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13.24 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 4120Ca and  2713Al from the following data:

m(4020Ca) = 39.962591 u

m(4120Ca) = 40.962278 u

m(2613Al) = 25.986895 u

m(2713Al) = 26.981541 u

Hint: The energy required is given by E=Δmc2.
Step 1: Find the neutron separation energies of the nucleus 4120Ca.
For C4120a:Separation energy=8.363007 MeV
For A2713l:Separation energy=13.059 MeV
If a neutron n10 is removed from C4120a, the corresponding nuclear reaction can be written as:
                           C4120aC4020a+n10
It is given that:
m(C4020a)=39.962591 u
m(C2713a)=40.962278 u
m(n10)=1.008665 u
The mass defect of this reaction is given as:

=39.962591+1.008665-40.962278=0.008978 u
But 1 u =931.5 MeV/C2
m=0.008978×931.5 MeV/c2
Hence, the energy required for neutron removal is calculated as:
E=mc2
=0.008978×931.5=8.363007 MeV
Step 2: Find the neutron separation energies of the nucleus { }_{13}^{27} \mathrm{Al}.
For { }_{13}^{27} \mathrm{Al}, the neutron removal reaction can be written as:{ }_{13}^{27} Al \rightarrow{ }_{13}^{26} Al+{ }_{0}^{1} n
It is given that:
mAl1326=25.986895 u
mAl1327=26.981541 u
The mass defect of this reaction is given as:
m=mA1326l+mn01-mA1327l
=25.986895+1.008665-26.981541
=0.014019 u
=0.01409×931.5 MeV/c2
Hence, the energy required for neutron removal is calculated as:
E=mc2
=0.014019×931.5=13.059 M e V