Silberberg6eISMChapter7 - CHAPTER 7 QUANTUM THEORY AND...

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CHAPTER 7 QUANTUM THEORY AND ATOMIC STRUCTURE The value for the speed of light will be 3.00x10 8 m/s except when more significant figures are necessary, in which cases, 2.9979x10 8 m/s will be used. TOOLS OF THE LABORATORY BOXED READING PROBLEMS B7.1 Plan: Plot absorbance on the y -axis and concentration on the x -axis. Since this is a linear plot, the graph is of the type y = mx + b , with m = slope and b = intercept. Any two points may be used to find the slope, and the slope is used to find the intercept. Once the equation for the line is known, the absorbance of the solution in part b) is used to find the concentration of the diluted solution, after which the dilution equation is used to find the molarity of the original solution. Solution: a) Absorbance vs. Concentration: 0 0.1 0.2 0.3 0.4 0.5 0 0.00001 0.00002 0.00003 0.00004 Concentration (M) Absorbance This is a linear plot, thus, using the first and last points given: m = 2 1 2 1 y y x x = 5 5 0.396 0.131 3.0x10 1.0x10 M = 13,250 = 1.3x10 4 / M Using the slope just calculated and any of the data points, the value of the intercept may be found. b = y mx = 0.396 – (13,250/ M )(3.0x10 –5 M ) = –0.0015 = 0.00 (absorbance has no units) b) Use the equation just determined: y = (1.3x10 4 / M ) x + 0.00. x = ( y – 0.00)/(1.3x10 4 / M ) = (0.236/1.3x10 4 ) M = 1.81538x10 –5 M = 1.8x10 5 M This value is M f in a dilution problem ( M i V i )= ( M f V f ) with V i = 20.0 mL and V f = 150. mL. M i =  f f i M V V = 5 1.81538x10 150. mL 20.0 mL M = 1.361538x10 –4 = 1.4x10 –4 M 7-1
B7.2 Plan: The color of light associated with each wavelength can be found from Figure 7.3. The frequency of each wavelength can be determined from the relationship c = or = c . The wavelength in nm must be converted to meters. Solution: a) red = 8 9 3.00x10 m/s 1 nm 671 nm 10 m = 4.4709x10 14 = 4.47x10 14 s –1 b) blue = 8 9 3.00x10 m/s 1 nm 453 nm 10 m = 6.6225x10 14 = 6.62x10 14 s –1 c) yellow-orange = 8 9 3.00x10 m/s 1 nm 589 nm 10 m = 5.0933786x10 14 = 5.09x10 14 s –1 END–OF–CHAPTER PROBLEMS 7.1 All types of electromagnetic radiation travel as waves at the same speed. They differ in both their frequency, wavelength, and energy. 7.2 Plan: Recall that the shorter the wavelength, the higher the frequency and the greater the energy. Figure 7.3 describes the electromagnetic spectrum by wavelength and frequency. Solution: a) Wavelength increases from left (10 –2 nm) to right (10 12 nm) in Figure 7.3. The trend in increasing wavelength is: x-ray < ultraviolet < visible < infrared < microwave < radio wave . b) Frequency is inversely proportional to wavelength according to the equation c = λν , so frequency has the opposite trend: radio wave < microwave < infrared < visible < ultraviolet < x-ray . c) Energy is directly proportional to frequency according to the equation E = h ν . Therefore, the trend in increasing energy matches the trend in increasing frequency: radio wave < microwave < infrared < visible < ultraviolet < x-ray . 7.3 a) Refraction is the bending of light waves at the boundary of two media, as when light travels from air into water. b) Diffraction is the bending of light waves around an object, as when a wave passes through a slit about as wide as its wavelength.
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