P237S05HW3sol

P237S05HW3sol - Physics 237 Spring 2005 Solution to...

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Physics 237 Spring 2005 Solution to Homework Assignment #3 3-14. We are given: u ! ( ) = 8 " hc 5 1 exp hc kT # $ % ( ) 1 . By definition, u ( λ ) d λ gives the energy density in the wavelength interval ( λ , λ +d λ ). Hence, u ( ) d = u f ( ) df " u f ( ) = u ( ) d df . This then yields: u f ( ) = 8 hc 5 1 exp hc kT # $ % ( ) 1 c f 2 = 8 hcf 5 c 5 1 exp hf kT # $ % ( ) 1 c f 2 = 8 f 2 c 3 hf exp hf kT # $ % ( ) 1 . (Here, we used f λ = c , and just took the absolute value of the derivative.) 3.15. (a) From Chapter 3, we know that λ m T = 2.898 mm. K. Hence, for T = 2.7 K, λ m = 1.07 mm. (b) Using f λ = c , we get f = 0.28 THz. (c) Treating the universe as a blackbody, the total power radiated per unit area = = 1 4 cU = 1 4 c u ( ) d 0 " # = 2 $ 5 k 4 15 h 3 c 2 T 4 . (You could also just use Stefan’s Law directly, but make sure you know where this comes from!) The total power incident on Earth is then obtained by multiplying this by the surface area of Earth: P earth = 2 5
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P237S05HW3sol - Physics 237 Spring 2005 Solution to...

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