Question

The energy released per fission of a ${{}_{92}U}^{235}$ nucleus is nearly

Answer 1

The fission process represented by the equation, ${}_{92}{U}^{235}{+}_0{n}^1{\to }_{56}B{a}^{144}{+}_{36}K{r}^{89}+3_0{n}^1$

Masses of reactants $=234.39+1.01=235.4amu$

Masses of products $=143.28+88.89+3\left(1.01\right)=235.2amu$

Energy released $=$ mass difference $=235.4-235.2=0.2amu=0.2\times 931\sim 200MeV$