MHT-CET Physics and Chemistry Full Test-5 Que-4 Solution

Q.4. 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

A. \frac{81}{16}

B. \frac{27}{32}

C. \frac{18}{10}

D. \frac{9}{4}

Answer: A

Explanation:-

From Stefan-Boltzmann’s law,

\mathrm{P}=\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{A} \sigma \mathrm{T}^{4}

Also, from Wien’s displacement law,

\lambda_{\max }=\frac{\mathrm{b}}{\mathrm{T}}(\mathrm{b} \rightarrow Wien’s constant )

\Rightarrow \mathrm{T}=\frac{\mathrm{b}}{\lambda}

\therefore \quad \mathrm{P}=\mathrm{A} \cdot \sigma\left(\frac{\mathrm{b}}{\lambda}\right)^{4} \Rightarrow \mathrm{P} \propto \frac{1}{(\lambda)^{4}}

\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 \lambda_{2}=\frac{2 \lambda}{3}

\therefore \quad \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{(\lambda)^{4}}{\left(\frac{2 \lambda}{3}\right)^{4}}=\frac{81}{16}