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 \( ^{239}_{ 94} Pu\), E_av = 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.