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The power radiated by a black body A is ...

The power radiated by a black body A is `E_(A)` and the maximum energy radiated was at the wavelength `lambda_(A)`.The power radiated by another black body B is `E_(B) = NE_(A)` and the radiated energy was at the maximum wavelength, `(3)/(4)lambda_(A)`. What is the value of N?

Text Solution

Verified by Experts

According to Wien.s displacement law
`lambda_("max")T="constant for both object A and B"`
`lambda_(A)T_(A)=lambda_(B)T_(B)`
`"Here "lambda_(B)=(1)/(2)lambda_(A)`
`(T_(B))/(T_(A))=(lambda_(A))/(lambda_(B))=(1)/((1)/(2))=2`
`T_(B)=2T_(A)`
From Stefan - Boltzmann law
`(E_(B))/(E_(A))=((T_(B))/(T_(A)))^(4)=(2)^(4)=16=N`
Object B has emitted at lower wavelength compared to A. So the object B would have emitted more energetic radiation than A.
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