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TS EAMCET · Physics · Thermal 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 \(\lambda_A\) and \(\lambda_B\) respectively. Difference in these two wavelengths is \(1 \mu \mathrm{m}\). If the temperature of the body \(A\) is \(5802 \mathrm{~K}\), then value of \(\lambda_B\) is

  1. A \(\frac{1}{2} \mu \mathrm{m}\)
  2. B \(1 \mu \mathrm{m}\)
  3. C \(2 \mu \mathrm{m}\)
  4. D \(\frac{3}{2} \mu \mathrm{m}\)
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Answer & Solution

Correct Answer

(D) \(\frac{3}{2} \mu \mathrm{m}\)

Step-by-step Solution

Detailed explanation

We knows from Stefan's law, Here, \[ E=e A \sigma T^4 \] \[ E_1=e_1 A \sigma T_1^4 \] \[ E_2=e_2 A \sigma T_2^4 \] so, \[ E_1=E_2 \]…
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