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 radiate energy at the same rate. The wavelength $λ_{B}$ , corresponding to the maximum spectral radiancy in the radiation from B, is shifted from the wavelength corresponding to the maximum spectral radiancy in the radiation from A by 1.00 mm. If the temperature of A is 5802 K, then

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the temperature of B is 1934 K

$λ_{B} = 1.5 m m$

the temperature of B is 11604 K

the temperature of B is 2901 K

answer is A, B.

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Detailed Solution

$P = e σ A T^{4} e T^{4} = c o n s t, T_{A} = 5802 k 0.01 \times 5802^{4} = 0.81 T_{B}^{4} T_{B} = {(\frac{0.01 \times 5802^{4}}{0.81})}^{\frac{1}{4}} = {(\frac{0.01}{0.81})}^{\frac{1}{4}} = {(\frac{0.01}{0.81})}^{\frac{1}{4}} \times 5802 T_{B} = 1934 K$

$A s T_{A} > T_{B} = λ_{A} < λ_{B} a s λ T = c o n s t λ_{B} - λ_{A} = 1 m m λ_{A} = λ_{B} - 1 \Rightarrow λ_{A} T_{A} = λ_{B} T_{B} (λ_{B} - 1) T_{A} = λ_{B} T_{B} λ_{B} T_{A} - T_{A} = λ_{B} T_{B} \Rightarrow λ_{B} (T_{A} - T_{B}) = T_{A} λ_{B} = \frac{T_{A}}{T_{A} - T_{B}} = \frac{5802}{5802 - 1920} = \frac{5802}{3882} ≃ 1.5 m m$