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A black body is at a temperature of 2880...

A black body is at a temperature of 2880 K. The energy of radiation emitted by this object with wavelength between 499 nm and 500 nm is`U_(1)`, between 999 nm and 1000 nm is `U_(2)` and between 1499 nm and 1500 nm is `U_(3)`. The Wein's constant `b = 2.88 xx 10^(6) "nm K"`. Then

A

`U_(1)=0`

B

`U_(3)=0`

C

`U_(1)gtU_(2)`

D

`U_(2)gtU_(1)`

Text Solution

Verified by Experts

The correct Answer is:
D

Wien’s displacement law is `lambda_(m)T=b` (b = Wien’s constant)
`lambda_(m)=(b)/(T)=(2.88xx10^(6)nm-K)/(2880K)`
`lambda=1000nm`
Energy distribution with wavelength will be as follows: From the graph it is clear that
`U_(2)gtU_(1)` (in fact `U_(2)` is maximum)
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