A black body radiates maximum energy at wavelength ' $\lambda$ and its emissive power is ' $E$ ' Now due to…
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
$\frac{81}{16}$
$\frac{27}{32}$
$\frac{18}{10}$
$\frac{9}{4}$
Solution
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 $\dot{\lambda}_2=\frac{2 \lambda}{3}$.
$\therefore \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{(\lambda)^4}{\left(\frac{2 \lambda}{3}\right)^4}=\frac{81}{16}$