$A$ nucleus ruptures into two nuclear parts which have their velocity ratio equal to $2 : 1$. What will be the ratio of their nuclear size (nuclear radius)?

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 range of the nuclear force is

State the dimension of the nucleus from the Rutherford experiment.

The radius of ${ }_{52} Te^{125}$ nucleus is $6 \text{ fermi}$. The radius of ${ }_{13} Al^{27}$ nucleus in meters is

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: The nuclear density of nuclides ${ }_{5}^{10} B, { }_{3}^{6} Li, { }_{26}^{56} Fe, { }_{10}^{20} Ne$ and ${ }_{83}^{209} Bi$ can be arranged as $\rho_{ Bi }^{ N } > \rho_{ Fe }^{ N } > \rho_{ Ne }^{ N } > \rho_{ B }^{ N } > \rho_{ Li }^{ N }$.
Reason $R$: The radius $R$ of a nucleus is related to its mass number $A$ as $R = R_0 A^{1/3}$,where $R_0$ is a constant.
In the light of the above statements,choose the correct answer from the options given below: