1144_Physics ProblemsTechnical Physics

# 1144_Physics ProblemsTechnical Physics - Chapter 40...

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Chapter 40 485 P40.61 (a) Starting with Planck’s law, IT hc e hc k T λ π , bg = 2 1 2 5 B the total power radiated per unit area d hc e d hc k T λλ , 0 2 5 0 2 1 ∞∞ zz = B . Change variables by letting x hc kT = B and dx hcd =− B 2 . Note that as varies from 0 →∞ , x varies from ∞→ 0. Then d hc x e dx x , ej 0 4 32 3 0 4 4 2 1 2 15 = F H G I K J B 4 B 4 . Therefore, d k TT σ , 0 5 44 2 15 z = F H G I K J = B 4 . (b) From part (a), == × ×⋅ × 2 15 2 1 38 10 15 6 626 10 3 00 10 5 52 3 4 34 3 8 2 k B 4 JK Js ms . .. e j 567 10 8 . W m K 24 . P40.62 Planck’s law states hc e hc e hc k T hc k T πλ , = 2 1 21 2 5 25 1 B B . To find the wavelength at which this distribution has a maximum, compute dI d hc e e e hc dI d hc e hc e e hc k T hc k T hc k T hc k T hc k T hc k T F H G I K J R S | T | U V | W | = = −+ R S | T | U V | W | = 1 1 0 2 1 5 1 0 26 1 5 2 2 2 6 BB B B B B B B Letting x hc = B , the condition for a maximum becomes xe e x x = 1 5. We zero in on the solution to this transcendental equation by iterations as shown in the table below. The solution is found to be x xe e xx 1 x xe e 1 4.000 00 4.074 629 4 4.964 50
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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