$A$ fundamental particle of mass $m$ and charge $+e$ is projected towards a much heavier charged particle of charge $+Ze$,where $Z > 0$. The distance of closest approach of the incident particle is ........

An $\alpha$-particle of energy $5 \text{ MeV}$ is scattered through $180^{\circ}$ by a gold nucleus. The distance of closest approach is of the order of:

State and explain Dalton's atomic theory.

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

In the Rutherford scattering experiment,what will be the correct scattering angle $\theta$ for an $\alpha$-particle with an impact parameter $b = 0$ (in $^o$)?

An alpha particle colliding with one of the electrons in a gold atom loses