The surface temperature of the sun is `T_(0)` and it is at average distance `d` from a planet. The radius of the sun is `R`. The temperature at which planet radiates the energy is

Options

A. `T_(0) sqrt((R)/(2d))`
B. `T_(0) sqrt((2R)/(d))`
C. `T_(0) sqrt((R)/(d))`
D. `T_(0) ((R)/(d))^(1//4)`

Answer

A

Explanation

`r =` Radius of planet
`T_(1)`= Temperature of planet at which energy is radiated
Power radiated by Sun, `P=4 pi R^(2)sigma T_(0)^(4)`
Energy recived by planet = `(P)/(4 pi d^(2)) xx pi r^(2) = (R^(2)T_(0)^(4)pi r^(2)sigma)/(d^(2))`
Energy radiated by planet = `4 pi r^(2)sigma T_(1)^(4)`
For thermal equilibrium,
`(R^(2)T_(0)^(4)pi r^(2)sigma)/(d^(2)) = 4 pi r^(2)sigma T_(1)^(4)`
`R^(2)T_(0)^(4) = 4d^(2)T_(1)^(4)`
`T_(1) = T_(0)sqrt((R)/(2d))`.