From the given data, the amount of energy required to break the nucleus of aluminium _13^27 Al is \ \ \ \ x × 10^-3 J. Mass of neutron =1.00866 u…

From the given data, the amount of energy required to break the nucleus of aluminium $\displaystyle { }_{13}^{27} \mathrm{Al}$ is \_\_\_\_ $\displaystyle x \times 10^{-3} \mathrm{~J}$.Mass of neutron $\displaystyle =1.00866 \mathrm{u}$ Mass of proton $\displaystyle =1.00726$ u Mass of Aluminium nucleus $\displaystyle =27.18846 \mathrm{u}$ (Assume $\displaystyle 1$ u corresponds to $\displaystyle x$ $\displaystyle J$ of energy) (Round off to the nearest integer)