# The fission properties of

Question:

The fission properties of ${ }_{94}^{239} \mathrm{Pu}$ are very similar to those of ${ }_{92}^{235} \mathrm{U}$.

The average energy released per fission is $180 \mathrm{MeV}$. How much energy, in MeV, is released if all the atoms in $1 \mathrm{~kg}$ of pure ${ }_{94}^{239} \mathrm{Pu}$ undergo fission?

Solution:

Average energy released per fission of ${ }_{94}^{239} \mathrm{Pu}, E_{a v}=180 \mathrm{MeV}$

Amount of pure $_{94} \mathrm{Pu}^{239}, m=1 \mathrm{~kg}=1000 \mathrm{~g}$

$N_{A}=$ Avogadro number $=6.023 \times 10^{23}$

Mass number of ${ }_{94}^{299} \mathrm{Pu}=239 \mathrm{~g}$

1 mole of ${ }_{94} \mathrm{Pu}^{239}$ contains $\mathrm{N}_{\mathrm{A}}$ atoms.

$\therefore m \mathrm{~g}$ of ${ }_{94} \mathrm{Pu}^{239}$ contains $\left(\frac{\mathrm{N}_{\mathrm{A}}}{\text { Mass number }} \times m\right)$ atoms

$=\frac{6.023 \times 10^{23}}{239} \times 1000=2.52 \times 10^{24}$ atoms

$\therefore$ Total energy released during the fission of $1 \mathrm{~kg}$ of ${ }_{94}^{239} \mathrm{Pu}$ is calculated as:

$E=E_{\alpha v} \times 2.52 \times 10^{24}$

$=180 \times 2.52 \times 10^{24}=4.536 \times 10^{26} \mathrm{MeV}$

Hence, $4.536 \times 10^{26} \mathrm{MeV}$ is released if all the atoms in $1 \mathrm{~kg}$ of pure ${ }_{94} \mathrm{Pu}^{239}$ undergo fission

Leave a comment

Click here to get exam-ready with eSaral