Question 8: Atoms & Nuclei

Question

A nuclear explosive is designed to deliver $1 \mathrm{~MW}$ power in the form of heat energy. If the explosion is designed with nuclear fuel consisting of $U^{235}$ to run a reactor at this power level for one year, then the amount of fuel needed is (Given energy per fission is $200 \mathrm{~MeV}$)

Accepted Answer

Energy released per fission is
$\begin{aligned}
E & =200 \mathrm{MeV} \
& =200 	imes 1.6 	imes 10^{-13} \
& =3.2 	imes 10^{-11} \mathrm{~J}
\end{aligned}$
Total energy required to run a $1 \mathrm{~MW}$ reactor for one year is
$\begin{aligned}
& =10^6 \mathrm{Js}^{-1} 	imes(365 	imes 24 	imes 60 	imes 60) \
& =3.15 	imes 10^{13} \mathrm{~J}
\end{aligned}$
Since, 1 fission ( 1 atom of $\mathrm{U}^{235}$ ) produces $3 \cdot 2 	imes 10^{-11} \mathrm{~J}$ of energy. So, total number of $\mathrm{U}^{235}$ atoms required is
$\frac{3.15 	imes 10^{13}}{3.2 	imes 10^{-11}}=9.84 	imes 10^{23}$
Now, $6.02 	imes 10^{23}$ atoms of $\mathrm{U}^{235}$ are contained in $235 \mathrm{~g}$ of $\mathrm{U}^{235}$
$	herefore$ Mass of $\mathrm{U}^{235}$ containing $9.84 	imes 10^{23}$ atoms is
$\begin{aligned}
M & =\frac{235}{6.02 	imes 10^{23}} 	imes 9.84 	imes 10^{23} \
& =384 \mathrm{~g}=0.384 \mathrm{~kg}
\end{aligned}$

Answer

$1 \mathrm{~kg}$

Answer

$0.01 \mathrm{~kg}$

Answer

$3.84 \mathrm{~kg}$

Answer

$0.384 \mathrm{~kg}$

Atoms & Nuclei

A nuclear explosive is designed to deliver 1 \mathrm{~MW} power in the form of heat energy. If the explosion is designed with nuclear fuel consisting of U^{235} to run a reactor at this power level for one year, then the amount of fuel needed is (Given energy per fission is 200 \mathrm{~MeV})

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