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The binding eneergy of 17CI^(35) nucleus...

The binding eneergy of `_17CI^(35)` nucleus is 298 MeV. Find its atomic mass. The mass of hyrogen atom `(_1H^(1))` is 1.008143 a.m.u and that of a neutron is 1.008986 a.m.u.Given 1 a.m.u=931MeV.

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The `_17CI^(35)` atom has 17 protons and 18 neutrons in its nucleus.Mass of `17(_1H^(1))` atom has 17 protons and 18 neutrons in its nucleus.Mass of 17 protons =`17xx 1.008143amu=17.138431amu`.Mass of 18 protons =`18xx 1.008986=18.161748`amu.Total =35.300179a.m.u.Mass defect `Delta m =(298)/(931)=0.320085a.m.u`.The atomic mass of `_17CI^(35)` would be the sum of equivalent of the binding energy of the nucleus.Hence atomic mass of `_17CI^(35)=35.300179-0.320085=34.980094a.m.u`.
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