Consider the fission of ${}^{235}\text{U}$ induced by a thermal neutron: ${}^{235}\text{U}{+}^1\text{n}{\to }^{92}\text{Kr}{+}^{141}\text{Ba}+3{,}^1\text{n}$ Atomic masses (in u) are: $m{(}^{235}\text{U})=235.0439$, $m{(}^1\text{n})=1.008665$, $m{(}^{92}\text{Kr})=91.9262$, $m{(}^{141}\text{Ba})=140.9144$.

Step-by-step solution

1. Write the total initial mass as the sum of the masses of ${}^{235}\text{U}$ and one neutron.
2. Write the total final mass as the sum of the masses of ${}^{92}\text{Kr}$, ${}^{141}\text{Ba}$ and three neutrons.
3. Subtract final mass from initial mass to obtain the mass defect $Deltam$.
4. Convert $\Delta m$ to energy using $E=\Delta m\times 931.5$ MeV to get the Q‑value per fission.
5. Compute the number of atoms in 1 g of ${}^{235}\text{U}$: $N=\frac{1,\text{g}}{235,{\text{g mol}}^{-1}}\times 6.022\times 10^{23}$.
6. Multiply the Q‑value (in MeV) by $1.602\times 10^{-13}$ J/MeV to get the energy per fission in joules.
7. Multiply energy per fission by $N$ to get the total energy released from 1 g.
8. For a power of 1 GW, divide the power by the energy per fission (in joules) to find the required fission rate.
9. Round the final figures to two significant digits.

Concept tested: Fission energy and power conversion

Published 21 September 2026 · How this solution was prepared and checked