The neutron separation energy is defined to be the energy required to remove a neutron form nucleus. Obtain the neutron separation energy of the nuclei `._(20)Ca^(41)` and `._(13)Al^(27)` from the following data : `m(._20Ca^(40))=39.962591u` and `m(._(20)Ca^(41))=40.962278u`
`m(._(13)Al^(26))=25.986895u` and `m(._(13)Al^(27))=26.981541u`

Explanation

For `""_(20^(41)Ca:` separtion energy = 8.363007 MeV
for `""_(13)^(27)Al:` separation energy =13.059 MeV
A nuetron `(""_(0)n^(1))` is removed from a `""_(20)^(41)Ca ` nucleus the corresponding nuclear reaction can be written as:
`""_(20)^(41)Ca to ""_(20)^(40)Ca +""_(0)^(1)n`
it is given that :
mass `m(""_(20^(41)Ca)=39.962591u`
mass `m(""_(20)^(41)Ca) = 40.962278u`
mass`m(""_(0)n^(1)) =1.008665 u`
the mass defect of this racation is given as :
`Delta m= m(""_(20)^(40)Ca)+(""_(0)^(1)n)-m(""_(20)^(41)Ca)`
39.962591+1.008665-40.962278=0.008978 u
But 1 `u= 931.5 MeV//c^(2)`
`therefore Delta m =0.008978 xx931.5 MeV//c^(2)`
Hence the energy reuired for neutron removal is calculated as :
`E=Delta mc^(2)`
`=0.008978xx931.5=8.363007 meV `
For` ""_(13)^(27)Al` the neutron removal rection can be written as:
`""_(13)^(27)Al to ""_(13)^(26)Al +""_(0)^(1)n`
it is given that :
mass `m(""_(13)^(27)Al)=26.981541 u`
mass `m(""_(13)^(26)Al)=25.986895 u`
the mass defect of this reaction isgiven as :
`Delta m=m(""_(13)^(26)Al)+m(""_(0)^1)n)-m(""_(13)^(27)Al)`
`=25.986895+1.008665-26.981541`
`=0.014019 u`
`=0.014019 xx931.2MeV.//C^(2)`
Hence the energy required for netron removal is calculated as :
`E=Delta mc^(2)`
`=0.014019xx931.5 =13.059 MeV`