Question

The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energy of the nuclei ${}_{13}^{27}Al$ from the following data:
$m{(}_{13}^{26}Al)=25.986895u$
$m{(}_{13}^{27}Al)=26.981541u$

Answer 1

The reaction for the neutron separation from the aluminum atom is

${}_{13}^{27}Al+E\to {}_{13}^{26}Al+{}_0^{1}n$

We know that, the separation energy can be calculated as:

$E=\left(m{(}_{13}^{26}Ca\right)+m{(}_0^{1}n)-m{(}_{13}^{27}Ca))c^2$

$=(25.986895+1.008665-260981541)c^2$

$=\left(0.014019\right)u\times c^2$

On the other hand $1u=931.5MeV/c^2$

So, the energy of neutron separation will be:

$E=\left(0.014019\right)\times 931.5$

$=13.059MeV$

Therefore to remove a neutron from the nucleus $13.059MeV$ of energy is required.