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Calculate the energy released when three...

Calculate the energy released when three alpha particles combine to form a `^12 C` nucleus. The atomic mass of `_2^4 He` is `4.002603 u`.

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The mass of `a^(12)C` atom is exactly `12 u`.
The energy released in the reaction `3(._(2)^(4)Hr)rarr._(6)^(12)C` is
`[3m(._(2)^(4)He)-m(._(6)^(12)C)]c^(2)`
`[3xx4.002603u-12 u](931 MeV//u)=7.27 MeV`.
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