Explain Rutherford's explanation for scattered $\alpha$-particles.

(N/A) Rutherford suggested that since a large number of $\alpha$-particles are scattered at very small angles,atoms must be largely hollow.
Since the gold foil is very thin,it can be assumed that $\alpha$-particles will suffer not more than one scattering during their passage through it.
$\alpha$-particles are nuclei of helium atoms and carry a charge of $+2e$ and have the mass of the helium atom.
For gold $Z=79$,the nucleus of gold is about $50$ times heavier than an $\alpha$-particle,so it is assumed to be stationary throughout the scattering process.
Under these assumptions,the trajectory of an $\alpha$-particle can be computed using Newton's second law of motion and Coulomb's law for the force of repulsion between the $\alpha$-particle and the positively charged nucleus.
The magnitude of Coulomb's repulsive force is given by:
$F = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{(Ze)(2e)}{r^{2}}$
where $r$ is the distance between the $\alpha$-particle and the nucleus,and $\epsilon_{0}$ is the permittivity of vacuum.
The force is directed along the line joining the $\alpha$-particle and the nucleus and varies continuously with the displacement of the $\alpha$-particle.