The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nu
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 `._(20)^(41)Ca` and `._(13)^(27)Al` from the following data :
`m(._(20)^(40)Ca)=39.962591 u`
`m(._(20)^(41)Ca)=40.962278 u`
`m(._(13)^(26)Al)=25.986895 u`
`m(._(13)^(27)Al)=26.981541 u`
1 Answers
When a neutron is seperated from `._(20)Ca^(41)` we are left with
`._(20)Ca^(40) i.e., ._(20)Ca^(41)to ._(20)Ca^(40)+ ._(0)n^(1)`
Now mass defect
`Delta M = m(m_(20)Ca^(40))+m_(n)-m(._(20)Ca^(4))`
`= 39.962591+1.008665-40.962278`
= 0.008978 a.m.u
`therefore` Neutron sepearation energy
`= 0.008978xx931 MeV = 8.362 MeV`
similarly `._(13)Al^(27)to ._(13)Al^(26)+._(0)n^(1)`
`therefore` Mass defect, `Delta M = m(._(13)Al^(26))+m_(n)-m(._(13)Al^(27))`
`= 25.986895+1.008665-26.981541`
= 0.013845 u
`therefore` Neutron seperation energy `= 0.0138454xx931 MeV`
= 12.89 MeV