    Question

# 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 λB corresponding to maximum spectral radiancy in the radiation from B shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A, by 1.00 μm. If the temperature of A is 5802 K

A

the temperature of B is 1934 K

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B

λB=1.5μm

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C

the temperature of B is 11604 K

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D

the temperature of B is 2901 K

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Solution

## The correct options are A the temperature of B is 1934 K B λB=1.5μm Power radiated and surface area is same for both A and B . Therefore, eAσT4AA=eBσT4BA ∴TATB=(eBeB)14=(0.810.01)14=3 thereforeTB=TA3=58023 = 1934 K TB=1934K According to Wien’ s displacement law, λmT=constant ∴λATA=λBTB or λATB=λB(TBTA)=λB3 Given, λB−λA=1μm ⇒λB−λB3=1μm or 23λB=1μm ⇒λB=1.5μm Note: λmT=B = Wien’s constant value of this constant for perfectly black body is 2.89×103 m-K. For other bodies this constant will have some different value. In the option (b) has different for different bodies. Option (b) is incorrect.  Suggest Corrections  0      Related Videos     