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STANDARD Physics Black Body Radiation MCQ Question

Type: MCQ-conceptual-Hard-Class 11

The power radiated by a black body is P and it radiates maximum energy at wavelength, λ₀. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength 34λ0\frac{3}{4} \lambda_0, the power radiated by it becomes nP. The value of n is

A

34\frac{3}{4}

B

43\frac{4}{3}

C

25681\frac{256}{81}

D

81256\frac{81}{256}

Correct Answer

Option C

Detailed Explanation

From Wien's law, λₘₐₓT = constant. So, λ0T=34λ0T′\lambda_0 T = \frac{3}{4} \lambda_0 T' implies T′=43TT' = \frac{4}{3} T. According to Stefan-Boltzmann law, P ∝ T⁴. So, P2P1=(T′T)4=(43)4=25681\frac{P_2}{P_1} = \left( \frac{T'}{T} \right)^4 = \left( \frac{4}{3} \right)^4 = \frac{256}{81}.

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