An alpha particle is accelerated through a potential difference of $10^6 \ V$. Its kinetic energy will be $MeV$.

Rutherford's $\alpha$-particle experiment showed that the atoms have

Choose the correct option from the following options given below.

An $\alpha$-particle of energy $5 \text{ MeV}$ is scattered through $180^{\circ}$ by a gold nucleus. The distance of closest approach is of the order of:

An $\alpha$ particle is accelerated through a potential difference of $10^6 \ V$. Then the kinetic energy $(K.E.)$ of the particle will be .......... $MeV$.

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?