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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 `_^4 He` is `4.002603 u`.

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The mass of a `.^(12)C` atom is exactly 1`2 u`. The energy released in the reaction is `3(_2^4He)rarr._6^(12)C)`
`[3m(._(2)^(4)He) -m(._(6)^(12)C) ]c^(2)=[3xx4.002603 u-12u](931 MeV//u)`
`=7.27 MeV` .
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