Modern Physics Ques 206
- 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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Answer:
Correct Answer: 206.$(3.847 \times 10^{4}$ $kg)$
Solution:
Formula:
- 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} $
BETA
