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(A) According to Wien's displacement law, $\lambda_m T = 	ext{constant}$.Let the initial temperature be $T_1$ and the final temperature be $T_2$. Giv

Vedclass · Accepted Answer

$256/81$

Vedclass · Answer

(A) According to Wien's displacement law, $\lambda_m T = 	ext{constant}$.Let the initial temperature be $T_1$ and the final temperature be $T_2$. Given $\lambda_{m1} = \lambda_0$ and $\lambda_{m2} = 3\lambda_0/4$.Using $\lambda_{m1} T_1 = \lambda_{m2} T_2$, we get:$T_2 = \frac{\lambda_{m1}}{\lambda_{m2}} T_1 = \frac{\lambda_0}{3\lambda_0/4} T_1 = \frac{4}{3} T_1$.According to the Stefan-Boltzmann law, the power radiated $P \propto T^4$.Therefore, the ratio of the powers is:$\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} ight)^4 = \left( \frac{4/3 T_1}{T_1} ight)^4 = \left( \frac{4}{3} ight)^4 = \frac{256}{81}$.

The energy spectrum of a black body has maximum energy at a wavelength of $\lambda_0$. Now, the temperature of the black body is increased such that the maximum energy is obtained at a wavelength of $3\lambda_0/4$. By what factor will the power radiated by the black body increase?

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