The neutron separation energy is defined as the energy required to remove a neutron from the nucleus.

Solution:

For ${ }_{20}^{41} \mathrm{Ca}$ : Separation energy $=8.363007 \mathrm{MeV}$

For ${ }_{13}^{27} \mathrm{Al}$ : Separation energy $=13.059 \mathrm{MeV}$

A neutron $\left({ }_{0} n^{1}\right)$ is removed from $\mathrm{a}_{20}^{41} \mathrm{Ca}$ nucleus. The corresponding nuclear reaction can be written as:

${ }_{20}^{41} \mathrm{Ca} \longrightarrow{ }_{20}^{40} \mathrm{Ca}+{ }_{0}^{1} \mathrm{n}$

It is given that:

Mass $m\left({ }_{20}^{40} \mathrm{Ca}\right)=39.962591 \mathrm{u}$

Mass $\left.m\left({ }_{20}^{41} \mathrm{Ca}\right)\right)=40.962278 \mathrm{u}$

Mass $m\left({ }_{0} n^{1}\right)=1.008665 \mathrm{u}$

The mass defect of this reaction is given as:

$\Delta m=m\left({ }_{20}^{40} \mathrm{Ca}\right)+\left({ }_{0}^{1} \mathrm{n}\right)-m\left({ }_{20}^{41} \mathrm{Ca}\right)$

$=39.962591+1.008665-40.962278=0.008978 \mathrm{u}$

But $1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2}$

∴Δm = 0.008978 × 931.5 MeV/c2

Hence, the energy required for neutron removal is calculated as:

$E=\Delta m c^{2}$

$=0.008978 \times 931.5=8.363007 \mathrm{MeV}$

For ${ }_{13}^{27} \mathrm{Al}$, the neutron removal reaction can be written as:

${ }_{13}^{27} \mathrm{Al} \longrightarrow{ }_{13}^{26} \mathrm{Al}+{ }_{0}^{1} \mathrm{n}$

It is given that:

Mass $m\left({ }_{13}^{27} \mathrm{Al}\right)=26.981541 \mathrm{u}$

Mass $m\left({ }_{13}^{26} \mathrm{Al}\right)=25.986895 \mathrm{u}$

The mass defect of this reaction is given as:

$\Delta m=m\left({ }_{13}^{26} \mathrm{Al}\right)+m\left({ }_{0}^{1} \mathrm{n}\right)-m\left({ }_{13}^{27} \mathrm{Al}\right)$

$=25.986895+1.008665-26.981541$

$=0.014019 \mathrm{u}$

$=0.014019 \times 931.5 \mathrm{MeV} / c^{2}$

Hence, the energy required for neutron removal is calculated as:

$E=\Delta m c^{2}$

$=0.014019 \times 931.5=13.059 \mathrm{MeV}$