Given that
Fission reactor power P = 1000 MW = 10^3* 10^{6} W = 10^{9} W
Half life of the ^{235}_{92} U, T_{\frac{1}{2}} = 5 years
Suppose 80% of the time the reactor operates = 80% of T_{\frac{1}{2}} = \frac{80}{100} * 5 = 4 years
\text{ 4 years } = 4*360*24 *60*60 . seconds
= 124416000 seconds
Step 1: Finding the number of atoms in the 1 kg fission matter
Avogadro number NA = 6.023 * 10^{23}
Atomic mass of ^{235}_{92} U =. 235
Mass of a Uranium-235 m = 1kg = 10^{3} g
Number of atoms in the 1 kg _{92}^{235}U,\ \ N=\frac{NA}{A}*m
N = \frac{6.023 * 10^{23}}{235} * 10^{3}
Step 2: Determine the total energy released by 1 kg mass of substance
The average amount of energy released per nuclear fusion = 200 MeV
E = \frac{6.023 * 10^{23}}{235} * 10^{3}* 200 MeV
E = 5.106 * 10^{26} MEV
E = 5.106 * 10^{26} * 1.6 * 10^{-13} Joules
E = 8.1696 * 10^{13} Joules
Step 3: Calculating the total energy consumed by substance in half life by fission reactor.
\text{ Energy consumed }= \frac{\text{ Energy of reactor released in 5 years }}{\text{ Energy release in 1 kg of substance }}
= \frac{124416000 * 10^{9}}{ 8.1696 * 10^{13} }
= 1522.9 kg
Hence, energy consumed by Uranium-235 = 1522.9 kg
Since the half life of Uranium-235 consumed = 1522.9 kg, the initially the amount will be doubled = 2 * 1522.9 = 3045.8 kg