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Calculate the energy released in fusion reaction : " "_(1)^(2)H + " "_(1)^(2)H to " "_(2)^(3)He +n, where BE of " "_(1)^(2)H = 2.23 MeV and of " "_(2)^(3)He = 7.73 MeV. |
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Answer» Solution :For the fusion reaction `" "_(1)^(2)H + " "_(1)^(2)H to " "_(2)^(3)He + n`, as per question binding energy of `" "_(1)^(2)H` is 2.23 MeV and that of `" "_(2)^(3)He` is 7.73 MeV. `THEREFORE` Total binding energy of reactants `E_(1) = 2.23 + 2.23 = 4.46 MeV` and total binding energy of products `E_(2) = 7.73 + 0 = 7.73 MeV``therefore` Energy released in fusion reaction `E = E_(2) - E_(1) = 7.73 - 4.46 = 3.27 MeV` |
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