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?

An $\alpha$-particle of energy $5 \text{ MeV}$ is moving forward for a head-on collision. The distance of closest approach from the nucleus of atomic number $Z=50$ is . . . . . . $\times 10^{-14} \text{ m}$.
$(k=9 \times 10^{9} \text{ SI}, e=1.6 \times 10^{-19} \text{ C}, 1 \text{ eV}=1.6 \times 10^{-19} \text{ J})$

Explain Rutherford's argument for scattered $\alpha$-particles.

What is the percentage of $\alpha$-particles that have more than $1^o$ scattering in the Geiger-Marsden experiment (in $\%$)?

Particles used in the Rutherford's scattering experiment to deduce the structure of atoms:

Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below $14\; K$.) What results do you expect?