Calculate the number of atoms in $39.4$ $g$ of gold. The molar mass of gold is $197$ $g$ $mol^{-1}$.

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:

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)

The $\alpha$-particle is the nucleus of an atom of

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?

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