A black body at a temperature of 3000 K cool | vedclass

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(A) According to Wien's displacement law,$\lambda_m T = b$,where $b = 3 	imes 10^{-3} \ m \cdot K$.Initially,$T_1 = 3000 \ K$. The initial wavelength

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$300$

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(A) According to Wien's displacement law,$\lambda_m T = b$,where $b = 3 	imes 10^{-3} \ m \cdot K$.Initially,$T_1 = 3000 \ K$. The initial wavelength is $\lambda_1 = \frac{b}{T_1} = \frac{3 	imes 10^{-3}}{3000} = 1 	imes 10^{-6} \ m = 1 \ \mu m$.As the body cools,the wavelength shifts by $\Delta \lambda = 9 \ \mu m$. Since the temperature decreases,the wavelength of maximum energy density increases: $\lambda_2 = \lambda_1 + \Delta \lambda = 1 \ \mu m + 9 \ \mu m = 10 \ \mu m = 10 	imes 10^{-6} \ m$.Now,calculate the new temperature $T_2$: $T_2 = \frac{b}{\lambda_2} = \frac{3 	imes 10^{-3}}{10 	imes 10^{-6}} = \frac{3 	imes 10^{-3}}{10^{-5}} = 300 \ K$.

$A$ black body at a temperature of $3000 \ K$ cools down. The wavelength corresponding to maximum energy density shifts by $\Delta \lambda = 9 \ \mu m$. What is the new temperature of the black body (in $K$)? (Given $b = 3 \times 10^{-3} \ m \cdot K$)

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