Given that
The amount of electricity that will be generated E = 200000 MW
Nuclear power plants are needed to provide the energy = 10% of E
= \frac{10}{100} * 200000 = 20000 MW
= 20000 * 10^{6} joules/sec
Nuclear power per year (Converting joules/sec to joules/year)
= 2 * 10^{10} * 365*24*60*60 = 6.307*10^{17} joules/year
The energy emitted by a single uranium nucleus fission is 200 MeV.
Per fusion, the amount of nuclear energy converted into electrical energy = 25 % of 200 MeV
\frac{25}{100} * 200 = 50 MeV = 50 * 10^{6} * 1.6* 10^{-19} = 8* 10^{-12} Joules
Step 1: Determining the number of atoms required to per fission per year
[math]
number of fissions per year = \frac{6.307*10^{17}}{8* 10^{-12}} = 78840* 10^{24} atoms
Step 2: Finding the mass of nuclear substance per year
1 g (1 mole) of the uranium contains 6.023* 10^{23} atoms
Atomic mass of uranium = 235 g = 235* 10^{-3} Kg
Mass of 78840* 10^{24} uranium atoms = \frac{235* 10^{-3}}{6.023* 10^{23}} * 78840* 10^{24}
= 3.016 * 10^{4} kg
As a results, the amount of uranium required each year = 3.016 * 10^{4} kg