When an $\alpha$-particle of mass $m$ moving with velocity $v$ bombards a heavy nucleus of charge $Ze$,its distance of closest approach from the nucleus depends on $m$ as

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

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:

Using Thomson's model of the atom,consider an atom consisting of two electrons,each of charge $-e$,embedded in a sphere of charge $+2e$ and radius $R$. In equilibrium,each electron is at a distance $d$ from the center of the atom. What is the equilibrium separation between the electrons?

In the gold foil experiment,the number of deflected $\alpha$-particles at an angle of $90^o$ is $63$. What is the number of $\alpha$-particles deflected at $120^o$?

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?