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 \({ }_{20}^{41} \mathrm{Ca}\) and  \({ }_{13}^{27} \mathrm{Al}\) from the following data:

m(\({ }_{20}^{40} \mathrm{Ca}\)) = 39.962591 u

m(\({ }_{20}^{41} \mathrm{Ca}\)) = 40.962278 u

m(\({ }_{13}^{26} \mathrm{Al}\)) = 25.986895 u

m(\({ }_{13}^{27} \mathrm{Al}\)) = 26.981541 u

Hint: The energy required is given by \(E=\Delta mc^2\).
Step 1: Find the neutron separation energies of the nucleus \({ }_{20}^{41} \mathrm{Ca}\).
ForC2041a:Separationenergy=8.363007MeV
ForA1327l:Separationenergy=13.059MeV
Ifaneutronn01isremovedfromC2041a,thecorrespondingnuclearreactioncanbewrittenas:
C2041aC2040a+n01
Itisgiventhat:
mC2040a=39.962591u
mC1327a=40.962278u
mn01=1.008665u
Themassdefectofthisreactionisgivenas:
m=mC2040a+n01-mC2041a
=39.962591+1.008665-40.962278=0.008978u
But1u=931.5MeV/C2
m=0.008978×931.5MeV/c2
Hence,theenergyrequiredforneutronremovaliscalculatedas:
E=mc2
=0.008978×931.5=8.363007MeV
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:\({ }_{13}^{27} Al \rightarrow{ }_{13}^{26} Al+{ }_{0}^{1} n\)
Itisgiventhat:
mAl1326=25.986895u
mAl1327=26.981541u
Themassdefectofthisreactionisgivenas:
m=mA1326l+mn01-mA1327l
=25.986895+1.008665-26.981541
=0.014019u
=0.01409×931.5MeV/c2
Hence,theenergyrequiredforneutronremovaliscalculatedas:
E=mc2
=0.014019×931.5=13.059MeV