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?

The diagram shows the path of four $\alpha$-particles of the same energy being scattered by the nucleus of an atom. Which of these is/are not physically possible?

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.

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 $... \times 10^5 \ m/s$.
$\left(\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \ SI \ unit, \text{mass of } \alpha \text{-particle } m = 6.72 \times 10^{-27} \ kg, e = 1.6 \times 10^{-19} \ C\right)$

What is shown by Thomson's experiments of electric discharge through gases? And explain the plum pudding model.

$A$ neutron collides head-on with a stationary hydrogen atom in its ground state. Which of the following statements are correct? (Assume that the hydrogen atom and neutron have the same mass):