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$\mathrm{AU}^{235}$ nuclear reactor generates energy at a rate of $3.70 \times 10^7 \mathrm{~J} / \mathrm{s}$. Each fission liberates $185 \mathrm{MeV}$ useful energy. If the reactor has to operate for $144 \times 10^4 \mathrm{~s}$, then, the mass of the fuel needed is (Assume Avogadro's number $\left.=6 \times 10^{23} \mathrm{~mol}^{-1}, 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)$
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Verified Answer
The correct answer is:
$0.705 \mathrm{~kg}$
In $1 \mathrm{~s}$, energy generated is $3.7 \times 10^7 \mathrm{~J}$
In $144 \times 10^4 \mathrm{~s}$, energy generated is
$=3.7 \times 10^7 \times 144 \times 10^4 \mathrm{~J}$
Also energy released in one fission is
$\begin{aligned}
& =185 \mathrm{meV} \\
& =185 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J}
\end{aligned}$
Number of fission $=\frac{3.7 \times 10^7 \times 144 \times 10^4}{185 \times 10^6 \times 1.6 \times 10^{-19}}$
$=1.8 \times 10^{24} \text { of } \mathrm{U}^{235} \text { atoms. }$
Mass contained in $1.8 \times 10^{24}$ atoms of $U^{235}$ is
$=\frac{235 \times 1.8 \times 10^{24}}{6.023 \times 10^{23}}=702.3 \mathrm{~g}=0.70 \mathrm{~kg}$
In $144 \times 10^4 \mathrm{~s}$, energy generated is
$=3.7 \times 10^7 \times 144 \times 10^4 \mathrm{~J}$
Also energy released in one fission is
$\begin{aligned}
& =185 \mathrm{meV} \\
& =185 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J}
\end{aligned}$
Number of fission $=\frac{3.7 \times 10^7 \times 144 \times 10^4}{185 \times 10^6 \times 1.6 \times 10^{-19}}$
$=1.8 \times 10^{24} \text { of } \mathrm{U}^{235} \text { atoms. }$
Mass contained in $1.8 \times 10^{24}$ atoms of $U^{235}$ is
$=\frac{235 \times 1.8 \times 10^{24}}{6.023 \times 10^{23}}=702.3 \mathrm{~g}=0.70 \mathrm{~kg}$
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