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)
$\Delta \mathrm{m}=\left(Z \mathrm{~m}_{\mathrm{P}}+(\mathrm{A}-\mathrm{Z}) \mathrm{m}_{\mathrm{n}}\right)-\mathrm{M}_{\mathrm{A} \ell}$
$=(13 \times 1.00726+14 \times 1.00866)-27.18846$
$=27.21562-27.18846$
$=0.02716 \mathrm{u}$
$\mathrm{E}=27.16 \mathrm{x} \times 10^{-3} \mathrm{~J}$
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