The exitance (power per unit area per unit wavelength) from a blackbody (Box 19-1) is

Chapter 19, Problem 19-15

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The exitance (power per unit area per unit wavelength) from a blackbody (Box 19-1) is given by the Planck distribution: where is wavelength, T is temperature (K), h is Plancks constant, c is the speed of light, and k is Boltzmanns constant. The area under each curve between two wavelengths in the blackbody graph in Box 19-1 is equal to the power per unit area (W/m2 ) emitted between those two wavelengths. We find the area by integrating the Planck function between wavelengths 1 and 2: For a narrow wavelength range, , the value of M is nearly constant and the power emitted is simply the product M . (a) Evaluate M at 2.00 m and at 10.00 m at T 1 000 K. (b) Calculate the power emitted per square meter at 1 000 K in the interval 1.99 m to 2.01 m by evaluating the product M , where 0.02 m. (c) Repeat part (b) for the interval 9.99 to 10.01 m. (d) The quantity [M ( 2 m)]/[M ( 10 m)] is the relative exitance at the two wavelengths. Compare the relative exitance at these two wavelengths at 1 000 K with the relative exitance at 100 K. What does your answer mean? 19-16.

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