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NCERT Class XII Chapter
Nuclei
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Question : 17 of 31
Marks: +1, -0
The fission properties of  94239Pu\,{}^{239}_{94}\mathrm{Pu} are very similar to those of  92235U\,{}^{235}_{92}\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  94239Pu\,{}^{239}_{94}\mathrm{Pu} undergo fission?
Solution:  
Number of atoms present in 1 mole i.e.,
239 g of  94239Pu\,{}^{239}_{94}\mathrm{Pu} = 6.023 × 102310^{23}
∴ Number of atoms present in 1000 g of  94239Pu\,{}^{239}_{94}\mathrm{Pu}
= 6.023×1023×1000239\frac{6.023 \times 10^{23} \times 1000}{239} = 2.52 × 102410^{24}
Energy released per fission = 180 MeV
Total energy released = 2.52 × 102410^{24} × 180 MeV = 4.54 × 102610^{26} MeV.
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