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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