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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`.

Text Solution

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The mass of a ``^(12)C` atom is exaxtly 12 u. The energy released in the reaction `3(`_(2)^(4)He) rarr `_(12)^(6)C` is
`[3m(_(2)^(4)He) - m(_(6)^(12)C)]c^2`
`=[3 xx 4*002603 u - 12u] (931 MeVu^(-1)) = 7*27MeV.`
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