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(D) According to Stefan-Boltzmann law,the rate of energy radiation is $E = e A \sigma T^4$. Since the surface areas $A$ and the rates of energy radiat

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$\frac{3}{2} \mu m$

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(D) According to Stefan-Boltzmann law,the rate of energy radiation is $E = e A \sigma T^4$. Since the surface areas $A$ and the rates of energy radiation $E$ are equal for both bodies,we have $e_A T_A^4 = e_B T_B^4$.Given $e_A = 0.01$,$e_B = 0.81$,and $T_A = 5802 \ K$,we get:$T_B^4 = \frac{e_A}{e_B} T_A^4 = \frac{0.01}{0.81} (5802)^4 = \frac{1}{81} (5802)^4$.Taking the fourth root,$T_B = \frac{5802}{3} = 1934 \ K$.According to Wien's displacement law,$\lambda_A T_A = \lambda_B T_B = b$ (constant).Thus,$\lambda_A = \lambda_B \frac{T_B}{T_A} = \lambda_B \frac{1934}{5802} = \frac{\lambda_B}{3}$.Given the difference $\lambda_B - \lambda_A = 1 \mu m$,we substitute $\lambda_A$:$\lambda_B - \frac{\lambda_B}{3} = 1 \mu m \Rightarrow \frac{2}{3} \lambda_B = 1 \mu m$.Therefore,$\lambda_B = \frac{3}{2} \mu m$.

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