CBSE 2024 · Region 2 · Set 2 · Q20 · 2 marks
Calculate the energy released/absorbed in the following nuclear reaction : \[{ }_{6}^{12} \mathrm{C}+{ }_{6}^{12} \mathrm{C} \longrightarrow{ }_{10}^{20} \mathrm{Ne}+{ }_{2}^{4} \mathrm{He} \] Given : $\displaystyle \mathrm{m}\left({ }_{6}^{12} \mathrm{C}\right)=12 \cdot 000000 \mathrm{u}$ \[\begin{aligned} & \mathrm{m}\left({ }_{10}^{20} \mathrm{Ne}\right)=19 \cdot 992439 \mathrm{u} \\ & \mathrm{~m}\left({ }_{2}^{4} \mathrm{He}\right)=4 \cdot 002603 \mathrm{u} \\ & 1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^{2} \end{aligned} \]
Marking-scheme solution
Energy $\displaystyle =$ mass defect $\displaystyle \times 931 \mathrm{MeV}$
Mass defect $\displaystyle =\Delta m=(2 \times 12.000000-19.992439-4.002603)$
$\displaystyle =0.004958 \mathrm{u}$
Energy released $\displaystyle =0.004958 \times 931 \mathrm{MeV}$
$\displaystyle =4.62 \mathrm{MeV}$
Asked in 2 different years
- 2024 · Region 2 · Set 2 · Q20
- 2025 · Region 6 · Set 2 · Q26
NucleiMass-Energy and Nuclear Binding EnergyApplyvery_short_answermedium
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Asked in 2024 & 2025. CBSE Class 12 Physics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.