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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.

Answer»



Solution :`""_(83)Bi^(209)` has 83 protons and 126 neutrons `(209-83=126)`
`:.` 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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