According to classical theory,the circular path of an electron in a Rutherford atom is

The distance of closest approach of an alpha particle to a nucleus when the alpha particle moves towards the nucleus with linear momentum $P$ is $d$. What is the distance of closest approach of the alpha particle to the nucleus if the linear momentum of the alpha particle is $1.5 P$?

An alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to

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

Give the relationship between impact parameter and scattering angle.

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