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A blackbody is at a temperature of 2880...

A blackbody is at a temperature of 2880K. The energy of radiation emitted by this object with wavelength between 499nm and 500nm is `U_1`, between 999nm and 1000nm is `U_2` and between 1499 nm and 1500 nm is `U_3`. The Wien constant `b=2.88xx10^6nmK`. Then

A

`U_1=0`

B

`U_2=0`

C

`U_1=U_2`

D

`U_2gtU_1`

Text Solution

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The correct Answer is:
D

From Wien's law, `lamda_m=T=` constant, where T is the temperature of black body and `lamda_m` is the wavelength corresponding to maximum energy of emission. Energy distribution of black body radiation given below: . `U_1` and `U_2` are not zero because a black body emits radiation of nearly all wavelengths.
ii. Since `U_1` corresponding to lower wavelength `U_3` corresponds to higher wavelength and `U_2` corresponds to medium wave length hence `U_2gtU_1`.
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