From the given data,
Question:

From the given data, the amount of energy required to break the nucleus of aluminium ${ }_{13}^{27} \mathrm{Al}$ is ____________$\mathrm{x} \times 10^{-3} \mathrm{~J}$.

Mass of neutron $=1.00866 \mathrm{u}$

Mass of proton $=1.00726 \mathrm{u}$

Mass of Aluminium nucleus $=27.18846 \mathrm{u}$

(Assume $1 \mathrm{u}$ corresponds to $\mathrm{x} \mathrm{J}$ of energy)

(Round off to the nearest integer)

Solution:

$\Delta \mathrm{m}=\left(Z \mathrm{~m}_{\mathrm{P}}+(\mathrm{A}-\mathrm{Z}) \mathrm{m}_{\mathrm{n}}\right)-\mathrm{M}_{\mathrm{A} \ell}$

$=(13 \times 1.00726+14 \times 1.00866)-27.18846$

$=27.21562-27.18846$

$=0.02716 \mathrm{u}$

$\mathrm{E}=27.16 \mathrm{x} \times 10^{-3} \mathrm{~J}$