$A$ free neutron decays into a proton,but a free proton does not decay into a neutron. This is because

If the radius of the $^{27}_{13}Al$ nucleus is taken to be $R_{Al}$,then the radius of the $^{125}_{53}Te$ nucleus is nearly

The mass of a neutron is approximately the same as that of:

If the radii of ${}_{13}^{27}Al$ and ${}_{30}^{64}Zn$ nuclei are $R_{1}$ and $R_{2}$ respectively,then $\frac{R_{1}}{R_{2}} = $

Ratio of specific charge of $\alpha$-particle to that of proton is

$STATEMENT-1$: The plot of atomic number ($y$-axis) versus number of neutrons ($x$-axis) for stable nuclei shows a curvature towards the $x$-axis from the line of $45^{\circ}$ slope as the atomic number is increased.
$STATEMENT-2$: Proton-proton electrostatic repulsions begin to overcome attractive nuclear forces in heavier nuclides,requiring more neutrons to maintain stability.