Given, m( C 20 40 a )=39.962591 u, m( C 20 41 a )= 40.962278 u, m( A 13 26 l ) =25.986895 u, and m( A 13 27 l ) = 26.981541 u.
The separation energy for C 20 41 a is 8.363007 MeV and the separation energy for A 13 27 l is 13.059 MeV.
When a neutron is removed from a C 20 41 a atom, then its nuclear reaction may be written as,
C 20 41 a→ C 20 41 a+ n 0 1
The mass defect of this reaction is given as,
Δ m Ca =m( C 20 40 a )+( n 0 1 )−m( C 20 41 a ) =39.962591+1..008665−40.962278 =0.008978 u
As 1 u= 931.5 c 2 MeV, so
Δ m Ca =( 0.008978 u )( 931.5 MeV/c 2 1 u ) = 8.363007 c 2 MeV
Thus, the energy required for neutron removal is calculated as:
E Ca =Δ m Ca c 2
Substituting the value of Δ m Ca in the above equation, we get:
E Ca =Δ m Ca c 2 = 8.363007 c 2 c 2 =8.363007 MeV
Hence, the neutron separation energies of the nuclei of C 20 41 a is 8.363007 MeV.
Now, when a neutron is removed from an A 13 27 l nucleus its nuclear reaction is given by,
A 13 27 l→ A 13 26 l+ n 0 1
The mass defect of this reaction is,
Δ m Al =m( A 13 26 l )+( n 0 1 )−m( A 13 27 l ) =25.986895+1.008665−26.981541 =0.014019 u
The expression for energy required for neutron removal is,
E Al =Δ m Al c 2
Substituting the value of Δ m Al in above equation, we get:
E=Δ m Al c 2 =0.014019×931.5 =13.059 MeV
Hence, the neutron separation energies of the nuclei of A 13 27 l is 13.059 MeV.