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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 `" "_(A)^(Z)X` is given by `S_(n) = [ m_(N)(" "_(A-1)^(Z)X) + m_(n) - m_(N) (" "_(A)^(Z)X)].c^(2)` Hence, in terms of atomic masses, the neutron separation energy of `" "_(20)^(41)Ca` is given by `S_(n)(" "_(20)^(41)Ca) = [m(" "_(20)^(40)Ca) + m_(n)-m(" "_(20)^(41)Ca)] xx 931.5 MEV = [39.962591 + 1.008665 - 40.962278] xx 931.5 MeV = 8.36 MeV` and neutron separation energy of `" "_(13)^(27)Al` is given by `S_(n)(" "_(13)^(27)Al) = [m(" "_(13)^(26)Al) + m_(n) -m(" "_(13)^(27)Al)] xx 931.5 MeV= [25.986895 + 1.008665 - 26.981541] xx 931.5 MeV = 13.06 MeV`. |
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