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 um. If the temperature of A is 5820 K, then

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$λ_{B} = 1 .5 μm$

the temperature of B is 1934 K

the temperature of B is 11604 K

the temperature of B is 2901 K

answer is A, B.

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Detailed Solution

According to Stefan's law

$\begin{array}{l} E = {eAσT}^{4} \Rightarrow E_{A} = e_{A} {AσT}_{A}^{4} and E_{B} = e_{B} {AσT}_{B}^{4} \\ ∵ E_{A} = E_{B} ∴ e_{A} T_{A}^{4} = e_{B} T_{B}^{4} \\ \Rightarrow T_{B} = {(\frac{e_{A}}{e_{B}} T_{A}^{4})}^{\frac{1}{4}} = {(\frac{1}{81} \times (5802)^{4})}^{\frac{1}{4}} \Rightarrow T_{B} = 1934 K \end{array}$

And, from Wien's law

$\begin{array}{l} λ_{A} \times T_{A} = λ_{B} \times T_{B} \\ \Rightarrow \frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{T_{A}} \Rightarrow \frac{λ_{B} - λ_{A}}{λ_{B}} = \frac{T_{A} - T_{B}}{T_{A}} \\ \Rightarrow \frac{1}{λ_{B}} = \frac{5802 - 1934}{5802} = \frac{3968}{5802} \Rightarrow λ_{B} = 1 .5 μm \end{array}$