Rutherford suggested that since 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 two $+2 e$ charge 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 it is assume that 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 the Coulomb's law for force of repulsion between the $\alpha$-particle and the positively charged nucleus.

The magnitude of Coulomb's repulsive force,

$\mathrm{F}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{(\mathrm{Z} e)(2 e)}{r^{2}}$

where $r$ 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 $\alpha$-particle.