detector. Radiation (a) is infrared and radiation (b) is radio frequency. (d) 10.5 fs 1 10⁻¹⁵ X 2.998 10⁸ m = 3.15 X 10⁻⁶ m (3.15 µm) S 6.19 Analyze/Plan. = change nm m. Solve. = 2.998 10⁸ m 532 1 nm X 1 1 10⁻⁹ nm m = 5.64 X 10¹⁴ s⁻¹ The color is green. Check. (3000 = 6 10¹⁴ s⁻¹; units are correct. 6.20 According to Figure 6.4, ultraviolet radiation has both higher frequency and shorter wavelength than infrared radiation. Looking forward to section 6.2, the energy of a photon is directly proportional to frequency (E = hv), so ultraviolet radiation yields more energy from a photovoltaic device. Quantized Energy and Photons (section 6.2) 6.21 Quantization means that energy changes can only happen in certain allowed increments. If the human growth quantum is one-foot, growth occurs instantaneously in one-foot increments. That is, a child experiences growth spurts of one-foot; her height changes by one-foot increments. 6.22 Planck's original hypothesis was that energy could only be gained or lost in discrete amounts (quanta) with a certain minimum size. The size of the minimum energy change is related to the frequency of the radiation absorbed or emitted, = hv, and energy changes occur only in multiples of hv. 143