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The wavelength of maximum intensity of radiation emitted by a star is $289.8 \mathrm{~nm}$. The radiation intensity of the star is
(Stefan's constant $=5.67 \times 10^{-8} \mathrm{Wm}^2 \mathrm{~K}^{-4}$, constant $b=2898 \mu \mathrm{mK}$ )
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(Stefan's constant $=5.67 \times 10^{-8} \mathrm{Wm}^2 \mathrm{~K}^{-4}$, constant $b=2898 \mu \mathrm{mK}$ )
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The correct answer is:
$128$
$\lambda_m=289.8 \mathrm{~nm}=289.8 \times 10^{-9} \mathrm{~m}$
From Wein's law,
$\begin{aligned} & \lambda_m T=b \quad \text { (constant) } \\ & \text { Temperature of star, } T=\frac{b}{\lambda_m}=\frac{2898 \times 10^{-6}}{289.8 \times 10^{-9}} \\ & =\frac{2898 \times 10^{-6}}{2898 \times 10^{-10}} \\ & =10^4 \mathrm{~K} \\ & \text { Radiation intensity } E=\sigma T^4 \\ & =5.67 \times 10^{-8}\left(10^4\right)^4 \\ & =5.67 \times 10^{-8} \times 10^{16} \\ & =5.67 \times 10^8 \mathrm{~W} / \mathrm{m}^2 \\ & \end{aligned}$
From Wein's law,
$\begin{aligned} & \lambda_m T=b \quad \text { (constant) } \\ & \text { Temperature of star, } T=\frac{b}{\lambda_m}=\frac{2898 \times 10^{-6}}{289.8 \times 10^{-9}} \\ & =\frac{2898 \times 10^{-6}}{2898 \times 10^{-10}} \\ & =10^4 \mathrm{~K} \\ & \text { Radiation intensity } E=\sigma T^4 \\ & =5.67 \times 10^{-8}\left(10^4\right)^4 \\ & =5.67 \times 10^{-8} \times 10^{16} \\ & =5.67 \times 10^8 \mathrm{~W} / \mathrm{m}^2 \\ & \end{aligned}$
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