Question

The fission properties of $^{239}_{ 94} Pu$ are very similar to those of $^{235}_{92} U$. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure. $^{239}_{ 94} Pu$ undergo fission?

Answer 1

Average energy released per fission of $^{239}_{ 94} Pu$, Eav = 180 Mev
Amount of pure $^{239}_{ 94} Pu$ , m = 1 kg = 1000 g
NA= Avogadro number = 6.023 × 1023
Mass number of $^{239}_{ 94} Pu$ = 239 g
1 mole of $^{239}_{ 94} Pu$ contains NA atoms.
mg of mg of contains contains ($\frac{NA}{Mass Number} \times m$)atoms
= $\frac{6.023\times 10^{23}}{239} \times 1000 = 2.52 \times 10^{24} atoms$
Total energy released during the fission of 1 kg of $^{239}_{ 94} Pu$ is calculated as:
$E = E_{av} \times 2.52 \times 10^{24}$
$E = 180 \times 2.52 \times 10^{24}$
$E = 4.536 \times 10^{26} MeV$
Hence, $4.536 \times 10^{26} MeV$ is released if all the atoms in 1 kg of pure $^{239}_{ 94} Pu$ undergo fission.