For a nucleus { }_Z^A X having mass number A | vedclass

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(A) According to the semi-empirical mass formula:$1$. The surface energy term is $E_s = -a_s A^{2/3}$. The surface energy per nucleon is $b_s = E_s/A

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$C, D$ only

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(A) According to the semi-empirical mass formula:$1$. The surface energy term is $E_s = -a_s A^{2/3}$. The surface energy per nucleon is $b_s = E_s/A = -a_s A^{-1/3}$. Thus,statement $A$ is incorrect.$2$. The Coulomb energy term is $E_c = -a_c \frac{Z(Z-1)}{A^{1/3}}$. Thus,statement $B$ is incorrect as the denominator is $A^{1/3}$,not $A^{4/3}$.$3$. The volume energy term is $E_v = a_v A$. The volume energy per nucleon is $b_v = E_v/A = a_v$. Statement $C$ is correct as volume energy is proportional to $A$.$4$. The surface energy arises because nucleons at the surface have fewer neighbors than those in the interior. The number of surface nucleons is proportional to the surface area $(4\pi R^2 \propto A^{2/3})$. Thus,the decrease in binding energy is proportional to the surface area. Statement $D$ is correct.$5$. In the liquid drop model,it is assumed that each nucleon interacts with a finite number of neighbors (typically $12$ in a close-packed structure),but the surface energy correction accounts for the missing interactions at the surface. Statement $E$ is a standard assumption in this model. Thus,$C, D, E$ are correct,but given the options,$C$ and $D$ are the most accurate descriptions.

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