An $\alpha$-particle of energy $5 \, MeV$ is scattered by a fixed uranium nucleus at $180^o$. What is the distance of closest approach between the particle and the uranium nucleus?

In a Rutherford scattering experiment,when a projectile of charge $Z_1$ and mass $M_1$ approaches a target nucleus of charge $Z_2$ and mass $M_2$,the distance of closest approach is $r_0$. The energy of the projectile is:

Choose the correct option from the following options given below.

Describe the Geiger-Marsden scattering experiment.

$A$ $7.9 \text{ MeV}$ $\alpha$-particle scatters from a target material of atomic number $Z = 79$. From the given data, the estimated diameter of the nucleus of the target material is (approximately) . . . . . . $\text{m}$.
$\left[\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2\right.$ and electron charge $\left.e = 1.6 \times 10^{-19} \text{ C}\right]$

$\sqrt{d_{1}}$ 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 . . . . . .