A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much ²³⁵₉₂U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of ²³⁵₉₂U and that this nuclide is consumed only by the fission process.

Question

Philoid · Accepted Answer

Half-life of the fuel in the fission reactor = 5 years

= 5 × 365 × 86400 s (∵ 1 day = 86400 seconds)

= 1.576 × 10⁸ sec

∵ 1 gm of Uranium fission gives 200 MeV energy [This can be worked out from the fission equation given below of Uranium].

∴ 1 gm of Uranium, =

Energy generated by Uranium fission =

= J

= 8.2 × 10¹⁰ J/gm

∵ The 1000MW reactor operates only 80% of it’s time,hence in 5 years amount of uranium consumed

=

= 1538 kg

∴ This amount is consumed within the half life time.

Hence, the initial amount will be = 2 × 1538Kg

= 3076Kg

A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much 23592U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of 23592U and that this nuclide is consumed only by the fission process.