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 energy has to be supplied
Mass defect: `Deltam = M (._20^(39)Ca) +M(._0^(1)n)-M(._20^(40)Ca)`
`= 38.970691 +1.008665- 39.962589`
`= 0.016767 "amu`
Equivalent energy `= 0.016767 xx 931`
`= 15.6 MeV`