The electric potential at the surface of an atomic nucleus $(Z=50)$ of radius $9 \times 10^{-15} \,m$ is

If the mass number of an atom is $A = 40$ and its electron configuration is $1s^2, 2s^2, 2p^6, 3s^2, 3p^6$,the number of neutrons and protons in its nucleus will be

Nuclei with magic numbers of protons $Z = 2, 8, 20, 28, 50, 82$ and magic numbers of neutrons $N = 2, 8, 20, 28, 50, 82, 126$ are found to be very stable.
$(i)$ Verify this by calculating the proton separation energy $S_p$ for $^{120}Sn$ $(Z = 50)$ and $^{121}Sb$ $(Z = 51)$. The proton separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given by $S_p = (M_{Z-1, N} + M_H - M_{Z, N})c^2$. Given:
$^{119}In = 118.9058 \ u, ^{120}Sn = 119.902199 \ u, ^{121}Sb = 120.903824 \ u, ^1H = 1.0078252 \ u$
$(ii)$ What does the existence of magic numbers indicate?

The radius of a nucleus of a mass number $A$ is directly proportional to

Which of the following is a fundamental force in nature?

$STATEMENT-1$: 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.