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The binding energy per nucleon number for deutron `H_(1)^(2)`and helium ` He_(2)^(4)` are `1.1 MeV` and `7.0 MeV` respectively . The energy released when two deuterons fase to form a belium nucleus `He_(2)^(4)` is …….. |
Answer» `._(1)^(2)H+._(1)^(2)H rarr._(2)^(4)He` Initial `B.E. = 2[2xx1.1] = 4.4 MeV` Final `B.E. = 4xx 7 = 28 MeV` Final `B.E. gt` Initial `B.E.` Energy released `= 28 - 4.4 = 23.6 MeV` |
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