A black body radiates maximum energy at wavel | vedclass

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(B) According to Wien's Displacement Law,$\lambda_{\max} T = b$,where $b$ is Wien's constant. Therefore,$T = \frac{b}{\lambda_{\max}}$.According to th

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$\frac{81 E}{16}$

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(B) According to Wien's Displacement Law,$\lambda_{\max} T = b$,where $b$ is Wien's constant. Therefore,$T = \frac{b}{\lambda_{\max}}$.According to the Stefan-Boltzmann Law,the total emissive power $E$ of a black body is proportional to the fourth power of its absolute temperature: $E = \sigma T^4$.Substituting the expression for $T$,we get $E = \sigma \left( \frac{b}{\lambda_{\max}} \right)^4$.Let the initial wavelength be $\lambda_1 = \lambda$ and the final wavelength be $\lambda_2 = \frac{2 \lambda}{3}$.Since $E \propto \frac{1}{\lambda_{\max}^4}$,we can write the ratio of the emissive powers as:$\frac{E'}{E} = \left( \frac{\lambda_1}{\lambda_2} \right)^4$.Substituting the values: $\frac{E'}{E} = \left( \frac{\lambda}{\frac{2 \lambda}{3}} \right)^4 = \left( \frac{3}{2} \right)^4$.Calculating the value: $\frac{E'}{E} = \frac{81}{16}$.Therefore,the new emissive power is $E' = \frac{81}{16} E$.

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

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