The maximum wavelength of radiation emitted b | vedclass

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(C) Given,maximum wavelength,$\lambda_m = 289.8 \ nm = 289.8 	imes 10^{-9} \ m = 2.898 	imes 10^{-7} \ m$.Stefan's constant,$\sigma = 5.67 	imes 10

Vedclass · Accepted Answer

$5.67 	imes 10^8 \ W \ m^{-2}$

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(C) Given,maximum wavelength,$\lambda_m = 289.8 \ nm = 289.8 	imes 10^{-9} \ m = 2.898 	imes 10^{-7} \ m$.Stefan's constant,$\sigma = 5.67 	imes 10^{-8} \ W \ m^{-2} \ K^{-4}$.Wien's constant,$b = 2898 \ \mu m \ K = 2898 	imes 10^{-6} \ m \ K$.According to Wien's displacement law,$\lambda_m = \frac{b}{T}$,so $T = \frac{b}{\lambda_m}$.Substituting the values,$T = \frac{2898 	imes 10^{-6}}{289.8 	imes 10^{-9}} = \frac{2898 	imes 10^{-6}}{2.898 	imes 10^{-7}} = 10^4 \ K$.According to Stefan-Boltzmann law,the intensity of radiation $I$ is given by $I = \sigma T^4$ (assuming the star acts as a black body,$e=1$).Substituting the values,$I = (5.67 	imes 10^{-8}) 	imes (10^4)^4 = 5.67 	imes 10^{-8} 	imes 10^{16} = 5.67 	imes 10^8 \ W \ m^{-2}$.

The maximum wavelength of radiation emitted by a star is $289.8 \ nm$. The intensity of radiation for the star is (Given: Stefan's constant $\sigma = 5.67 \times 10^{-8} \ W \ m^{-2} \ K^{-4}$,Wien's constant $b = 2898 \ \mu m \ K$)

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