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 at which the energy density is maximum

Vedclass · Accepted Answer

$\frac{16}{81}$

Vedclass · Answer

(B) According to Wien's displacement law, $\lambda_m T = 	ext{constant}$, where $\lambda_m$ is the wavelength at which the energy density is maximum and $T$ is the absolute temperature.Initially, $\lambda_1 = \lambda_0$, so $T_1 = \frac{C}{\lambda_0}$.Finally, $\lambda_2 = \frac{3\lambda_0}{2}$, so $T_2 = \frac{C}{\lambda_2} = \frac{C}{3\lambda_0/2} = \frac{2C}{3\lambda_0} = \frac{2}{3} T_1$.The power radiated by a black body is given by the Stefan-Boltzmann law, $P = \sigma A T^4$.The ratio of power radiated is $\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} ight)^4$.Substituting the values, $\frac{P_2}{P_1} = \left( \frac{2/3 T_1}{T_1} ight)^4 = \left( \frac{2}{3} ight)^4 = \frac{16}{81}$.

The energy spectrum of a black body exhibits a maximum of wavelength $\lambda_0$. The temperature of the black body is now changed such that the energy is maximum at wavelength $\frac{3\lambda_0}{2}$. The ratio of power radiated by the black body will be

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