When $\alpha$-particles with kinetic energy $\frac{1}{2} mv^2$ are bombarded on a heavy nucleus of charge $Ze$,the distance of closest approach is 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 $\left.e = 1.6 \times 10^{-19} \text{ C}\right]$

An $\alpha$-particle of energy $4 \text{ MeV}$ is scattered through $180^\circ$ by a fixed uranium nucleus. The distance of the closest approach is of the order of

$A$ proton is fired from very far away towards a nucleus with charge $Q=120 \ e$,where $e$ is the electronic charge. It makes a closest approach of $10 \ fm$ to the nucleus. The de Broglie wavelength (in units of $fm$) of the proton at its start is: (take the proton mass,$m_0 = (5/3) \times 10^{-27} \ kg$,$h/e = 4.2 \times 10^{-15} \ J \cdot s/C$,$\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$,$1 \ fm = 10^{-15} \ m$)

Rutherford's $\alpha$-particle experiment showed that the atoms have

What is the plum pudding model of the atom?