From the given data, the amount of energy required break the nucleus of aluminium ${ }_{13}^{27} \mathrm{Al}$ is $\_\_\_\_$ $x \times 10^{-3} \mathrm{~J}$. Mass of neutron $=1.00566 \mathrm{u} \mid$,br> Mass of proton $=1.00726 \mathrm{u}$ Mass of Aluminium nucleus = 27.18846 u (Assume 1 u corresponds to xJ of energy) (Round off to the nearest integer)