Q: In the process of nuclear fission of 1 gram uranium, the mass lost is 0.92 milligram. The efficiency of powder house run by it is 10 %. To obtain 400 megawatt powder from the powder house, how much uranium will be required per hour ? (c = 3 x 10^{8} m/s )

Sol: Powder to be obtained from powder house = 400 mega watt/ Energy obtained per hour

= 400 megawatt x 1 hour = (400 x 10^{6} watt) x 3600 second

= 144 x 10^{10} joule.

Here only 10% of input is utilized .

In order to obtain 144 x 10^{10} joule of useful energy, the output energy from the powder house

(10 E)/100 = 144 x 10^{10} J

E = 144 x 10^{11} joule

Let, this energy is obtained form a mass- loss of Δm kg.

Then (Δm)c^{2} = 144 × 10^{11} joule

Δm = (144 × 10^{11})/(3 × 10^{8})^{2} = 16 × 10^{-5} kg = 0.16 g

Since 0.92 milligram (= 0.92 x 10^{-3} g) mass is lost in 1 g uranium, hence for mass loss of 0.16 g, the

uranium required is = (1 × 0.16 )/(0.92 × 10^{-3} )= 174 g

Thus to rum the powder house, 174 gm uranium is required per hour.