chap34 - PARTICLES AND WAVES 34 EXERCISES Section 34.2...

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PARTICLES AND WAVES 34 EXERCISES Section 34.2 Blackbody Radiation 15. INTERPRET This is a problem about blackbody radiation. We want to explore the connection between temperature and the radiated power. DEVELOP From the Stefan-Boltzmann law (Equation 34.1), we see that the total radiated power, or luminosity, of a blackbody is proportional to T 4 , PA T σ = 4 . EVALUATE Doubling the absolute temperature increases the luminosity by a factor of 2 4 16. = 2 T == 3 s ASSESS A blackbody is a perfect absorber of electromagnetic radiation. As the temperature of the blackbody increases, its radiated power also goes up. 16. (a) To a good approximation, the surface of Rigel radiates like a blackbody, with emissive power given by Equation 34.1, (b) Equation 34.2a gives 48 2 4 4 4 8 / (5.67 10 W/m K )(10 K) 5.67 10 W/m . 4 max 2.898 mm K/10 K 290 nm, λ =⋅ = in the ultraviolet. 17. INTERPRET We are given the temperature of the blackbody, and asked to find the wavelengths that correspond to peak radiance and median radiance. DEVELOP The wavelength at which a blackbody at a given temperature radiates the maximum power is given by Wien’s displacement law (Equation 34.2a): peak 2.898 mm K T = Similarly, the median wavelength, below and above which half the power is radiated, is given by Equation 34.2b: median 4.11 mm K T = EVALUATE Using the above formulas, we obtain peak median (2.898 mm K)/(288 K) 10.06 m (4.11 mm K)/(288 K) 14.27 m μ λμ = ⋅= = ASSESS The wavelengths are in the infrared. Note that median peak . > 18. From Equation 34.2a, the temperature corresponding to the given peak wavelength in the blackbody spectrum is 2.898 mm K/40 m 72.5 K. T = 19. INTERPRET We find the wavelength for the peak radiance of solar blackbody radiation, and the median wavelength. In both cases, we’ll use the per-unit-wavelength basis, Equations 34.2a and 34.2b. DEVELOP Wien’s law gives us the peak wavelength: peak 2.898 mm K. T = The median wavelength is given by median 4.11 mm K. T The temperature of the Sun is T 5800 K, = so we can use these equations to solve for the desired wavelengths. EVALUATE (a) 4 peak 2.898 mm K/5800 K 5.00 10 mm 500 nm =⋅= × = (b) 4 median 4.11 mm K/5800 K 7.09 10 mm 709 nm = ×= ASSESS The peak wavelength is near the center of the visible spectrum (green) and the median wavelength is just beyond the visible in the near infrared region. Section 34.3 Photons 34.1
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34.2 Chapter 34 20. From Equation 34.6, which equals (a) (b) 2.07 eV, and (c) 12.4 keV for the given frequencies. , Eh f γ = 15 9 (4.136 10 eV s)(1 MHz) 4.14 10 eV, −− ×⋅ = × 21. INTERPRET This problem is about the relationship between the wavelength and energy of a photon. DEVELOP Using / cf λ = and Equation 34.6, the wavelength of a photon can be written as ch c f E == EVALUATE With and 6.5 eV E = 1240 eV nm, hc = we find the wavelength to be 1240 eV nm 191 nm 6.5 eV hc E = This photon is in the ultraviolet region of the electromagnetic spectrum.
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.

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chap34 - PARTICLES AND WAVES 34 EXERCISES Section 34.2...

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