# An apha particle is projected from infinity with the velocity Vo towards the nucleus of an atom having atomic number equal to Z then find out (i)closest distance of approach (R). (ii) what is the velocity of the alpha particle at the distance R1 (R1>R) from the nucleus

Dear student

Distance of closest approach:
When alpha particle moves towards the nucleus, its kinetic energy keeps on decreasing because kinetic energy is being consumed in doing work against the force of repulsion of the nucleus, the distance from the nucleus at which velocity of the alpha particle becomes zero,is known as the closest approach.
The distance of closest approach' d' is the distance at which kinetic energy is equal to the coulomb energy.

Here d = R
so R = $\frac{q\alpha }{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{E}}$

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