The spectral emissive power E_lambda for a bo | vedclass

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(D) According to the Stefan-Boltzmann law,the total emissive power (which is the area under the spectral emissive power curve) is directly proportiona

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$\sqrt{3}$

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(D) According to the Stefan-Boltzmann law,the total emissive power (which is the area under the spectral emissive power curve) is directly proportional to the fourth power of the absolute temperature.$E = \sigma T^4$Given that the area $A \propto T^4$,we have:$A_1 = k T_1^4 = A$$A_2 = k T_2^4 = 9A$Dividing the two equations:$\frac{A_2}{A_1} = \left(\frac{T_2}{T_1}\right)^4 = 9$Taking the square root twice:$\left(\frac{T_2}{T_1}\right)^2 = 3$$\frac{T_2}{T_1} = \sqrt{3}$According to Wien's displacement law,the product of the wavelength corresponding to maximum emission and the absolute temperature is constant:$\lambda_1 T_1 = \lambda_2 T_2$Therefore,the ratio is:$\frac{\lambda_1}{\lambda_2} = \frac{T_2}{T_1} = \sqrt{3}$

The spectral emissive power $E_\lambda$ for a body at temperature $T_1$ is plotted against the wavelength and the area under the curve is found to be $A$. At a different temperature $T_2$,the area is found to be $9A$. Then $\lambda_1/\lambda_2 =$

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