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(A) Given: $\lambda_m = 289.8 \ nm = 289.8 	imes 10^{-9} \ m$.Wien's constant $b = 2898 \ \mu m \ K = 2898 	imes 10^{-6} \ m \ K$.Stefan's constant

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

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(A) Given: $\lambda_m = 289.8 \ nm = 289.8 	imes 10^{-9} \ m$.Wien's constant $b = 2898 \ \mu m \ K = 2898 	imes 10^{-6} \ m \ K$.Stefan's constant $\sigma = 5.67 	imes 10^{-8} \ W m^{-2} K^{-4}$.From Wien's displacement law,$\lambda_m T = b$.Therefore,the temperature of the star is $T = \frac{b}{\lambda_m} = \frac{2898 	imes 10^{-6}}{289.8 	imes 10^{-9}} = \frac{2898 	imes 10^{-6}}{289.8 	imes 10^{-9}} = 10^4 \ K$.The radiation intensity (emissive power) $E$ is given by Stefan-Boltzmann law: $E = \sigma T^4$.$E = (5.67 	imes 10^{-8}) 	imes (10^4)^4$.$E = 5.67 	imes 10^{-8} 	imes 10^{16}$.$E = 5.67 	imes 10^8 \ W/m^2$.

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 $= 5.67 \times 10^{-8} \ W m^{-2} K^{-4}$,Wien's constant $b = 2898 \ \mu m \ K$).

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