In Rutherford's experiment,the number of particles scattered at a $90^{\circ}$ angle is $x$ per second. The number of particles scattered per second at an angle of $60^{\circ}$ is:

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

If an alpha particle with energy $7.7 \text{ MeV}$ is bombarded on a thin gold foil,the closest distance from the nucleus it can reach is . . . . . . $\text{m}$. (Atomic number of gold $= 79$ and $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9$ in $SI$ units)

Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below $14\; K$.) What results do you expect?

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

In the Rutherford's nuclear model of the atom,the nucleus (radius about $10^{-15} \; m$) is analogous to the sun about which the electron moves in an orbit (radius $\approx 10^{-10} \; m$) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom,would the earth be closer to or farther away from the sun than it actually is? The radius of the earth's orbit is about $1.5 \times 10^{11} \; m$. The radius of the sun is taken as $7 \times 10^{8} \; m$.