According to classical theory,the Rutherford atom was:

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

$A$ fundamental particle of mass $m$ and charge $+e$ is projected towards a much heavier charged particle of charge $+Ze$,where $Z > 0$. The distance of closest approach of the incident particle is ........

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?

In a Geiger-Marsden experiment,what is the distance of closest approach to the nucleus of a $7.7 \;MeV$ $\alpha$-particle before it comes momentarily to rest and reverses its direction?

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?