The energy spectrum of a black body exhibits | vedclass

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Question

(B) According to Wien's displacement law, $\lambda_m T = 	ext{constant}$, where $\lambda_m$ is the wavelength corresponding to maximum energy and $T$

Vedclass · Accepted Answer

$81/16$

Vedclass · Answer

(B) According to Wien's displacement law, $\lambda_m T = 	ext{constant}$, where $\lambda_m$ is the wavelength corresponding to maximum energy and $T$ is the absolute temperature.Initially, $\lambda_1 = \lambda$, so $T_1 = \frac{C}{\lambda}$.Finally, $\lambda_2 = \frac{2\lambda}{3}$, so $T_2 = \frac{C}{2\lambda/3} = \frac{3C}{2\lambda} = \frac{3}{2} T_1$.The power radiated by a black body is given by the Stefan-Boltzmann law, $P = \sigma A T^4$.Therefore, the ratio of the new power $P_2$ to the initial power $P_1$ is $\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} ight)^4$.Substituting the values, $\frac{P_2}{P_1} = \left( \frac{3/2 T_1}{T_1} ight)^4 = \left( \frac{3}{2} ight)^4 = \frac{81}{16}$.

The energy spectrum of a black body exhibits a maximum around a wavelength '$\lambda$'. The temperature of the black body is now changed such that the energy is maximum around a wavelength $2\lambda/3$. The power radiated by the black body will now increase by a factor

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