The wavelength of maximum intensity of radiat | vedclass

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Question

(A) Given: $\lambda_m = 289.8 \, nm = 289.8 	imes 10^{-9} \, m$, $b = 2898 \, \mu m K = 2898 	imes 10^{-6} \, m K$, $\sigma = 5.67 	imes 10^{-8} \,

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$5.67 	imes 10^8 \, W/m^2$

Vedclass · Answer

(A) Given: $\lambda_m = 289.8 \, nm = 289.8 	imes 10^{-9} \, m$, $b = 2898 \, \mu m K = 2898 	imes 10^{-6} \, m K$, $\sigma = 5.67 	imes 10^{-8} \, W m^{-2} K^{-4}$.Using Wien's displacement law: $\lambda_m T = b$.$T = \frac{b}{\lambda_m} = \frac{2898 	imes 10^{-6}}{289.8 	imes 10^{-9}} = \frac{2898 	imes 10^{-6}}{289.8 	imes 10^{-9}} = 10 	imes 10^3 = 10^4 \, K$.Radiation intensity (emissive power) $E = \sigma T^4$.$E = (5.67 	imes 10^{-8}) 	imes (10^4)^4$.$E = 5.67 	imes 10^{-8} 	imes 10^{16} = 5.67 	imes 10^8 \, W/m^2$.

The wavelength of maximum intensity of radiation emitted by a star is $289.8 \, nm$. The radiation intensity of the star is (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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