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

A

256/81256/81

B

43\frac{4}{3}

C

34\frac{3}{4}

D

81256\frac{81}{256}

Step-by-Step Solution

λmaxT=constant(Wien’s law)\lambda_{\text{max}} T = \text{constant} \quad \text{(Wien's law)}

So, λmax1T1=λmax2T2\textbf{So, } \lambda_{\text{max}_1} T_1 = \lambda_{\text{max}_2} T_2

⇒λ0T=3λ04T′\Rightarrow \lambda_0 T = \frac{3\lambda_0}{4} T'

⇒T′=43T\Rightarrow T' = \frac{4}{3}T

So, P2P1=(T′T)4=(43)4=25681\textbf{So, } \frac{P_2}{P_1} = \left(\frac{T'}{T}\right)^4 = \left(\frac{4}{3}\right)^4 = \frac{256}{81}

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