A black body has a maximum wavelength lambda_ | vedclass

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(B) According to Wien's displacement law,the product of the wavelength of maximum emission and the absolute temperature is a constant:$\lambda_m T = \

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$\frac{2}{3} \lambda_m$

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(B) According to Wien's displacement law,the product of the wavelength of maximum emission and the absolute temperature is a constant:$\lambda_m T = 	ext{constant}$Therefore,$\lambda_m \propto \frac{1}{T}$.Given $T_1 = 2000 \ K$ and $T_2 = 3000 \ K$,and the initial wavelength is $\lambda_m$:$\frac{\lambda_2}{\lambda_m} = \frac{T_1}{T_2}$$\frac{\lambda_2}{\lambda_m} = \frac{2000}{3000} = \frac{2}{3}$$\lambda_2 = \frac{2}{3} \lambda_m$Thus,the corresponding wavelength at $3000 \ K$ is $\frac{2}{3} \lambda_m$.

$A$ black body has a maximum wavelength $\lambda_{m}$ at a temperature of $2000 \ K$. Its corresponding wavelength at a temperature of $3000 \ K$ will be:

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