Calculate the energy required to remove the least tightly neutron form `.^20(Ca^(40))`. Given that
Mass of `.^20(Ca^(40)) = 39.962589 amu`
Mass of `.^20(Ca^(39)) = 38.970691 amu`
Mass of neutron `= 1.008665 amu`

Explanation

Here in order to remove a neutron energy has to be supplied
Mass defect ,` Delta m=m(""_(20)^(39)Ca)+M*""_(0)^(1)n)-M(""_(20)^(40)Ca )`
`=38.970691 +1.008665-39.962589`
`=0.016767 amu`
Equivalent energy `=0.6767xx931 (:' amu =931Me v)`
`=15.6 Me V `