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:

In Rutherford's alpha scattering experiment, a thin gold foil was bombarded with alpha particles. Most of the alpha particles passed straight through the foil. This observation suggests that:

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

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 mass 'm' and kinetic energy 'K' approaches a heavy nucleus of charge +Ze. What is its distance of closest approach?

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

Which of the following modifications to Rutherford's alpha scattering experiment would result in a decrease in the number of alpha particles scattered through large angles?

An alpha particle of mass 'm' and velocity 'v' approaches a gold nucleus (charge +79e). How does the distance of closest approach depend on the mass of the alpha 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

The scattering of alpha particles at very small angles in Rutherford's experiment is primarily due to:

Which observation from Rutherford's experiment provided evidence for the existence of a dense, positively charged nucleus?