The distribution of relative intensity I(lamb | vedclass

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(D) According to Wien's displacement law,the product of the wavelength corresponding to maximum intensity $(\lambda_m)$ and the absolute temperature $

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$1500$

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(D) According to Wien's displacement law,the product of the wavelength corresponding to maximum intensity $(\lambda_m)$ and the absolute temperature $(T)$ of the blackbody is a constant,known as Wien's constant $(b)$.$\lambda_m T = b$From the given graph,the peak intensity occurs at $\lambda_m = 2 \ \mu m = 2 	imes 10^{-6} \ m$.The Wien displacement law constant is given as $b = 2.9 	imes 10^{-3} \ mK$.Substituting these values into the formula:$T = \frac{b}{\lambda_m} = \frac{2.9 	imes 10^{-3} \ mK}{2 	imes 10^{-6} \ m}$$T = 1.45 	imes 10^3 \ K = 1450 \ K$.Rounding to the nearest given option,the approximate temperature is $1500 \ K$.

The distribution of relative intensity $I(\lambda)$ of blackbody radiation from a solid object versus the wavelength $\lambda$ is shown in the figure. If the Wien displacement law constant is $2.9 \times 10^{-3} \ mK$,what is the approximate temperature of the object in $K$?

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