 Solution

In the supplied signed convention, and omitting pairing, the four contributions per nucleon areThe volume contribution is the constant $-15.7\mathrm{MeV}$. The positive surface correction decreases with $A$. At fixed proton fraction $Z/A$, the positive Coulomb correction grows as $A^{2/3}$, while the asymmetry correction is constant. Along the beta-stable valley the fraction $Z/A$ itself falls, and the asymmetry correction consequently grows.

Because the formula depends on both $A$ and $Z$, no unique plot against $A$ exists until a path $Z(A)$ is chosen. The original sketch below uses the continuous minimum of the nuclear Coulomb-plus-asymmetry terms,ignoring the small neutron-proton-electron rest-mass difference in locating that path. This is a stated smooth approximation to the valley of stability, not an assertion that nuclei have noninteger $Z$.

The four individual terms and their sum appear in one plot. A typical well-bound nucleus has $E_{\mathrm{bind}}/A\sim -8\mathrm{MeV}$ in the paper's convention, equivalently conventional positive binding energy per nucleon $B/A\sim +8\mathrm{MeV}$. If a positive-binding-energy plot is preferred, every plotted contribution must have its sign reversed. The competition between the surface and Coulomb penalties gives the familiar maximum of positive binding per nucleon near the iron-group nuclei.