An alpha particle of energy $K \text{ MeV}$ is moving towards a nucleus of atomic number $Z$. The distance of closest approach of the alpha particle to the nucleus in metres is

In the Geiger-Marsden scattering experiment,the thickness of the thin gold foil is . . . . . . m.

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:

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 . . . . . . $\times 10^{-14} \text{ m}$.
$(k=9 \times 10^{9} \text{ SI}, e=1.6 \times 10^{-19} \text{ C}, 1 \text{ eV}=1.6 \times 10^{-19} \text{ J})$

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 $1.5 P$?

What is shown by Thomson's experiments of electric discharge through gases? And explain the plum pudding model.