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 4120Ca and 2713Al from the following data: m(4020Ca ) = 39.962591 u m(4120Ca ) = 40.962278 u m(2613Al ) = 25.986895 u m(2713Al ) = 26.981541 u
The reaction for the neutron separation from the aluminum atom is 1327Al+E→ 1326Al+ 01n We know that, the separation energy can be calculated as: E=(m(1326Ca)+m(01n)−m(1327Ca))c2 =(25.986895+1.008665−260981541)c2 =(0.014019)u×c2 On the other hand 1u=931.5MeV/c2 So, the energy of neutron separation will be: E=(0.014019)×931.5 =13.059MeV Therefore to remove a neutron from the nucleus 13.059MeV of energy is required.
