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The B.E. per nucleon of ""_(1)H^(2) and ""_(2)He^(4) are 1.1 MeV and 7 MeV respectively. If two deuteron (""_(1)H^(2)) nuclei react to form single helium nucleus, the energy released |
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Answer» 13.9 MEV `E="energy of "H_(2)^(4)-"energy of "2_(1)H^2` `=(4 xx 7-2 xx 1.1 xx 2)MeV=(28-4.4)MeV` `E=23.6 MeV` |
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