The temperature of a black body is 2880 , K. | vedclass

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(D) According to Wien's displacement law,the wavelength corresponding to maximum spectral emissive power is given by $\lambda_m T = b$.Substituting th

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$U_2 > U_1$

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(D) According to Wien's displacement law,the wavelength corresponding to maximum spectral emissive power is given by $\lambda_m T = b$.Substituting the given values: $\lambda_m = \frac{b}{T} = \frac{2.88 	imes 10^6 \, nm \cdot K}{2880 \, K} = 1000 \, nm$.Since the peak of the black body radiation curve occurs at $\lambda_m = 1000 \, nm$,the spectral energy density $E_\lambda$ is maximum at this wavelength.The energy $U$ in a small wavelength interval $\Delta \lambda$ is proportional to $E_\lambda \Delta \lambda$. Since the intervals are equal $(\Delta \lambda = 1 \, nm)$,the energy is directly proportional to the height of the curve at that wavelength.Comparing the heights at the given intervals: the interval around $1000 \, nm$ (where $U_2$ is defined) is at the peak,while the intervals around $500 \, nm$ and $1500 \, nm$ are on the slopes.Thus,$U_2$ is the maximum,and $U_2 > U_1$.

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