What is the order of magnitude of the dimension of the nucleus as estimated by Rutherford?

In the nucleus of ${ }_{11} Na^{23}$,the number of protons,neutrons and electrons are

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

Given below are two statements $:$ one is labelled as Assertion $(A)$ and other is labelled as Reason $(R).$
Assertion $(A) :$ Neutrons penetrate matter more readily as compared to protons.
Reason $(R) :$ Neutrons are slightly more massive than protons.

If the nuclear forces between two protons,two neutrons,and a proton-neutron are denoted by $F_{pp}$,$F_{nn}$,and $F_{pn}$ respectively,then:

Nuclear reactions are given as $(i)$ $(n, p) {}_{16}S^{32} \to {}_{15}P^{32}$,$(ii)$ $(p, \alpha) {}_{8}O^{16} \to {}_{9}F^{19}$,$(iii)$ ${}_{7}N^{14} + ? \to {}_{6}C^{14} + {}_{1}H^{1}$. The missing particles or nuclides in these reactions are respectively: