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:
$m\left( {{}_{13}^{26}Al} \right) = 25.986895u$
$m\left( {{}_{13}^{27}Al} \right) = 26.981541u$

Answer 1

Hint: The aluminium loses one neutron from its nucleus and the difference in the mass of the atom releases an energy. The neutron separation energy is the term given to this energy. The unit of this energy is generally written as eV.
Formula used:
\[E = \left\{ {\left( {m\left( {{}_{13}^{27}Al} \right)} \right) + \left( {m\left( {{}_0^1n} \right)} \right) - \left( {m\left( {{}_{13}^{27}Al} \right)} \right)} \right\}{c^2}\]
Where $m\left( {{}_{13}^{26}Al} \right)$ is the mass of aluminium after the removal of neutrons.
$m\left( {{}_{13}^{27}Al} \right)$ is the mass of aluminium before the removal of the neutron.
\[m\left( {{}_0^1n} \right)\] is the mass of a neutron.
C is the speed of light in vacuum
E is the separation energy

Complete step by step solution:
In the following nuclear reaction, the neutron of the aluminium atom is released by applying an external force, normally the neutron is strongly bound to the nucleus, when the structure of the nucleus is changed by removal of neutrons. A lot of energy is released because the binding energy of the ${}_{13}^{26}Al$ is less than the binding energy of the ${}_{13}^{27}Al$, the difference in this energy can be calculated by the famous equation by Einstein-
 $E = \Delta m{c^2}$
Where $\Delta m$is the overall change in the mass of the system,
And c is the speed of light.
To calculate the difference in the total mass of the system, the masses of the ${}_{13}^{26}Al$and a neutron are added up. Then the mass of the ${}_{13}^{27}Al$,which is the mass when the neutron resides inside the atomic nucleus is subtracted.
This difference in the total mass is then multiplied with the square of the speed of light in vacuum and the separation energy is calculated.
We have,
$m\left( {{}_{13}^{26}Al} \right) = 25.986895u$
$m\left( {{}_{13}^{27}Al} \right) = 26.981541u$
It is known that mass of a neutron is \[1.008665u\]and
Therefore the equation,
\[E = \left\{ {\left( {m\left( {{}_{13}^{27}Al} \right)} \right) + \left( {m\left( {{}_0^1n} \right)} \right) - \left( {m\left( {{}_{13}^{27}Al} \right)} \right)} \right\}{c^2}\]
On putting the values-
$E = (25.986895 + 1.008665 - 26.981541){c^2}$
$E = (0.014019){c^2}$
Converting the units from u to eV-
We know that,
$1u = 931.5MeV/{c^2}$
Hence,
\[E = (0.014019){c^2} \times \dfrac{{931.5MeV}}{{{c^2}}}\]
$E = 13.059MeV$
The separation energy of $13.059MeV$ is required to separate the neutron from the nucleus.

Note: The value of separation energy is positive, which means to separate the electron from the nucleus, this energy has to be provided to the nucleus. For heavier unstable elements like Uranium and Thorium, this energy is negative which signifies release of energy.