Question

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.

Answer 1

235 g of uranium contains 6.02 × 10²³ atoms.
1 g of uranium = $\frac{1}{235}\times 6.023\times {10}^{23}$ atoms
∴ 0.7 g of uranium = $\frac{1}{235}\times 6.023\times {10}^{23}\times 0.007$ atoms

1 atom gives 200 MeV.

∴ Total energy released = $\frac{6.023\times {10}^{23}\times 0.007\times 200\times {10}^6\times 1.6\times {10}^{-19}}{235}\mathrm{J}$ = 5.74$\times$10⁸ J