Question

A $U^{235}$ nuclear reactor generates energy at a rate of $3.70\times {10}^7$ J/s. Each fission liberates $185MeV$ 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 $=6\times {10}^{23}mo{l}^{-1},1eV=1.6\times {10}^{-19}J$)

• A. $70.5kg$
• B. $0.705kg$
• C. $13.1kg$
• D. $1.31kg$

Answer 1

Answer: (B)

$0.705kg$

Total energy generated in given time will be: $(3.7\times {10}^7)\times (144\times {10}^4)J$. = $532.8\times {10}^{11}J$

Hence, number of particles of $U^{235}$ required will be

$N=\frac{532.8\times {10}^{11}}{185\times {10}^6\times 1.6\times {10}^{-19}\times 6\times {10}^{23}}$

Mass of $U^{235}$ required will be $235N=0.705Kg$