The fission of a ${}^{235}\text{U}$ nucleus after capturing a thermal neutron can be represented as ${}^{235}\text{U}{+}^1\text{n}{\to }^{141}\text{Ba}{+}^{92}\text{Kr}+3{}^1\text{n}.$ Given the atomic masses: $m{(}^{235}\text{U})=235.0439\text{u}$, $m{(}^1\text{n})=1.008665\text{u}$, $m{(}^{141}\text{Ba})=140.9144\text{u}$, $m{(}^{92}\text{Kr})=91.9262\text{u}$. Calculate the total energy released in this fission reaction. (1 u = $931.5\text{MeV}/c^2$).

Step-by-step solution

1. Compute the total mass of the reactants: $m_{\text{react}}=m{(}^{235}\text{U})+m{(}^1\text{n})=235.0439+1.008665=236.052565\text{u}$.
2. Compute the total mass of the products: $m_{\text{prod}}=m{(}^{141}\text{Ba})+m{(}^{92}\text{Kr})+3m{(}^1\text{n})=140.9144+91.9262+3(1.008665)=235.866595\text{u}$.
3. Find the mass defect: $\Delta m=m_{\text{react}}-m_{\text{prod}}=236.052565-235.866595=0.185970\text{u}$.
4. Convert the mass defect to energy: $E=\Delta m\times 931.5\text{MeV}=0.185970\times 931.5\approx 173.3\text{MeV}$.
5. Round to an appropriate number of significant figures: $E\approx 1.73\times 10^2\text{MeV}$.

Concept tested: Energy released in nuclear fission (mass‑energy equivalence)

Published 29 August 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

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