When an ${\rm{\alpha }}$ - particle of mass ‘m’ moving with the velocity ‘v’ bombards on a heavy nucleus of charge ‘Ze’, its distance of closest approach from the nucleus depends on m as:

$\alpha -$ particles of energy $400\,\,KeV$ are bombarded on nucleus of $_{82}Pb$. In scattering of $\alpha -$ particles, its minimum distance from nucleus will be

An elementary particle of mass $m$ and charge $+ e$ is projected with velocity $v$ at a much more massive particle of charge $Ze,$ where $Z > 0.$ What is the closest possible approach of the incident particle

An $\alpha$- particle of energy 5 MeV is scattered through ${180^0}$ by a fixed uranium nucleus. The distance of the closest approach is of the order of

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

$\alpha -$ particles of energy $400\,\,KeV$ are bombarded on nucleus of $_{82}Pb$. In scattering of $\alpha -$ particles, its minimum distance from nucleus will be

An $\alpha$- particle of energy 5 MeV is scattered through ${180^0}$ by a fixed uranium nucleus. The distance of the closest approach is of the order of

An alpha nucleus of energy $\frac{1}{2}m{v^2}$ 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 $5\,MeV$ energy strikes with a nucleus of uranium at stationary at an scattering angle of ${180^0}$. The nearest distance upto which $\alpha$-particle reaches the nucleus will be of the order of

In a hydrogen atom, the distance between the electron and proton is $2.5\,\, imes \,\,{10^{ - 11}}\,\,m$. The electrical force of attraction between them will be

In Davisson - Germer experiment maximum intensity is observed at