An alpha particle colliding with one of the electrons in a gold atom loses

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:

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

Particles used in the Rutherford's scattering experiment to deduce the structure of atoms:

An alpha particle of energy $K \text{ MeV}$ is moving towards a nucleus of atomic number $Z$. The distance of closest approach of the alpha particle to the nucleus in metres is

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