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

$81/16$

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, $\lambda_1 = \lambda$ and $T_1 = T$.Finally, $\lambda_2 = 2\lambda/3$ and $T_2 = T'$.Using $\lambda_1 T_1 = \lambda_2 T_2$, we get $\lambda T = (2\lambda/3) T'$, which implies $T' = 3T/2$.The power radiated by a black body is given by the Stefan-Boltzmann law, $P = \sigma A T^4$, where $\sigma$ is the Stefan-Boltzmann constant and $A$ is the surface area.Initially, $P_1 = \sigma A T^4$.Finally, $P_2 = \sigma A (T')^4 = \sigma A (3T/2)^4 = \sigma A (81/16) T^4$.Therefore, the ratio of the power radiated is $P_2/P_1 = (81/16) (\sigma A T^4) / (\sigma A T^4) = 81/16$.The power radiated increases by a factor of $81/16$.

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

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