13.17 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 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure \({ }_{94}^{239} \mathrm{Pu}\) undergo fission?

Hint: One atom of \({ }_{94}^{239} \mathrm{Pu}\) gives 180 MeV of energy.
Step 1: Find the number of atoms in 1kg of \({ }_{94}^{239} \mathrm{Pu}\).
Average energy released per fission of P94239u, Eav=180 MeV
Amount of pure Atomic mass of  P94239u, m=1kg=1000 g
NA=Avogadro number=6.023×1023
Mass number of P94239u=239 g
1 mole of P94239u contains NA atoms.
m g of P94239u  containsNAMass number×matoms
1000 g of P94239u contains NAMass number×1000atoms=2.52×1024 atoms
 
Step 2: Find the total energy released.
Total energy released during the fission of 1 kg of P94239u is calculated as:
E=Eav×2.52×1024
=180×2.52×1024=4.536×1026 MeV
Hence,  4.536×1024 MeV has released if all the atoms in 1 kg of pure P94239u undergo fission.