Calculate the energy released/absorbed in the following nuclear reaction: _6^12 C +_6^12 C →_10^20 Ne +_2^4 He Given: m (_6^12 C )=12 · 000000 u m…

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