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`.

Explanation

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`