Question 862: Thermodynamics & KTG

Question

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

Accepted Answer

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

Answer

$$\frac{81}{16}$$

Answer

$$\frac{27}{32}$$

Answer

$$\frac{18}{10}$$

Answer

$$\frac{9}{4}$$

Thermodynamics & KTG

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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