Sum
Calculate the energy released by 1g of natural uranium assuming 200 MeV is released in each fission event and that the fissionable isotope 235U has an abundance of 0.7% by weight in natural uranium.
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Solution
235 g of uranium contains 6.02 × 1023 atoms.
1 g of uranium = `1/235 xx 6.023 xx 10^23` atoms
∴ 0.7 g of uranium = `1/235 xx 6.023 xx 10^23 xx 0.007` atoms
1 atom gives 200 MeV.
∴ Total energy released = `(6.023 xx 10^23 xx 0.007 xx 200 xx 10^6 xx 1.6 xx 10^-19)/235 "J" = 5.74 xx 10^8 "J"`
Concept: Nuclear Energy - Fission
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