Question 869: Thermodynamics & KTG

Question

A black body radiates maximum energy at wavelength '$$\lambda$$' and its emissive power is $$\mathrm{E}$$. 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 Wien's Displacement law,
$$\lambda_{\max }=\frac{b}{T} \Rightarrow T=\frac{b}{\lambda_{\max }}$$
From Stefan-Boltzmann law
$$E=\sigma T^4=\sigma\left(\frac{b}{\lambda_{\max }}ight)^4$$
Let the new emissive power be $$\mathrm{E}^{\prime}$$.
$$\begin{aligned}
	herefore \quad & E^{\prime}=\sigma\left(\frac{b}{\frac{2 \lambda_{\max }}{3}}ight)^4 \
& E^{\prime}=\frac{81}{16} E
\end{aligned}$$

Answer

$$\frac{51 \mathrm{E}}{8}$$

Answer

$$\frac{81 \mathrm{E}}{16}$$

Answer

$$\frac{61 E}{27}$$

Answer

$$\frac{71 \mathrm{E}}{19}$$

Thermodynamics & KTG

A black body radiates maximum energy at wavelength '$\lambda' and its emissive power is \mathrm{E}. Now due to change in temperature of that body, it radiates maximum energy at wavelength \frac{2 \lambda}{3}$. At that temperature emissive power is

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