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Calculate binding energy per n ucleon of ""_(83)Bi^(209). Given mass of ""_(83)Bi^(209)=208.980388 a.m.u, mass of neutron =1.008665 a.m.u. and mass of proton =1.007825 a.m.u. Take 1 a.m.u. =931 M eV. |
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Answer» `:.` Mass of `""_(83)Bi^(209)=83xx1.007825+126xx1.008665` `=210.741265` a.m.u. Mass DEFECT of `""_(83)Bi^(209)=210.741265-208.980388` `=1.760877` a.m.u. `:.` Binding energy of `""_(83)Bi^(209)` `=1.760877xx931` M EV So B.E./nucleons `=(1.760877xx931)/(209)` `=7.844` M eV/A. |
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