Chapter 40485P40.61(a)Starting with Planck’s law,IThcehck Tλπ,bg=−2125Bthe total power radiated per unit areadhcedhck Tλλ,025021∞∞zz=−B.Change variables by lettingxhckT=Banddxhcd=−B2.Note that as varies from 0→∞, xvaries from ∞→0.Thendhcxedxx,ej0432304421215∞∞−=FHGIKJB4B4.Therefore,dkTTσ,0544215∞z=FHGIKJ=B4.(b)From part (a),==××⋅×−−21521 38 1015 6 626 103 00 105523434382kB4JKJsms...ej=×⋅−567 108. WmK24.P40.62Planck’s law states hcehcehck Thck Tπλ,=−−−212125251BB.To find the wavelength at which this distribution has a maximum, computedIdhceeehcdIdhcehceehck Thck Thck Thck Thck Thck T−−−−FHGIKJRS|T|UV|W|==−−+−RS|T|UV|W|=−−−−1102151026152226BBBBBBBBLetting xhc=B, the condition for a maximum becomes xeexx−=15.We zero in on the solution to this transcendental equation by iterations as shown in the table below.The solution is found to bexxeexx−1xxee−14.000 004.074 629 44.964 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 .