Chapter 30: Quantum PhysicsAnswers to Even-Numbered Conceptual Questions2. If energy is quantized, as suggested by Planck, the amount of energy for even a single high-frequencyphoton can be arbitrarily large. The finite energy in a blackbody simply can’t produce such high-frequencyphotons, and therefore the infinite energy implied by the “ultraviolet catastrophe” cannot occur. In classicalphysics, any amount of energy can be in the form of high-frequency light—the energy does not have to besupplied in discrete, large lumps as in Planck’s theory. Therefore, classical physics implies that allfrequencies of light have the same amount of energy, no matter how high the frequency. This is what leads tothe “catastrophe.”4. Planck’s theory of blackbody radiation implies a one-to-one relationship between the absolute temperatureof a blackbody and the frequency of light at the peak of its radiated energy spectrum. This relationship isgiven by Wien’s displacement law (equation 30-1). Therefore, by measuring the peak in the radiated energyfrom a star, we can tell its temperature. In broad terms, a blue star is very hot, a red star much less so, and ayellowish star like our Sun is intermediate in temperature.6. A monochromatic source of light means—literally—that it emits light of a single color. This means that allthe photons emitted by the source have the same frequency, and hence they also have the same energy.8. (a) A photon from a green light source always has less energy than a photon from a blue light source. (b) Aphoton from a green light source always has more energy than a photon from a red light source. The reasonfor these results is that the energy of a photon depends linearly on the frequency of light; that is, E = hƒ.10. Classically, it should be possible to eject electrons with light of any frequency—all that is required is toincrease the intensity of the beam of light sufficiently. The fact that this is not the case means that theclassical picture is incorrect. In addition, the fact that there is a lowest frequency that will eject electronsimplies that the energy of the photon is proportional to its frequency, in agreement with E = hƒ.12. Yes. An electron and a proton have the same de Broglie wavelength λ =h p( ) if they have the samemomentum.Solutions to Problems and Conceptual Exercises1. Picture the Problem: The blackbody spectrum of blackbody A peaks at a longer wavelength than that of blackbody B.Strategy: Use Wien’s Displacement Law (equation 30-1) to answer the conceptual question.Solution: 1. (a) Because blackbody A has the longer wavelength it also has the lower frequency. We know from equation 30–1, however, that the peak frequency of a blackbody spectrum is proportional to the absolute temperature of the blackbody. Therefore, the temperature of blackbody A is lower than the temperature of blackbody B. 2. (b) The best explanation is II. Blackbody B has the higher temperature because an increase in temperature means an increase in frequency, which corresponds to a decrease in wavelength. Statement I is false.Insight: Statement I is false because increasing the temperature of a blackbody shifts the peak emission to greater frequency, not greater wavelength.2. Picture the Problem: The star Betelgeuse emits a peak frequency that is proportional to its surface temperature.Strategy: Solve Wien’s Displacement Law (equation 30-1) for the temperature of Betelgeuse.Solution: Solve equation 30-1 for the temperature: fpeak= 5.88×1010 s−1⋅K−1( )TT =1.82×1014 Hz5.88×1010 s−1⋅K−1= 3100 KInsight: Betelgeuse, like all red-giant stars, has a cooler surface temperature than our own Sun. The first printing of the Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.30 – 1Chapter 30: Quantum Physics James S. Walker, Physics, 4th Editionthird edition text had incorrectly listed the peak frequency as 3.09×1014 Hz, which would correspond to an incorrect surface temperature of 5260 K, much closer to the 5800 K surface temperature of our Sun.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.30 – 2Chapter 30: Quantum Physics James S. Walker, Physics, 4th Edition3. Picture the Problem: The human body can be considered a blackbody radiator with a surface temperature of 95F. Thepeak radiation frequency and wavelength are functions of that temperature.Strategy: Solve Wien’s Displacement Law (equation 30-1) for the peak frequency. Remember to convert the Fahrenheittemperature to Kelvin. Use equation 14-1 to calculate the wavelength, where the wave speed is the speed of light.Solution: 1. Calculate the peak frequency: fpeak= 5.88×1010s−1⋅K−1( )T= 5.88×1010s−1⋅K−1( )5995−32( )+273.15⎡⎣⎤⎦Kfpeak=1.81×1013 Hz2. Calculate the wavelength: λ =cf=3.00×108 m/s1.812×1013 Hz=16.6 μmInsight: This wavelength falls in the infrared portion of the electromagnetic spectrum (see Chapter 25).4. Picture the Problem: The initial Big Bang can be considered as a blackbody radiator whose peak frequency is a function of its temperature.Strategy: Solve Wien’s Displacement Law (equation 30-1) for the peak frequency. Use equation 14-1 to calculate the wavelength, where the wave speed is the speed of light.Solution: 1. (a) Calculate the peak frequency: fpeak= 5.88×1010s−1⋅K−1( )T= 5.88×1010s−1⋅K−1( )2.7 K( )=1.6×1011 Hz2. (b) Calculate the wavelength: λ =cf=3.00×108m/s1.6×1011 Hz=1.9 mmInsight: This wavelength falls in the microwave portion of the electromagnetic spectrum (see Chapter 25).5. Picture the Problem: The Sun is a blackbody radiator with a surface temperature of 5800 K.Strategy: Solve Wien’s Displacement Law (equation 30-1) for the peak frequency.Solution: Write the frequency using equation 30-1: fpeak= 5.88×1010s−1⋅K−1( )T= 5.88×1010s−1⋅K−1( )5800 K( )= 3.4×1014 HzInsight: This peak corresponds to a wavelength of 880 nm and falls in the infrared portion of the electromagnetic spectrum (see Chapter 25). We feel
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