A 1000 MW fission reactor consumes half of its fuel in 500 y How much u92235 did it contain - Physics - Nuclei

Question

A 1000 MW fission reactor consumes half of its fuel in 500 y How
much u 92 235 did it contain initially? Assume the reactor operates
80% of the time, that all the energy generated arises from the
fission of u 92 235 and that this nuclide is consumed only by the
fission process

Vijay Shankar Yadav · Accepted Answer

Dear Student, Half life of the fuel of the fission reactor,

years = 5 × 365 × 24 × 60 × 60 s We know that in the fission of 1 g of

nucleus, the energy released is equal to 200 MeV 1 mole, i e , 235 g of

contains 6 023 × 1023 atoms ∴1 g

contains

The total energy generated per gram of

is calculated as:

The reactor operates only 80% of the time Hence, the amount of

consumed in 5 years by the 1000 MW fission reactor is calculated as:

∴Initial amount of

= 2 × 1538 = 3076 kg Regards

A 1000 MW fission reactor consumes half of its fuel in 500 y. How much u 92 235 did it contain initially? Assume the reactor operates 80% of the time, that all the energy generated arises from the fission of u 92 235 and that this nuclide is consumed only by the fission process.

A 1000 MW fission reactor consumes half of its fuel in 500 y. How much ${}_{92}^{235}u$ did it contain initially? Assume 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.