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 graph which depicts the results of the Rutherford gold foil experiment with $\alpha$-particles is:
$\theta$: Scattering angle
$Y$: Number of scattered $\alpha$-particles detected
(Plots are schematic and not to scale)

What radioactive source did Geiger and Marsden use in the scattering experiment?

In $1911$,the physicist Ernest Rutherford discovered that atoms have a tiny,dense nucleus by shooting positively charged particles at a very thin gold foil. $A$ key physical property which led Rutherford to use gold was that it was

Statement $(A)$: As one considers orbits with higher values of $n$ in a hydrogen atom,the electric potential energy of the atom increases.
Statement $(B)$: In Thomson's model,an atom is a spherical cloud of positive charges with electrons embedded in it.
Statement $(C)$: The orbital picture in Bohr's model of the hydrogen atom was consistent with the uncertainty principle.

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