A black body is at a temperature of 5760 K. | vedclass

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(C) According to Wien's displacement law,the wavelength corresponding to maximum spectral emissive power is given by:$\lambda_m = \frac{b}{T} = \frac{

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$U_2>U_1$

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(C) According to Wien's displacement law,the wavelength corresponding to maximum spectral emissive power is given by:$\lambda_m = \frac{b}{T} = \frac{2.88 	imes 10^6 \ nm \ K}{5760 \ K} = 500 \ nm$This means the peak of the black body radiation curve occurs at $\lambda = 500 \ nm$,where the energy emitted is $U_2$.Comparing the values at different wavelengths from the spectral distribution curve:At $\lambda = 250 \ nm$,the energy is $U_1$.At $\lambda = 500 \ nm$,the energy is $U_2$ (maximum).At $\lambda = 1000 \ nm$,the energy is $U_3$.From the graph,it is clear that $U_2$ is the maximum value,so $U_2 > U_1$ and $U_2 > U_3$. Also,comparing $U_1$ and $U_3$ from the graph,$U_3 > U_1$. Thus,the correct relation is $U_2 > U_1$.

$A$ black body is at a temperature of $5760 \ K$. The energy of radiation emitted by the body at wavelength $250 \ nm$ is $U_1$,at wavelength $500 \ nm$ is $U_2$ and that at $1000 \ nm$ is $U_3$. Wien's constant,$b = 2.88 \times 10^6 \ nm \ K$. Which of the following is correct?

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