In the process of nuclear fission of `1 g` uranium, the mass lost is `0.92 mg`. The efficiency of power house run by the fission reactor is `10%`.To obtain `400` megawatt power from the power house, how much uranium will be required per hour? `(c=3xx10^(8) m s^(-1))`.

Explanation

Power to be obtained from power house = 400 mega watt
`therefore` Energy obtained per hour = 400 mega watt x 1hour = (`400xx10^6` watt) x 3600 second = `1.44xx10^10` joule
Here only 10% of input is utilized. In order to obtain `144 xx 10^10` joule of useful energy, the output energy from the power house `(10E)/100=144xx10^10` J, E=`1.44xx10^11`joule
Let, this energy is obtained from a mass-loss of `Deltam` kg.
Then `(Deltam)c^2=144xx10^11` joule
`Deltam=(144xx10^11)/(3xx10^8)^2=16xx10^(-5)` kg = 0.16 gm
Since 0.92 milli gram (= `0.92xx10^(-3)` gm) mass is lost in 1 gm uranium , hence for a mass loss of 0.16 gm, the uranium required is `=(1xx0.16)/(0.92xx10^(-3))` = 174 gm
Thus to run the power house, 174 gm uranium is required per hour.