$A$ nucleus with energy $\frac{1}{2}mv^2$ is fired at a target nucleus of charge $Ze$. The distance of closest approach to the $Ze$ nucleus is proportional to which of the following?

In an alpha particle scattering experiment,the distance of closest approach for the $\alpha$-particle is $4.5 \times 10^{-14} \ m$. If the target nucleus has an atomic number $Z = 80$,then the maximum velocity of the $\alpha$-particle is approximately $... \times 10^5 \ m/s$.
$\left(\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \ SI \ unit, \text{mass of } \alpha \text{-particle } m = 6.72 \times 10^{-27} \ kg, e = 1.6 \times 10^{-19} \ C\right)$

The graph which depicts the results of the Rutherford gold foil experiment with $\alpha$-particles is:
$\theta$: Scattering angle
$Y$: Number of scattered $\alpha$-particles detected
(Plots are schematic and not to scale)

Based on which experiment did the Rutherford nuclear model come from?

An $\alpha$ particle is accelerated through a potential difference of $10^6 \ V$. Then the kinetic energy $(K.E.)$ of the particle will be .......... $MeV$.

Assertion $(A)$: The impact parameter for $\alpha$-particles scattered by $180^{\circ}$ is zero.
Reason $(R)$: Zero impact parameter means that the $\alpha$-particles tend to hit the centre of the nucleus.