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A 1000 MW fission reactor consumes half ...

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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Energy generated per gram of `""_(92)^(235)U = (6 xx 10^(23) xx 200 xx 1.6 xx 10^(-13))/(235) J g^(-1)`
The amount of `""_(92)^(235)U` consumed in 5y with 80% on - time.
`=(5 xx 0.8 xx 3.154 xx 10^(16) xx 235)/(1.2 xx 1.6 xx 10^(13)) g = 1544 g`
The initial amount of `""_(92)^(235)U = 3088 kg. `
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