A proton is projected toward a fixed nucleus of charge +Ze with velocity vo.Initially the two particles are very far apart.When the proton is a distance R from the nucleus its velocity has decreased to 1/2 v_o.How far from the nucleus will the proton be when its velocity has dropped to 1/4 v_o?

Question

Quizplus · Accepted Answer

The problem involves conservation of energy. Initially, the kinetic energy of the proton is (1/2)m(vo)^2. As it approaches the nucleus, some kinetic energy is converted into electric potential energy. When the velocity is 1/2 vo, the kinetic energy is (1/2)m(1/2 vo)^2. When the velocity is 1/4 vo, the kinetic energy is (1/2)m(1/4 vo)^2. By setting up the conservation of energy equations for these two scenarios and solving for the distance, we find that the distance is 4/5 R.

A Proton Is Projected Toward a Fixed Nucleus of Charge