If in Rutherford's experiment,the number of particles scattered at $90^o$ angle are $28$ per min,then the number of scattered particles at an angle $60^o$ and $120^o$ will be:

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)

An $\alpha$-particle of $5 \; MeV$ energy strikes a stationary uranium nucleus at a scattering angle of $180^o$. The distance of closest approach of the $\alpha$-particle to the nucleus will be of the order of:

$A$ beam of fast-moving alpha particles was directed towards a thin gold foil. The parts $A, B$,and $C$ of the incident beam and their corresponding transmitted or reflected parts $A^{\prime}, B^{\prime}$,and $C^{\prime}$ are shown in the adjoining diagram. The number of alpha particles in:

An $\alpha$-particle of energy $5 \, MeV$ is scattered by a fixed uranium nucleus at $180^o$. What is the distance of closest approach between the particle and the uranium nucleus?

In an atom,for the electron to revolve around the nucleus,the necessary centripetal force is obtained from which of the following forces exerted by the nucleus on the electron?