Question

A proton is fired from far away toward the nucleus of a mercury atom. Mercury is element number 80, and the diameter of the nucleus is 14.0 fm.If the proton is fired at a speed of 3.2×10⁷ m/s , what is its closest approach to the surface of the nucleus? Assume the nucleus remains at rest.

Answer 1

Solution) Distance of closest approach is defined as the minimum distance of the charged particle from the nucleus at which the initial kinetic energy of the particle is equal to potential energy due to charged nucleus .

Here we use law of conservation of energy

Initial total energy = final total energy

Total energy = kinetic energy + potential energy

At close distance kinetic energy =0

At far distance (r tends to infinity ) so potential energy =0