AP 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
- A \(\frac{1}{2} \mu \mathrm{m}\)
- B \(1 \mu \mathrm{m}\)
- C \(2 \mu \mathrm{m}\)
- D \(\frac{3}{2} \mu \mathrm{m}\)
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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