Question 838: Thermodynamics & KTG

Question

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

Accepted Answer

$$	herefore \quad$$ Emissive power of black body is, From Wien's Law:
$$\begin{aligned}
\mathrm{E} & =\sigma \mathrm{T}^4 \
\lambda & =\frac{\mathrm{b}}{\mathrm{T}} \
	herefore \quad \mathrm{E} & =\frac{\sigma \mathrm{b}^4}{\lambda^4} \
	herefore \quad \frac{\mathrm{E}_2}{\mathrm{E}_1} & =\frac{\lambda_1^4}{\lambda_2^4}=\frac{\lambda^4}{\left(\frac{\lambda}{3}ight)^4}=\frac{81}{1}
\end{aligned}$$

Answer

$$16: 1$$

Answer

$$256: 1$$

Answer

$$81: 1$$

Answer

$$128: 1$$

Thermodynamics & KTG

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

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