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

From Stefan-Boltzmann's law,
\(\mathrm{P}=\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{A \sigma T}^4\)
Also, from Wien's displacement law,
\(\begin{aligned}
& \lambda_{\max }=\frac{\mathrm{b}}{\mathrm{T}}(\mathrm{b} \rightarrow \text { Wien's constant }) \\
& \Rightarrow \mathrm{T}=\frac{\mathrm{b}}{\lambda}
\end{aligned}\)
\(\begin{aligned}
\therefore \quad P & =A \cdot \sigma\left(\frac{b}{\lambda}\right)^4 \\
\Rightarrow P & \propto \frac{1}{(\lambda)^4}
\end{aligned}\)
\(\therefore \quad\) Ratio of power dissipated is \(\frac{\mathrm{P}_2}{\mathrm{P}_1}=\left(\frac{\lambda_1}{\lambda_2}\right)^4\) Given \(\lambda_1=\lambda\) and \(\dot{\lambda}_2=\frac{2 \lambda}{3}\).
\(\therefore \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{(\lambda)^4}{\left(\frac{2 \lambda}{3}\right)^4}=\frac{81}{16}\)