A black body at a high temperature T radiates energy at the rate of U (in `Wm^(-2)` ). When the temperature falls to half (I e . `(T)/(2)` ) the radiated energy (in `Wm^(-2)` ) will be

Options

A. `(U)/(8)`
B. `(U)/(16)`
C. `(U)/(4)`
D. `(U)/(2)`

Answer

B

Explanation

According to Stefa's law of radiation
`UpropT^(4)`" " (because"U is the energy")
`therefore" " (U_(1))/U_(2)=(T_(1)/(T_(2)))^(4)=((T)/(T//2))^(4)" " (becauseT_(2)=(T)/(2))`
`"or " (U_(1))/(U_(2))=((2)/(1))^(4)((16)/(1))`
`"or " U_(2)=(U_(1))/(16)rArrU_(2)=(U)/(16) " " (becauseU_(1)=U)`