Question

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

Answer 1

Total energy produced by the reactor in time $t=10$ years,

$E=1000\times {10}^6\times 10\times 3.15\times {10}^{7}J$
$=3.15\times {10}^{17}J$

$Efficiency=\frac{\text{output energy}}{\text{input energy}}$

$\Rightarrow$ Input energy caused by fission $=\frac{\text{output energy}}{efficiency}$$=\frac{3.15\times {10}^{17}}{\left(10/100\right)}=3.15\times {10}^{18}J$

Energy produced by 1 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{\text{Total energy}}{\text{Energy per fission}}$$=\frac{3.15\times {10}^{18}}{3.2\times {10}^{-11}}$$\approx 9.8\times {10}^{28}$

Hence, mass of uranium required is given by:

$m=\frac{N}{N_0}\times 235kg=\frac{9.8\times {10}^{28}}{6.02\times {10}^{26}}$$=38.4\times {10}^{3}kg$