1.

Calculate binding energy per nucleon of deuteron whosemass is 2.013554 a.m.u. Take mass of proton and neutron as 1.007825 a.m.u. and 1.008665 a.m.u. respectively and 1 a.m.u. =931 M eV.

Answer»



Solution :Mass defect of `""_(1)H^(2)=m_(p)+m_(n)-m_("nucleus")[""_(1)H^(2)]`
`=[1.007825+1.008665-2.013554]` a.m.u.
`=[2.01649-2.013554]` a.m.u.
`=0.002936 XX 931=2.733` M eV
`:.` Binding ENERGY of DEUTERON
`=0.002936xx931=2.733` M eV
`:.` Binding energy PER nucleon of deuteron
`=(2.733)/(2)=1.36` M eV


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