Two bodies P and Q have thermal emissivities of 0.04 and 0.64 and equal outer surface areas. They emit total radiant power at the same rate. The wavelength of maximum spectral radiance in the radiatio

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Two bodies P and Q have thermal emissivities of 0.04 and 0.64 and equal outer surface areas. They emit total radiant power at the same rate. The wavelength of maximum spectral radiance in the radiation from Q is longer than that from P by 1.2 μ m. The temperature of P is 3000 K. Then:

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ϵ P ​ T P 4 ​ = ϵ Q ​ T Q 4 ​ gives T Q ​ = 3000 × ( 0.64 0.04 ​ ) to the power one quarter = 3000 × 2 1 ​ = 1500 K. Wien: λ Q ​ T Q ​ = λ P ​ T P ​ , so λ Q ​ = 2 λ P ​ . With λ Q ​ − λ P ​ = 1.2 μ m, λ P ​ = 1.2 μ m and λ Q ​ = 2.4 μ m. Options A and B.

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Options:(select one or more)

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