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Binding energy per nucleus of .(1)^(2)H ...

Binding energy per nucleus of `._(1)^(2)H` and `._(2)^(4)He` are 1.1 MeV and 7 MeV respectively. Calculate the amount of energy released in the following process:
`._(1)^(2)H + ._(1)^(2)H to ._(2)^(4)He`

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Amount of energy released
`= sum` Binding energy of products
`- sum` Binding energy of reactions
`=[4 xx 7] - [4 xx 1.1] = 23.6 MeV`
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