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(D) According to Stefan-Boltzmann law,the power radiated is $P = e A \sigma T^4$. Since the bodies have equal surface area and radiate at the same rat

Vedclass · Accepted Answer

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

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(D) According to Stefan-Boltzmann law,the power radiated is $P = e A \sigma T^4$. Since the bodies have equal surface area and radiate at the same rate $(P_A = P_B)$,we have $e_A T_A^4 = e_B T_B^4$.Given $e_A = 0.01$,$e_B = 0.81$,and $T_A = 5802 \ K$.Substituting the values: $0.01 	imes (5802)^4 = 0.81 	imes T_B^4$.Taking the fourth root on both sides: $T_B = T_A 	imes (0.01 / 0.81)^{1/4} = 5802 	imes (1/81)^{1/4} = 5802 / 3 = 1934 \ K$.From Wien's displacement law,$\lambda_A T_A = \lambda_B T_B = b$ (where $b \approx 2898 \ \mu m \cdot K$).Thus,$\lambda_A = b / T_A = 2898 / 5802 \approx 0.5 \ \mu m$.Given $\lambda_B - \lambda_A = 1 \ \mu m$,so $\lambda_B = 1 + 0.5 = 1.5 \ \mu m = \frac{3}{2} \ \mu m$.

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