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

According to classical theory,the Rutherford atom was:

In Rutherford's experiment,the number of particles scattered at a $90^{\circ}$ angle is $x$ per second. The number of particles scattered per second at an angle of $60^{\circ}$ is:

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 . . . . . . .

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.

An $\alpha$-particle of energy $4 \text{ MeV}$ is scattered through $180^\circ$ by a fixed uranium nucleus. The distance of the closest approach is of the order of