Question

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

Answer 1

Number of atoms in 1 kg of pure Pu=$\frac{6.023\times {10}^{23}}{239}\times 1000=2.52\times {10}^{24}$

As average energy released in fission is $180MeV$, the total energy released is

$\Rightarrow 2.52\times {10}^{24}\times 180MeV$

$\Rightarrow 4.53\times {10}^{26}MeV$.