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Calculate the energy in fusion reaction: ""_(1)H^(2) + ""_(1)H^(2) to ""_(2)He^(3) + ""_(0)n^(1), where B.E. Of ""_(1)H^(2) = 2.23 MeV and ""_(2)He^(3) = 7.73 MeV. |
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Answer» SOLUTION :INITIAL binding energy `BE_1 = (2.23 + 2.23) = 4.46 MEV` Final binding energy `BE_2 = 7.73 MeV` Energy released = (7.73 – 4.46) MeV= 3.27 MeV |
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