Question

An electron with charge −e and mass m moves in a circular orbit of radius r around a nucleus of charge Ze, where Z is the atomic number of the nucleus. Ignore the gravitational force between the electron and the nucleus. Find an expression in terms of these quantities for the speed of the electron in this orbit. (Use any variable or symbol stated above along with the following as necessary: k for Coulomb's constant.)

Answer 1

Answer:

$v=\sqrt{\frac{kZe^2}{mr}}$

Explanation:

The electrostatic attraction between the nucleus and the electron is given by:

$F=k\frac{(e)(Ze)}{r^2}=k\frac{Ze^2}{r^2}$ (1)

where

k is the Coulomb's constant

Ze is the charge of the nucleus

e is the charge of the electron

r is the distance between the electron and the nucleus

This electrostatic attraction provides the centripetal force that keeps the electron in circular motion, which is given by:

$F=m\frac{v^2}{r}$ (2)

where

m is the mass of the electron

v is the speed of the electron

Combining the two equations (1) and (2), we find

$k\frac{Ze^2}{r^2}=m\frac{v^2}{r}$

And solving for v, we find an expression for the speed of the electron:

$v=\sqrt{\frac{kZe^2}{mr}}$