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The neutron separation energy is defined...

The neutron separation energy is defined to be the energy required to remove a neutron form nucleus. Obtain the neutron separation energy of the nuclei `._(20)Ca^(41)` and `._(13)Al^(27)` from the following data : `m(._20Ca^(40))=39.962591u` and `m(._(20)Ca^(41))=40.962278u`
`m(._(13)Al^(26))=25.986895u` and `m(._(13)Al^(27))=26.981541u`

Text Solution

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Neutron separation energy `S_(n)` of a nucleus `""_(X)X^(A)` is given by
`S_(n) =[m_(N)(""_(Z)X^(A-1)) + m_(n) -m_(N)(""_(Z)X^(A))]c^(2)`
Adding and subtracting the term Zme in the bracket above, ignoring mass defects due to electronic binding energies, we get `S_(n)` in terms of atomic masses.
`S_(n) =[m(""_(z)X^(A-1))+m_(n)-m(""_(z)X^(A))]c^(2)`
When a neutron is separated from `(""_(20)Ca^(41))`, we get
`""_(20)Ca^(41) to ""_(20)Ca^(40) + ""_(0)n^(1)`
`therefore S_(n)(""_(20)Ca^(41)) = (39.962591 + 1.008665- 40.962278)u`.
`=0.008978 xx 931.5=8.363` MeV
When a neutron is separated energy from `""_(13)Al^(27)`, we get
`S_(n)(""_(13)Al^(27)) =[m_(N)(""_(13)Al^(26)) + m_(n)-m_(N)(""_(13)Al^(27))]c^(2)`
=13.06 MeV
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