An elementary particle of mass $m$ and charge $+e$ is projected with velocity $v$ at a much more massive particle of charge $Ze,$ where $Z > 0.$ What is the closest possible approach of the incident particle?

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

An $\alpha$-particle of $5 \; MeV$ energy strikes a stationary uranium nucleus at a scattering angle of $180^o$. The distance of closest approach of the $\alpha$-particle to the nucleus will be of the order of:

Find the number of atoms in $39.4 \ g$ of gold. The molar mass of gold is $197 \ g \ mol^{-1}$.

The distance of closest approach of an alpha particle to a nucleus when the alpha particle moves towards the nucleus with linear momentum $P$ is $d$. What is the distance of closest approach of the alpha particle to the nucleus if the linear momentum of the alpha particle is $1.5 P$?

Statement $(A)$: As one considers orbits with higher values of $n$ in a hydrogen atom,the electric potential energy of the atom increases.
Statement $(B)$: In Thomson's model,an atom is a spherical cloud of positive charges with electrons embedded in it.
Statement $(C)$: The orbital picture in Bohr's model of the hydrogen atom was consistent with the uncertainty principle.