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Unformatted text preview: PARTICLES AND WAVES E XERCISES Section 34.2 Blackbody Radiation 15. I NTERPRET This is a problem about blackbody radiation. We want to explore the connection between temperature and the radiated power. D EVELOP From the Stefan-Boltzmann law (Equation 34.1), 4 , P AT σ = we see that the total radiated power, or luminosity, of a blackbody is proportional to 4 . T E VALUATE Doubling the absolute temperature increases the luminosity by a factor of 4 2 16. = A SSESS 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, 4 8 2 4 4 4 8 2 / (5.67 10 W/m K )(10 K) 5.67 10 W/m . P A T- = = ⋅ = × 3 s (b) Equation 34.2a gives 4 max 2.898 mm K/10 K 290 nm, λ = ⋅ = in the ultraviolet. 17. I NTERPRET We are given the temperature of the blackbody, and asked to find the wavelengths that correspond to peak radiance and median radiance. D EVELOP 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 λ = ⋅ 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 λ = ⋅ E VALUATE 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 λ μ λ μ = ⋅ = = ⋅ = A SSESS The wavelengths are in the infrared. Note that median peak . λ λ 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 μ = ⋅ = 19. I NTERPRET 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. D EVELOP Wien’s law gives us the peak wavelength: peak 2.898 mm K. T λ = ⋅ The median wavelength is given by median 4.11 mm K. T λ = ⋅ The temperature of the Sun is 5800 K, T = so we can use these equations to solve for the desired wavelengths. E VALUATE (a) 4 peak 2.898 mm K/5800 K 5.00 10 mm 500 nm λ- = ⋅ = × = (b) 4 median 4.11 mm K/5800 K 7.09 10 mm 709 nm λ- = ⋅ = × = A SSESS 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 34 34.2 Chapter 34 20. From Equation 34.6, , E hf γ = which equals (a) 15 9 (4.136 10 eV s)(1 MHz) 4.14 10 eV,-- × ⋅ = × (b) 2.07 eV, and (c) 12.4 keV for the given frequencies....
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This note was uploaded on 09/12/2009 for the course PH PH101 taught by Professor Prof.kim during the Spring '09 term at Korea Advanced Institute of Science and Technology.

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