The energy liberated on complete fission of $1\, kg$ of $_{92}{U^{235}}$ is (Assume $200\, MeV$ energy is liberated on fission of $1$ nucleus).

  • A
    $8.2 \times 10^{10}\, J$
  • B
    $8.2 \times 10^{9}\, J$
  • C
    $8.2 \times 10^{13}\, J$
  • D
    $8.2 \times 10^{16}\, J$

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Complete the reaction: $n + _{92}^{235}U \to _{56}^{144}Ba + .... + 3n$

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$(2)$ The rest mass of a stable nucleus is greater than the sum of the rest masses of its constituent nucleons.
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$A$ $1000 \; MW$ fission reactor consumes half of its fuel in $5.00 \; y$. How much $_{92}^{235} U$ (in $kg$) 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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