Question

An interesting (but oversimplified) model of an atom pictures an electron “in orbit” around a proton. Suppose this electron is moving in a circular orbit of radius 0.10 nm (1.0 x 10-10) and the force that makes this circular motion possible is the electric force exerted by the proton on the electron. Find the speed of the electron.

Answer 1

Let us calculate the electric force exerted by the proton on the electron:

$F = \dfrac{k_eq_pq_e}{r^2}$ . Module of charges of proton and electron is $q = 1.6\cdot10^{-19}\,\mathrm{C}.$ Therefore, the force is

$F = \dfrac{9\cdot10^9\cdot (1.6\cdot10^{-19})^2}{(1.0\cdot10^{-10})^2} = 2.3\cdot10^{-8}\,\mathrm{N}.$ The acceleration can be calculated as

$a = \dfrac{F}{m_e}$ , but also the acceleration is $a = \dfrac{v^2}{r}.$ Therefore, velocity is

$v = \sqrt{\dfrac{Fr}{m_e}} = \sqrt{\dfrac{2.3\cdot10^{-8}\,\mathrm{N}\cdot1.0\cdot10^{-10}\,\mathrm{m}}{9.1\cdot10^{-31}\,\mathrm{kg}}} = 1.6\cdot10^3\,\mathrm{km/s}.$