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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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Answer» 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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