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A $U^{235}$ nuclear reactor generates energy at a rate of $3.70 \times 10^7\; J/s$ Each fission liberates $185\;MeV$ useful energy. If the reactor has to operate for $144 \times 10^4$ seconds, then the mass of the fuel needed is (Assume Avogadro's number $=6 \times 10^{23} \;mol^{-1},1 eV=1.6 \times 10^{-19}\;J)$

\[(a)\;70.5\;kg \quad (b)\;0.705\;kg \quad (c)\;13.1\;kg \quad (d)\;1.31\;kg \]
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answered Nov 7, 2013 by pady_1

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