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A black body radiates poer P and maximum...

A black body radiates poer P and maximum energy is radiated by it around a wavelength `lambda_(0)`. The temperature of the black body is now changed such that it radiates maximum energy around the wavelength `(3lambda_(0))/4`. The power radiated by it now is

A

`(256)/(81)`

B

`4/3`

C

`3/4`

D

`(81)/(256)`

Text Solution

Verified by Experts

The correct Answer is:
A
We know ,
`lambda_("max")=T`= constant (Wien's law)
So, `lambda_("max"_(1)), T_(1)=lambda_("max"_(2))T_(2)`
`implies lambda_(0) T=(3lambda_(0))/(4)T'`
`implies T'=(4)/(3)T`
So, `(P_(2))/(P_(1))=((T')/(T))^(4)=((4)/(3))^(4)=(256)/(81)`
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