A black body radiates maximum energy at wavel | vedclass

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Question

(A) According to Wien's displacement law,$\lambda_{\max} T = b$ (constant),which implies $T \propto \frac{1}{\lambda}$.According to Stefan-Boltzmann's

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$\frac{81}{16} E$

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(A) According to Wien's displacement law,$\lambda_{\max} T = b$ (constant),which implies $T \propto \frac{1}{\lambda}$.According to Stefan-Boltzmann's law,the emissive power $E$ (or power radiated per unit area) is given by $E = \sigma T^4$.Substituting $T \propto \frac{1}{\lambda}$ into the Stefan-Boltzmann law,we get $E \propto \left(\frac{1}{\lambda}\right)^4$,or $E \propto \lambda^{-4}$.Let the initial state be $(\lambda_1, E_1)$ and the final state be $(\lambda_2, E_2)$.Given $\lambda_1 = \lambda$ and $\lambda_2 = \frac{2\lambda}{3}$.Then,$\frac{E_2}{E_1} = \left(\frac{\lambda_1}{\lambda_2}\right)^4 = \left(\frac{\lambda}{\frac{2\lambda}{3}}\right)^4 = \left(\frac{3}{2}\right)^4 = \frac{81}{16}$.Therefore,$E_2 = \frac{81}{16} E$.

$A$ 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 new temperature,the emissive power is:

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