Question

The neutron separation energy is defined as the energy required toremove a neutron from the nucleus. Obtain the neutron separationenergies of the nuclei 4120Ca and 2713Al from the following data:m(4020Ca ) = 39.962591 um(4120Ca ) = 40.962278 um(2613Al ) = 25.986895 um(2713Al ) = 26.981541 u

Answer 1

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.