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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## This note was uploaded on 10/15/2008 for the course PHYS 237 taught by Professor Stephonalexander during the Spring '08 term at Penn State.

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P237S05HW3sol - Physics 237 Spring 2005 Solution to...

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