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(D) Given: $\lambda_1 = 0.26 \mu m$,$\lambda_2 = 0.13 \mu m$. According to Wien's displacement law,$\lambda_m T = 	ext{constant}$. Therefore,$\lambda

Vedclass · Accepted Answer

$1:16$

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(D) Given: $\lambda_1 = 0.26 \mu m$,$\lambda_2 = 0.13 \mu m$. According to Wien's displacement law,$\lambda_m T = 	ext{constant}$. Therefore,$\lambda_1 T_1 = \lambda_2 T_2$. $\frac{T_1}{T_2} = \frac{\lambda_2}{\lambda_1} = \frac{0.13}{0.26} = \frac{1}{2}$. According to the Stefan-Boltzmann law,the total emissive power $E$ of a black body is proportional to the fourth power of its absolute temperature,i.e.,$E \propto T^4$. Thus,the ratio of emissive powers is $\frac{E_1}{E_2} = \left(\frac{T_1}{T_2}ight)^4$. Substituting the values,$\frac{E_1}{E_2} = \left(\frac{1}{2}ight)^4 = \frac{1}{16}$. Hence,the ratio is $1:16$.

When the temperature of a black body increases,it is observed that the wavelength corresponding to maximum energy changes from $0.26 \mu m$ to $0.13 \mu m$. The ratio of the emissive powers of the body at the respective temperatures is

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