Chapter7ISM

# Chapter7ISM - CHAPTER 7 QUANTUM THEORY AND THE ELECTRONIC...

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CHAPTER 7 QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS Problem Categories Biological : 7.110, 7.122, 7.125, 7.132, 7.140. Conceptual : 7.25, 7.26, 7.27, 7.28, 7.59, 7.60, 7.68, 7.94, 7.98, 7.99, 7.101, 7.112, 7.116, 7.140, 7.142. Descriptive : 7.127. Environmental : 7.109, 7.125. Difficulty Level Easy : 7.7, 7.8, 7.9, 7.10, 7.11, 7.12, 7.15, 7.16, 7.19, 7.20, 7.26, 7.28, 7.30, 7.33, 7.39, 7.40, 7.41, 7.42, 7.55, 7.56, 7.60, 7.63, 7.65, 7.67, 7.69, 7.70, 7.75, 7.76, 7.77, 7.89, 7.90, 7.101, 7.115. Medium : 7.17, 7.18, 7.21, 7.22, 7.25, 7.27, 7.29, 7.31, 7.32, 7.34, 7.57, 7.58, 7.59, 7.61, 7.62, 7.64, 7.66, 7.78, 7.87, 7.88, 7.91, 7.92, 7.93, 7.94, 7.95, 7.96, 7.97, 7.98, 7.100, 7.107, 7.108, 7.110, 7.113, 7.114, 7.116, 7.117, 7.118, 7.119, 7.120, 7.121, 7.124, 7.128, 7.129, 7.130, 7.132, 7.138, 7.140, 7.142, 7.143. Difficult : 7.68, 7.99, 7.102, 7.103, 7.104, 7.105, 7.106, 7.109, 7.111, 7.112, 7.122, 7.123, 7.125, 7.126, 7.127, 7.129, 7.131, 7.133, 7.134, 7.135, 7.136, 7.137, 7.139, 7.141. 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 97 11 0 m 456 nm 456 10 m 4.56 10 m 1nm −− × ×= × = × 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 × 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.

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CHAPTER 7: QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS 200 (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 × = = ν × c The problem asks for the wavelength in units of nanometers. Recall that 1 nm = 1 × 10 9 m. 9 1nm 0.122 m 11 0 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) 1mi 1km 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) 71 0 1.61 km 1000 m ? distance (m) (2.8 10 mi) 4.5 10 m × × = × 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 1s (4.5 10 m) 1.5 10 s 3.00 10 m × = × = × ?min 2 .5min 7.11 8 2 1 3.00 10 m/s 3.26 10 m 9192631770 s × == = × = ν 7 3.26 10 nm c This radiation falls in the microwave region of the spectrum. (See Figure 7.4 of the text.) 7.12 The wavelength is: 7 1m 6.05780211 10 m 1,650,763.73 wavelengths = × 8 7 3.00 10 m/s = 6.05780211 10 m × λ × 14 1 4.95 10 s c ν×
CHAPTER 7: QUANTUM THEORY AND THE ELECTRONIC STRUCTURE OF ATOMS 201 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.

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## This note was uploaded on 12/16/2011 for the course CHM 100 taught by Professor Cs during the Spring '11 term at SUNY Albany.

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Chapter7ISM - CHAPTER 7 QUANTUM THEORY AND THE ELECTRONIC...

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