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The binding energies per nucleon for deuterium and helium are `1.1 MeV` and `7.0 MeV` respectively. What energy in joules will be liberated when `2` deuterons take part in the reaction. |
Answer» `._(1)^(2)H + ._(1)^(2)H rarr ._(2)^(4)He + Q` Binding energy per nucleon of helium `(._(2)^(4)He) = 7 MeV` Binding energy `= 4 xx 7 =28 Me V` Binding energy nucleon of deuterium `(._(2)^(1)H) = 1.1 MeV` Binding energy `= 2 xx 1.1 = 2.2 MeV` Energy liberated `(Q) = (28 - (2.2)2] = 23.6 MeV` i.e., `Q = 23.6 xx 10^(6) xx 6 xx 10^(-19), Q = 37.76 xx 10^(-13)J` |
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