CBSE 2025, Set 1

Question

When a neutron collides with ${}^{235}_{92}	ext{U}$, the nucleus gives ${}^{140}_{54}	ext{Xe}$ and ${}^{94}_{38}	ext{Sr}$ as fission products and two neutrons are ejected. Calculate the mass defect and the energy released (in MeV) in the process.

Given:
$$m\!\left({}^{235}_{92}	ext{U}ight) = 235\cdot04393 	ext{ u}, \quad m\!\left({}^{140}_{54}	ext{Xe}ight) = 139\cdot92164 	ext{ u}$$
$$m\!\left({}^{94}_{38}	ext{Sr}ight) = 93\cdot91536 	ext{ u}, \quad m\!\left({}^{1}_{0}	ext{n}ight) = 1\cdot00866 	ext{ u}$$
$$1 	ext{ u} = 931 	ext{ MeV}/c^2$$

Edvaya · Accepted Answer

The correct answer is Subjective. **Nuclear reaction:** $${}^{235}_{92}	ext{U} + {}^{1}_{0}	ext{n} ightarrow {}^{140}_{54}	ext{Xe} + {}^{94}_{38}	ext{Sr} + 2\,{}^{1}_{0}	ext{n}$$ **Mass of reactants:** $$m_{	ext{reactants}} = m({}^{235}_{92}	ext{U}) + m({}^{1}_{0}	ext{n}) = 235\cdot04393 + 1\cdot00866 = 236\cdot05259 	ext{ u}$$ **Mass of products:** $$m_{	ext{products}} = m({}^{140}_{54}	ext{Xe}) + m({}^{94}_{38}	ext{Sr}) + 2 	imes m({}^{1}_{0}	ext{n})$$ $$= 139\cdot92164 + 93\cdot91536 + 2 	imes 1\cdot00866$$ $$= 139\cdot92164 + 93\cdot91536 + 2\cdot01732 = 235\cdot85432 	ext{ u}$$ **Mass defect:** $$\Delta m = m_{	ext{reactants}} - m_{	ext{products}} = 236\cdot05259 - 235\cdot85432 = \boxed{0\cdot19827 	ext{ u}}$$ **Energy released:** $$E = \Delta m 	imes 931 	ext{ MeV} = 0\cdot19827 	imes 931 = \boxed{184\cdot59 	ext{ MeV}}$$ > **Significance:** Each fission of U-235 releases approximately **184-200 MeV** of energy. This is about $3 	imes 10^7$ times more energy per reaction than burning a carbon atom ($\approx 4$ eV). This enormous energy release per fission is the basis of nuclear reactors and weapons.

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