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`.

Explanation

`3._2He^4 to ._6C^12 + Q `
`therefore Q=m(3 ._2He^4)-m(._6C^12)`
=12.0078-12
=0.007809
therefore` Energy released =0.007809 x 931 MeV
=7.27 MeV