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Chapter 7

Chapter 7 - CHAPTER 7 QUANTUM THEORY AND THE ELECTRONIC...

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CHAPTER 7 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS 7.7 (a) 8 6 13 3.00 10 m/s 3.5 10 m 8.6 10 /s × = = = × = ν × 3 3.5 10 nm c λ × (b) 8 14 9 3.00 10 m/s 5.30 10 /s 566 10 m × = = = × = λ × 14 5.30 10 Hz c ν × 7.8 (a) Strategy: We are given the wavelength of an electromagnetic wave and asked to calculate the frequency. Rearranging Equation (7.1) of the text and replacing u with c (the speed of light) gives: ν = λ c Solution: Because the speed of light is given in meters per second, it is convenient to first convert wavelength to units of meters. Recall that 1 nm = 1 × 10 9 m (see Table 1.3 of the text). We write: 9 9 7 1 10 m 456 nm 456 10 m 4.56 10 m 1 nm × × = × = × Substituting in the wavelength and the speed of light (3.00 × 10 8 m/s), the frequency is: 8 7 m 3.00 10 s or 4.56 10 m × = = = λ × 14 1 14 6.58 10 s 6.58 10 Hz c ν × × Check: The answer shows that 6.58 × 10 14 waves pass a fixed point every second. This very high frequency is in accordance with the very high speed of light. (b) Strategy: We are given the frequency of an electromagnetic wave and asked to calculate the wavelength. Rearranging Equation (7.1) of the text and replacing u with c (the speed of light) gives: λ = ν c Solution: Substituting in the frequency and the speed of light (3.00 × 10 8 m/s) into the above equation, the wavelength is: 8 9 m 3.00 10 s 0.122 m 1 2.45 10 s × λ = = = ν × c

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CHAPTER 7: QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS 195 The problem asks for the wavelength in units of nanometers. Recall that 1 nm = 1 × 10 9 m. 9 1 nm 0.122 m 1 10 m = × = × 8 1.22 10 nm λ × 7.9 Since the speed of light is 3.00 × 10 8 m/s, we can write 8 8 1.61 km 1000 m 1 s (1.3 10 mi) 1 mi 1 km 3.00 10 m × × × × = × 2 7.0 10 s × Would the time be different for other types of electromagnetic radiation? 7.10 A radio wave is an electromagnetic wave, which travels at the speed of light. The speed of light is in units of m/s, so let’s convert distance from units of miles to meters. (28 million mi = 2.8 × 10 7 mi) 7 10 1.61 km 1000 m ? distance (m) (2.8 10 mi) 4.5 10 m 1 mi 1 km = × × × = × Now, we can use the speed of light as a conversion factor to convert from meters to seconds ( c = 3.00 × 10 8 m/s). 10 2 8 1 s (4.5 10 m) 1.5 10 s 3.00 10 m = × × = × = × ? min 2.5 min 7.11 8 2 1 3.00 10 m/s 3.26 10 m 9192631770 s × = = = × = ν 7 3.26 10 nm c λ × This radiation falls in the microwave region of the spectrum. (See Figure 7.4 of the text.) 7.12 The wavelength is: 7 1 m 6.05780211 10 m 1,650,763.73 wavelengths λ = = × 8 7 3.00 10 m/s = 6.05780211 10 m × = = λ × 14 1 4.95 10 s c ν × 7.15 34 8 9 (6.63 10 J s)(3.00 10 m/s) 624 10 m × × = ν = = = λ × 19 3.19 10 J hc h E × 7.16 (a) Strategy: We are given the frequency of an electromagnetic wave and asked to calculate the wavelength. Rearranging Equation (7.1) of the text and replacing u with c (the speed of light) gives: λ = ν c Solution: Substituting in the frequency and the speed of light (3.00 × 10 8 m/s) into the above equation, the wavelength is:
CHAPTER 7: QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS 196 8 7 14 m 3.00 10 s 4.0 10 m 1 7.5 10 s × = = × = × 2 4.0 10 nm λ × Check: The wavelength of 400 nm calculated is in the blue region of the visible spectrum as expected.

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Chapter 7 - CHAPTER 7 QUANTUM THEORY AND THE ELECTRONIC...

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