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

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

$A$ 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 ........

State the limitations of the plum pudding model of the atom.

An alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to

$A$ $7.9 \text{ MeV}$ $\alpha$-particle scatters from a target material of atomic number $Z = 79$. From the given data, the estimated diameter of the nucleus of the target material is (approximately) . . . . . . $\text{m}$.
$\left[\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2\right.$ and electron charge $\left.e = 1.6 \times 10^{-19} \text{ C}\right]$