Stability Curve and Nuclear Forces in Heavy Nuclides

To determine the correct option, let us analyze both statements in detail:

Analysis of STATEMENT-1:

• Consider a Cartesian coordinate system where the number of neutrons $N$ is plotted along the $x$-axis and the atomic number (number of protons) $Z$ is plotted along the $y$-axis.
• A line with a $45^\circ$ slope passing through the origin is represented by the equation:
  $y = x \implies Z = N$
• For light stable nuclei (typically $Z \le 20$), the number of neutrons is approximately equal to the number of protons ($N \approx Z$), so the stability curve lies along the line of $45^\circ$ slope.
• As the atomic number $Z$ increases, stable nuclides require a higher proportion of neutrons to maintain stability, resulting in $N > Z$ (i.e., the neutron-to-proton ratio $\frac{N}{Z} > 1$).
• Since the $x$-coordinate ($N$) increases more rapidly than the $y$-coordinate ($Z$), the plot of $Z$ versus $N$ bends or curves towards the $x$-axis away from the line $Z = N$.
• Therefore, STATEMENT-1 is True.

Analysis of STATEMENT-2:

• The strong nuclear force is attractive and acts between all nucleons (proton-proton, neutron-neutron, and proton-neutron). However, it is a short-range force and exhibits saturation, meaning a nucleon interacts only with its nearest neighbors. Consequently, the total attractive nuclear energy is roughly proportional to the mass number $A$.
• In contrast, the electrostatic (Coulomb) repulsive force between protons is long-range and acts between every pair of protons, scaling approximately as $Z(Z-1) \propto Z^2$.
• In heavier nuclides (large $Z$), this cumulative long-range proton-proton electrostatic repulsion increases significantly and begins to compete with and threaten to overcome the attractive nuclear binding forces.
• To maintain stability and prevent the nucleus from flying apart, an excess of neutrons is required. These extra neutrons provide additional attractive strong nuclear forces without contributing to electrostatic repulsion, diluting the proton density and stabilizing the heavy nucleus.
• Thus, the onset of strong proton-proton electrostatic repulsions overcoming attractive forces in heavier nuclides directly explains why more neutrons ($N > Z$) are needed, which in turn causes the stability curve to bend towards the neutron ($x$) axis.
• Therefore, STATEMENT-2 is True, and it is the correct explanation for STATEMENT-1.

Hence, the correct option is (A).