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Calculate the energy of the following nuclear reaction: `._(1)H^(2)+._(1)H^(3) to ._(2)He^(4) + ._(0)n^(1)+Q` Given: `m(._(1)H^(2))=2.014102u, m(._(1)H^(3))=3.016049u, m(._(2)He^(4))=4.002603u, m(._(0)n^(1))=1.008665u` |
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Answer» Correct Answer - `17.58MeV` In the given nuclear reaction, Mass of reactants = `m(._(1)H^(2))+m(._(1)H^(3))` `=2.0141202+3.016049` `=5.030151u` Mass of products = `m(._(2)H^(4))+m(._(0)n^(1))` `=4.002603+1.008665` `=5.011268u` Mass defect, `Deltam=5.030151u-5.011268u` `=0.018883u` Energy released `=0.018883xx931MeV` `=17.58MeV` |
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