# A 1000 MW fission reactor consumes half of its fuel in 5 yr. How much \${}_{92}^{235}U\$ did it contain initially?

A 1000 MW fission reactor consumes half of its fuel in 5 yr. How much \${}{92}^{235}U\$ did it contain initially?
Assume that the reactor operates 80% of the time that all the energy generated - arises from the fission of \${}
{92}^{235}U\$ and that this nuclide is consumed only by the fission process.

Given power of reactor P = 1000MW
We will use concept that the energy generated in one fission of \${}{92}^{235}U\$ is 200 MeV.
Number of \${}
{92}^{235}U\$ attom in 1 g = 1/235 x 6.023 x \${{10}^{23}}\$
Energy generated per gram of \${}{92}^{235}U\$ =(1/235 x 6.023x\${{10}^{23}}\$ x 200 x 1.6x\${{10}^{-13}}\$
Total energy generated in 5 year with 80% of the time = 1000 x \${{10}^{6}}\$x5x365x24x60x60x80/100
Mass of \${}
{92}^{235}U\$ consumed in 5 yr,

m = Total energy/Energy consumed per gram
= 1.538 x \${{10}^{6}}\$ g = 1538 kg

Initial amount of \${}_{92}^{235}U\$ = 1544 x 2 kg = 3076 kg