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

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(C) \(27\)

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Detailed explanation

\(\Delta m =\left( Zm m _{ P }+( A - Z ) m _{ n }\right)- M _{ Al}\)

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

\(=27.21562-27.18846\)

\(=0.02716 u\)

\(E =27.16 \times 10^{-3} J\)

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