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(C) The mass defect $\Delta m$ is calculated as: $\Delta m = (Z m_p + (A - Z) m_n) - M_{Al}$.Here,$Z = 13$ (number of protons) and $A - Z = 14$ (numbe

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$27$

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(C) The mass defect $\Delta m$ is calculated as: $\Delta m = (Z m_p + (A - Z) m_n) - M_{Al}$.Here,$Z = 13$ (number of protons) and $A - Z = 14$ (number of neutrons).$\Delta m = (13 	imes 1.00726 + 14 	imes 1.00866) - 26.98154$.$\Delta m = (13.09438 + 14.12124) - 26.98154$.$\Delta m = 27.21562 - 26.98154 = 0.23408 \, {u}$.Given that $1 \, {u}$ corresponds to $1 \, {J}$ of energy,the binding energy $E = 0.23408 \, {J}$.To express this in the form $x 	imes 10^{-3} \, {J}$,we have $E = 234.08 	imes 10^{-3} \, {J}$.Note: The provided mass of the aluminium nucleus in the prompt $(27.18846 \, {u})$ is physically incorrect as it exceeds the sum of its constituent nucleons. Using the standard mass of ${ }_{13}^{27} {Al}$ $(26.98154 \, {u})$,the result is approximately $234$. If we follow the prompt's specific (though incorrect) mass value,$\Delta m = 27.21562 - 27.18846 = 0.02716 \, {u}$,which gives $27.16 	imes 10^{-3} \, {J}$. Rounding to the nearest integer,we get $27$.

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Binding energy per nucleon of a fixed nucleus $X^A$ is $8 \ MeV$. It absorbs a neutron moving with kinetic energy $2 \ MeV$ and converts into $Y$,emitting two photons of energy $1 \ MeV$ and $4 \ MeV$ respectively one after the other. The binding energy per nucleon of $Y$ (in $MeV$) is:

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