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A black body radiates maximum energy at wavelength ' $\lambda$ ' and its emissive power is $E$. Now due to change in temperature of that body, it radiates maximum energy at wavelength $\frac{2 \lambda}{3}$. At that temperature emissive power is
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Verified Answer
The correct answer is:
$\frac{81 \mathrm{E}}{16}$
From Wien's Displacement law,
$\lambda_{\max }=\frac{\mathrm{b}}{\mathrm{T}} \Rightarrow \mathrm{T}=\frac{\mathrm{b}}{\lambda_{\max }}$
From Stefan-Boltzmann law
$\mathrm{E}=\sigma \mathrm{T}^4=\sigma\left(\frac{\mathrm{b}}{\lambda_{\max }}\right)^4$
Let the new emissive power be $\mathrm{E}^{\prime}$.
$\begin{aligned}
\therefore \quad E^{\prime} & =\sigma\left(\frac{b}{\frac{2 \lambda_{\max }}{3}}\right)^4 \\
\mathrm{E}^{\prime} & =\frac{81}{16} \mathrm{E}
\end{aligned}$
$\lambda_{\max }=\frac{\mathrm{b}}{\mathrm{T}} \Rightarrow \mathrm{T}=\frac{\mathrm{b}}{\lambda_{\max }}$
From Stefan-Boltzmann law
$\mathrm{E}=\sigma \mathrm{T}^4=\sigma\left(\frac{\mathrm{b}}{\lambda_{\max }}\right)^4$
Let the new emissive power be $\mathrm{E}^{\prime}$.
$\begin{aligned}
\therefore \quad E^{\prime} & =\sigma\left(\frac{b}{\frac{2 \lambda_{\max }}{3}}\right)^4 \\
\mathrm{E}^{\prime} & =\frac{81}{16} \mathrm{E}
\end{aligned}$
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