Question

In a nuclear reactor, ${}^{235}U$ undergoes fission liberating $200MeV$ of energy. The reactor has a $10$% efficiency and produces $1000MeV$ power. If the reactor is to function for $10$ yrs then find the total mass of uranium required.

• A. $36.5\times {10}^{3}kg$
• B. $36\times {10}^{2}kg$
• C. $39.5\times {10}^{3}kg$
• D. $38.2\times {10}^{3}kg$

Answer 1

The correct option is D $38.2\times {10}^{3}kg$

Total energy produced by the reactor in time $t=10yr$.
$E=1000\times {10}^6\times 10\times 3.15\times {10}^{7}J$
$=3.15\times {10}^{17}J$
$\because$ Efficiency $=\frac{outputenergy}{inputenergy}$
Input energy caused by fission $=\frac{outputenergy}{efficiency}$
$=\frac{3.15\times {10}^{17}}{\left(10/100\right)}=3.15\times {10}^{18}J$
Energy produced by one fission of ${}^{235}U=200MeV$
$=200\times 1.6\times {10}^{-13}J$
$=3.2\times {10}^{-11}J$
Therefore, number of fissions required
$=\frac{totalenergy}{energyperfission}$
$=\frac{3.15\times {10}^{18}}{3.2\times {10}^{-11}}=9.8\times {10}^{28}$
Hence, mass of uranium required is given by
$m=\frac{N}{N_a}\times 235=\frac{9.8\times {10}^{28}}{6.02\times {10}^{26}}$
$=38.2\times {10}^{3}kg$