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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 ……..A. 20.8 MeVB. 16.6MeVC. 25.2MeVD. 23.6MeV |
Answer» Correct Answer - d Energy released would be `Delta E` =total energy of `._(2)He^(4)` `- 2xx` (total binding energy of `._(1)_(He)^(4)` ` 4 xx 7.0 -2(1.1)(2)` `=23.6 MeV` . |
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