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The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei ""_(20)^(41)Ca and ""_(13)^(27)Al from the following data: m(""_(20)^(40)Ca) = 39.962591 u m(""_(20)^(41)Ca) = 40.962278 u m(""_(13)^(26)Al) = 25.986895 u m (""_(13)^(27)Al) = 26.981541 u. |
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Answer» Solution :NEUTRON SEPARATION energy `S_(n)` of a nucleus `""_(Z)^(A)X` is `S(n)=[m_(N)(""_(Z)^(A-1)X)+m(n)-m_(N)(""_(Z)^(A)X)]c^(2)` From given data, `S(n)(""_(20)^(51)Ca)=8.36MeV, S(n)(""_(13)^(27)Al)=13.06MeV` |
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