Modern Physics 5 Question 31
34. In a nuclear reactor ${ }^{235} U$ undergoes fission liberating $200 MeV$ of energy. The reactor has a $10 %$ efficiency and produces $1000 MW$ power. If the reactor is to function for $10 yr$, find the total mass of uranium required.
(2001, 5M)
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Solution:
- The reactor produces $1000 MW$ power or $10^{9} J / s$. The reactor is to function for $10 yr$. Therefore, total energy which the reactor will supply in $10 yr$ is
$$ \begin{aligned} E & =(\text { power })(\text { time }) \\ & =\left(10^{9} J / s\right)(10 \times 365 \times 24 \times 3600 s) \\ & =3.1536 \times 10^{17} J \end{aligned} $$
But since the efficiency of the reactor is only $10 %$, therefore actual energy needed is 10 times of it or $3.1536 \times 10^{18} J$. One uranium atom liberates $200 MeV$ of energy or $200 \times 1.6 \times 10^{-13} J$ or $3.2 \times 10^{-11} J$ of energy. So, number of uranium atoms needed are
$$ \frac{3.1536 \times 10^{18}}{3.2 \times 10^{-11}}=0.9855 \times 10^{29} $$
or number of kg-moles of uranium needed are
$$ n=\frac{0.9855 \times 10^{29}}{6.02 \times 10^{26}}=163.7 $$
Hence, total mass of uranium required is
$$ \begin{aligned} & m=(n) M=(163.7)(235) kg \\ & \text { or } \quad m \approx 38470 kg \\ & \text { or } \quad m=3.847 \times 10^{4} kg \end{aligned} $$
