The energy spectrum of a black body exhibits a maximum around a wavelength `lamda_(0)`, the temperature of the black body is now changed such that the energy is maximum around a wavelength `3lamda_(0)//4`. The power radiated by the black body will now increase by a factor of

Options

A. `64//27`
B. 256/81
C. `4//3`
D. `16//9`

Answer

B

Explanation

Let `T_(1) and T_(2)` be the initial and final temperatures of the black body.
`thereforeT_(1)lamda_(0)=T_(2)xx(3)/(4)lamda_(0)` (From Wein.s displacement law)
`therefore T_(1)=(3)/(4)T_(2) or (T_(2))/(T_(1))=(4)/(3)`
Now, initial radiated power`=sigmaT_(1)^(4)`
Final radiated power`=sigmaT_(2)^(4)`
(where `sigma` is Stefan Boltzmann constant)
`therefore` The factor by whichthe radioactive power increases
`=(sigmaT_(2)^(4))/(sigmaT_(1)^(4))=((T_(2))/(T_(1)))^(4)=(4)/(3)=(256)/(81)`.