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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)=U_(2)`

D

`U_(2)gtU_(1)`

Text Solution

Verified by Experts

From Wien.s Law `lamda_(m)T=` constant where T is the temperature of black body and `lamda_(m)` is the wavelength corresponding to the maximum energy of emission
Energy distribution of blackbody radiation is given below:
1. `U_(1)` and `U_(2)` are not zero because a blackbody radiates at all wavelengths.
2. Since `U_(1)` corresponds lower wavelength and `U_(3)` corresponding higher wavelength and `U_(2)` corresponds meidum wavelength. Thus `U_(2)gtU_(1)`
Hence d is correct.
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