Question

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

Answer 1

Average energy released per fission of ${\displaystyle {}_{94}^{239}\text{Pu},{\text{E}}_{\text{av}}=180\text{MeV}}$

Amount of pure ${\displaystyle {}_{94}{\text{Pu}}^{239}}$ m = 1 kg = 1000 g

N_A= Avogadro number = 6.023 × 10²³

Mass number of ${\displaystyle {}_{94}^{239}\text{Pu}}$= 239 g

1 mole of ${\displaystyle {}_{94}{\text{Pu}}^{239}}$  contains N_A atoms.

∴ mg of ${\displaystyle {}_{94}{\text{Pu}}^{239}}$ contain ${\displaystyle (\frac{{\text{N}}_{\text{A}}}{\text{Mass number}}\times m)}$ atom

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

∴ Total energy released during the fission of 1 kg of ${\displaystyle {}_{94}^{239}\text{Pu}}$ is calculated as:

${\displaystyle \text{E}={\text{E}}_{\text{av}}\times 2.52\times {10}^{24}}$

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

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