The wavelength of maximum intensity of radiat | vedclass

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Question

(A) According to Wien's displacement law,$\lambda_{\max} T = b$.Given $\lambda_{\max} = 289.8 \, nm = 289.8 	imes 10^{-9} \, m$ and $b = 2898 \, \mu

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) According to Wien's displacement law,$\lambda_{\max} T = b$.Given $\lambda_{\max} = 289.8 \, nm = 289.8 	imes 10^{-9} \, m$ and $b = 2898 \, \mu m \cdot K = 2898 	imes 10^{-6} \, m \cdot K$.Calculating the temperature $T$:$T = \frac{b}{\lambda_{\max}} = \frac{2898 	imes 10^{-6}}{289.8 	imes 10^{-9}} = 10 	imes 10^3 = 10^4 \, K$.According to Stefan-Boltzmann Law,the radiation intensity (emissive power) $E$ is given by $E = \sigma T^4$.Substituting the values:$E = (5.67 	imes 10^{-8}) 	imes (10^4)^4 = 5.67 	imes 10^{-8} 	imes 10^{16} = 5.67 	imes 10^8 \, W/m^2$.

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

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