Question

In a nuclear reactor, fission is produced in 1 g of ${}^{235}U\left(235.0349amu\right)$. In assuming that ${}_{53}^{92}Kr\left(91.8673amu\right)$ and ${}_{36}^{141}Ba\left(140.9139amu\right)$ are produced in all reactions and no energy is lost, calculate the total energy produced in kilowatt hour. Given: 1 amu $=931MeV$.

Answer 1

Mass defect $=△m=235.0349-(91.8673+140.9139)=2.2539amu$

Energy released $=2.2539\times 931=2098.1947MeV$