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Physics · 2025 · 2 marks
CBSE 2025 · Region 5 · Set 1 · Q20
When a neutron collides with _92^235 U, the nucleus gives _54^140 Xe and _38^94 Sr as fission products and two neutrons are ejected. Calculate the…
When a neutron collides with $\displaystyle { }_{92}^{235} \mathrm{U}$, the nucleus gives $\displaystyle { }_{54}^{140} \mathrm{Xe}$ and $\displaystyle { }_{38}^{94} \mathrm{Sr}$ as fission products and two neutrons are ejected. Calculate the mass defect and the energy released (in MeV ) in the process. Given: \[\begin{aligned} & \mathrm{m}\left({ }_{92}^{235} \mathrm{U}\right)=235 \cdot 04393 \mathrm{u}, \quad \mathrm{~m}\left({ }_{54}^{140} \mathrm{Xe}\right)=139 \cdot 92164 \mathrm{u} \\ & \mathrm{~m}\left({ }_{38}^{94} \mathrm{Sr}\right)=93 \cdot 91536 \mathrm{u}, \quad{ }_{0}^{1} \mathrm{n}=1 \cdot 00866 \mathrm{u} \\ & 1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^{2} \end{aligned} \]
Marking-scheme solution
$\displaystyle { }_{92}^{235} \mathrm{U}+{ }_{0}^{1} n \rightarrow{ }_{54}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+2\left({ }_{0}^{1} n\right)$
$\displaystyle \Delta m=m\left({ }_{0}^{1} n\right)+m\left({ }_{92}^{235} \mathrm{U}\right)-\left(m\left({ }_{54}^{140} \mathrm{Xe}\right)+m\left({ }_{38}^{94} \mathrm{Sr}\right)+2 \times m\left({ }_{0}^{1} n\right)\right)$
$\displaystyle =1.00866+235.04393-139.92164-93.91536-2 \times 1.00866$
$\displaystyle =0.19827 \mathrm{u}$
Energy released $\displaystyle =\Delta m \times 931 \mathrm{MeV}$
$\displaystyle =0.19827 \times 931 \mathrm{MeV}$
$\displaystyle =184.59 \mathrm{MeV}$
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CBSE Class 12 Physics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.