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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 : $\displaystyle \mathrm{m}\left({ }_{6}^{12} \mathrm{C}\right)=12 \cdot 000000 \mathrm{u}$ \[\begin{aligned} & \mathrm{m}\left({ }_{10}^{20} \mathrm{Ne}\right)=19 \cdot 992439 \mathrm{u} \\ & \mathrm{~m}\left({ }_{2}^{4} \mathrm{He}\right)=4 \cdot 002603 \mathrm{u} \\ & 1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^{2} \end{aligned} \]

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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