The power radiated by a black body is P and i | vedclass

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(D) According to $Wien's$ displacement law,$\lambda_{\max} T = 	ext{constant}$.Let the initial temperature be $T$ and the final temperature be $T'$.G

Vedclass · Accepted Answer

$\frac{256}{81}$

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(D) According to $Wien's$ displacement law,$\lambda_{\max} T = 	ext{constant}$.Let the initial temperature be $T$ and the final temperature be $T'$.Given $\lambda_{\max, 1} = \lambda_0$ and $\lambda_{\max, 2} = \frac{3}{4}\lambda_0$.Using the law: $\lambda_0 T = \left(\frac{3}{4}\lambda_0ight) T' \Rightarrow T' = \frac{4}{3}T$.According to the $Stefan-Boltzmann$ law,the power radiated by a black body is $P = A \sigma T^4$,which implies $P \propto T^4$.Therefore,$\frac{P_2}{P_1} = \left(\frac{T'}{T}ight)^4$.Substituting the values: $n = \left(\frac{4/3 T}{T}ight)^4 = \left(\frac{4}{3}ight)^4 = \frac{256}{81}$.

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

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