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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_3=0`

C

`U_1gtU_2`

D

`U_2gtU_1`

Text Solution

Verified by Experts

The correct Answer is:
D
(d) According to wein's
displacement law,
`lambda_mT=2.88xx10^6nmK`
The wavelength at the
peak of the spectrum
becomes
`lambda_m=(2.88xx10^6nmK)/(2880K)=10^3nm`
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