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(B) According to Wien's displacement law,the product of the wavelength of maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ is a const

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$58$

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(B) According to Wien's displacement law,the product of the wavelength of maximum emission $(\lambda_m)$ and the absolute temperature $(T)$ is a constant $(b)$.$\lambda_m T = b$Given:$T = 5 	imes 10^4 \ K$$b = 0.0029 \ mK$Substituting the values:$\lambda_m = \frac{b}{T} = \frac{0.0029}{5 	imes 10^4}$$\lambda_m = \frac{2.9 	imes 10^{-3}}{5 	imes 10^4} = 0.58 	imes 10^{-7} \ m$$\lambda_m = 58 	imes 10^{-9} \ m$Since $1 \ nm = 10^{-9} \ m$,we get:$\lambda_m = 58 \ nm$.

$A$ particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^4 \ K$. The wavelength in nanometers at which its radiation becomes maximum is $(b = 0.0029 \ mK)$.

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