Question

An electron moves in a circular orbit about a stationary proton, the centripetal force is provided by the electrostatic force of attraction between the proton and the electron, assuming that the electron has a kinetic energy of 3×10-18J, what is the speed of the electron and the radius of the orbit of the electron.

Answer 1

Speed of the electron:

$$v=\sqrt{2E_K/m}=2.57·10^6\text{ m/s}.$$

The radius:

$$m\frac {v^2}r=k\frac{q_1q_2}{r^2},\\\space\\ r=\frac{kq_1q_2}{mv^2}=\frac{kq_1q_2}{2E_K}=\frac{ke^2}{2E_K}=38.5\text{ pm}.$$