Draw conclusions from the two main features of the graph of binding energy per nucleon versus the atomic mass number $(A)$.

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(N/A) The two main features of the binding energy per nucleon curve lead to the following conclusions:
$(i)$ The nuclear force is attractive and sufficiently strong to produce a binding energy per nucleon of approximately $8 \text{ MeV}$ for nuclei with $30 < A < 170$. This indicates that the nuclear force is short-ranged and exhibits the property of saturation.
$(ii)$ For very heavy nuclei $(A > 170)$,the binding energy per nucleon decreases. This occurs because the long-range Coulomb repulsion between protons becomes significant,reducing the stability. Consequently,if a heavy nucleus $(A = 240)$ splits into two lighter nuclei $(A = 120)$,the nucleons become more tightly bound,releasing energy. This process is known as nuclear fission.
$(iii)$ For very light nuclei $(A \leq 10)$,the binding energy per nucleon is low. When two light nuclei fuse to form a heavier nucleus,the resulting nucleus is more tightly bound,releasing energy. This process is known as nuclear fusion,which powers the Sun and hydrogen bombs.

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Which of the following statements is correct?

For a nucleus ${ }_Z^A X$ having mass number $A$ and atomic number $Z$:
$A.$ The surface energy per nucleon $(b_s) = a_1 A^{2/3}$
$B.$ The Coulomb contribution to the binding energy $b_c = -a_2 \frac{Z(Z-1)}{A^{4/3}}$
$C.$ The volume energy $b_v = a_3 A$
$D.$ Decrease in the binding energy is proportional to surface area.
$E.$ While estimating the surface energy,it is assumed that each nucleon interacts with $12$ nucleons,($a_1, a_2$ and $a_3$ are constants)
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