A black body is at a temperature of 2880;K. T | vedclass

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(D) According to Wien's displacement law,the wavelength corresponding to the maximum spectral emissive power is given by ${\lambda _m}T = b$.Substitut

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${U_2} > {U_1}$

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(D) According to Wien's displacement law,the wavelength corresponding to the maximum spectral emissive power is given by ${\lambda _m}T = b$.Substituting the given values,we get ${\lambda _m} = \frac{b}{T} = \frac{2.88 	imes {10^6}}{2880} = 1000\;nm$.This means the peak of the black body radiation curve occurs at $1000\;nm$.The energy $U$ emitted in a small wavelength interval $\Delta \lambda$ is proportional to the area under the $E_{\lambda} - \lambda$ curve in that interval.Since the peak of the curve is at $1000\;nm$,the spectral emissive power $E_{\lambda}$ is maximum at $1000\;nm$.Comparing the intervals,the interval for ${U_2}$ ($999\;nm$ to $1000\;nm$) is closest to the peak wavelength,while the interval for ${U_1}$ ($499\;nm$ to $500\;nm$) is further away.Therefore,the area under the curve for ${U_2}$ is greater than the area for ${U_1}$,implying ${U_2} > {U_1}$.

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