Two bodies $A$ and $B$ have thermal emissivities of $0.01$ and $0.81$ respectively. The outer surface areas of the two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength $λ_{B}$ corresponding to maximum spectral radiancy in the radiation from $B$ is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from $A$ by $1.00 μ m$ . If the temperature of $A$ is $5802 K$ .

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The temperature of $B$ is $11604 K$

The temperature of $B$ is $2901 K$

$λ_{B} = 1.5 μ m$

The temperature of $B$ is $1934 K$

answer is A, B.

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Detailed Solution

$ε_{A} σ T_{A}^{4} = ε_{B} σ T_{B}^{4} \frac{T_{A}}{T_{B}} = {[\begin{matrix} ε_{B} \\ ε_{A} \end{matrix}]}^{\frac{1}{4}} = {[\begin{matrix} 0.81 \\ 0.01 \end{matrix}]}^{\frac{1}{4}} = 3 T_{B} = \frac{T_{A}}{3} = \frac{5802}{3} = 1934 K$

From Wein's displacement law,

$λ_{A} T_{A} = λ_{B} T_{B} \frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{T_{A}} = \frac{1934}{5802} = \frac{1}{3}$

Also $λ_{B} - λ_{A} = 1.00$

After solving above equations, we get

$λ_{B} = 1.5 μ m$

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