If the number of scattered alpha particles is $56$ at a scattering angle of ${90^o}$,then the number of scattered particles at a ${60^o}$ angle will be:

$\sqrt{d_{1}}$ and $\sqrt{d_{2}}$ are the impact parameters corresponding to scattering angles $60^{\circ}$ and $90^{\circ}$ respectively,when an $\alpha$-particle is approaching a gold nucleus. For $d_{1} = x d_{2}$,the value of $x$ will be . . . . . .

What is a head-on collision? For that,state the impact parameter.

The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy $K$ is $r_{0}$. The distance of the closest approach when the $\alpha$-particle is fired at the same nucleus with kinetic energy $2K$ will be . . . . . . .

An alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to

Given below are two statements:
$Statement$ $I$: Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it,is Rutherford's model.
$Statement$ $II$: An atom is a spherical cloud of positive charges with electrons embedded in it,is a special case of Rutherford's model.
In the light of the above statements,choose the most appropriate from the options given below.