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Obtain the binding energy (in MeV) of a nitrogen nucleus (" "_(7)^(14)N). Given m(" "_(7)^(14)N) = 14.00307 u. |
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Answer» Solution :SINCE a nucleus of `" "_(7)^(14)N` CONTAINS 7 protons and 7 NEUTRONS, hence mass DEFECT of `" "_(7)^(14)N` `DeltaM = 7(m_(H))+7m_(n) - m( " "_(7)^(14)N ) = 7 xx 1.007825 + 7 xx 1.008665 u - 14.00307 u = 0.11236 u` `therefore` Binding energy `E_(b) =Delta M xx 931.5 MeV =0.11236 xx 931.5 MeV =104.7` MeV. |
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