(a) Explanation of information on the size of nucleus (b) Showing the independence of density on mass number (a) Many of the $\alpha$-particles pass through the foil. A few particles deflect by more than $90^{\circ}$. Rutherford argued that to deflect the $\alpha$-particles backward, it must experience a large repulsive force. It shows that most of the part of an atom is the empty space and its positive charge is concentrated tightly at its centre and its size is very small as compared to the size of atom. (nearly $\;\frac{1}{10,000}$ to $\;\frac{1}{10,000}$ times the size of atom) Alternatively, In Rutherford experiment, the calculation of distance of closest approach provides information about the size of the nucleus. Let $K$ be the initial kinetic energy of the alpha particle. At the distance of closest approach $\;\; \frac{1}{4\pi\epsilon_0} \; \frac{(Ze)(2e)}{a^2} = k$ $\; \therefore \;\; a = \frac{2Ze^2}{4\pi\epsilon_0 k}$ (b) Radius of the nucleus of mass number $A$, $R = R_0 A^{\frac{1}{3}}$, where, $R_0$ is constant. Volume of the nucleus, $V \;=\; \frac{4}{3} \pi R^3$ $\;=\; \frac{4}{3} \pi \left(R_0 A^{\frac{1}{3}}\right)^3$ $\;=\; \frac{4}{3} \pi A R_0^3$ $\text{Density}(\rho) \;=\; \frac{\; \text{ mass } \;}{\; \text{ volume } \;} \;=\; \frac{m A}{\left(\; \frac{4}{3} \pi R_0^3 A\right)}$ $\;=\; \frac{3m}{4\pi R_0^3}$ i.e., independent of mass number $A$.