13.18 A 1000 MW fission reactor consumes half of its fuel in 5.00 y. 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.

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Payal Gupta | Contributor-Level 10

8 months ago

13.18 Half life of the fuel in the fission reactor, $T_{1 / 2}$
= 5 years = 5 $* 365 * 24 * 60 * 60 s$

= 157.68 $* 10^{6} s$

We know that in the fission of 1 g of ${}_{92}^{235}U$
, the energy released = 200 MeV

1 mole i.e. 235 gm of ${}_{92}^{235}U$ contains 6.023 $* 10^{23}$ atoms

Therefore 1 gm of ${}_{92}^{235}U$
contains = $\frac{6.023 * 10^{23}}{235}$ = 2.563 $* 10^{21}$ atoms

The total energy Q generated per gm of ${}_{92}^{235}U$ = 200 $*$ 2.563 $* 10^{21}$ MeV/g = 5.12

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13.18 A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much U 92 235 did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of U92235 and that this nuclide is consumed only by the fission process.

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