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

In a Rutherford scattering experiment,when a projectile of charge $Z_1$ and mass $M_1$ approaches a target nucleus of charge $Z_2$ and mass $M_2$,the distance of closest approach is $r_0$. The energy of the projectile is:

According to classical theory,the circular path of an electron in a Rutherford atom is

In the Rutherford experiment,for a head-on collision of $\alpha$-particles with a gold nucleus,the impact parameter is:

Why are atoms electrically neutral?

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