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AP EAMCET2013PhysicsThermal Properties of Matter

Two bodies A and B of equal surface area have thermal emissivities of 0.01 and 0.81 respectively. The two bodies are radiating energy at the same rate. Maximum energy is radiated from the two bodies A and B at wavelengths _A and _B respectively. Difference in these two wavelengths is 1 m . If the temperature of the body A is 5802 ~K , then value of _B is

Options

  1. A1 2 m
  2. B1 m
  3. C2 m
  4. D3 2 m

Correct answer

D. 3 2 m

Step-by-step solution

We knows from Stefan's law, Here, E=e A T^4 E₁=e₁ A T₁^4 E₂=e₂ A T₂^4 so, E₁=E₂ array lll & e₁ T₁^4=e₂ T₂^4 & T₂= ( e₁ e₂ T₁^4 )^ 1 / 4 = ( 1 81 (5802)^4 )^ 1 / 4 & T_B=1934 ~K array From Wein's law, _ A T_A = _B T_B array llrl & _A _B & = T_B T_A & _B- _A _B & = T_B-T_B T_A & 1 _B & = 5802-1934 5802 = 3968 5802 & _B & = 3 2 m array

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