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A black body radiates maximum energy at wavelength ' $\lambda$ ' and its emissive power is ' $E$ '. Now due to a change in temperature of that body, it radiates maximum energy at wavelength $\frac{\lambda}{3}$. At that temperature emissive power is
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The correct answer is:
$81: 1$
$\therefore \quad$ Emissive power of black body is, From Wien's Law:
$$
\begin{aligned}
\mathrm{E} & =\sigma \mathrm{T}^4 \\
\lambda & =\frac{\mathrm{b}}{\mathrm{T}} \\
\therefore \quad \mathrm{E} & =\frac{\sigma \mathrm{b}^4}{\lambda^4} \\
\therefore \quad \frac{\mathrm{E}_2}{\mathrm{E}_1} & =\frac{\lambda_1^4}{\lambda_2^4}=\frac{\lambda^4}{\left(\frac{\lambda}{3}\right)^4}=\frac{81}{1}
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{E} & =\sigma \mathrm{T}^4 \\
\lambda & =\frac{\mathrm{b}}{\mathrm{T}} \\
\therefore \quad \mathrm{E} & =\frac{\sigma \mathrm{b}^4}{\lambda^4} \\
\therefore \quad \frac{\mathrm{E}_2}{\mathrm{E}_1} & =\frac{\lambda_1^4}{\lambda_2^4}=\frac{\lambda^4}{\left(\frac{\lambda}{3}\right)^4}=\frac{81}{1}
\end{aligned}
$$
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