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:
$m(_{20}^{40} Ca) = 39.962591 \; u$
$m(_{20}^{41} Ca) = 40.962278 \; u$
$m(_{13}^{26} Al) = 25.986895 \; u$
$m(_{13}^{27} Al) = 26.981541 \; u$
(Given mass of neutron $m_n = 1.008665 \; u$)

(N/A) The neutron separation energy $S_n$ is the energy required to remove a neutron from a nucleus,given by $S_n = [m(A-1, Z) + m_n - m(A, Z)] \times 931.5 \; MeV/u$.
For $_{20}^{41} Ca$:
Reaction: $_{20}^{41} Ca \rightarrow _{20}^{40} Ca + _{0}^{1} n$
Mass defect $\Delta m = m(_{20}^{40} Ca) + m_n - m(_{20}^{41} Ca)$
$\Delta m = 39.962591 + 1.008665 - 40.962278 = 0.008978 \; u$
$S_n = 0.008978 \times 931.5 \; MeV \approx 8.363 \; MeV$.
For $_{13}^{27} Al$:
Reaction: $_{13}^{27} Al \rightarrow _{13}^{26} Al + _{0}^{1} n$
Mass defect $\Delta m = m(_{13}^{26} Al) + m_n - m(_{13}^{27} Al)$
$\Delta m = 25.986895 + 1.008665 - 26.981541 = 0.014019 \; u$
$S_n = 0.014019 \times 931.5 \; MeV \approx 13.059 \; MeV$.