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 electron and the nucleus is given by Coulomb's Law

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

where we have

k is the Coulomb's constant

e is the charge of the electron

Ze is the charge of the nucleus

r is radius of the electron's orbit, which corresponds to the distance between the electron and the nucleus

The electron is moving by uniform circular motion, so the centripetal force acting on it is given by

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

where

m is the mass of the electron

v is the speed of the electron

r is the radius of the orbit

Since the centripetal force is provided by the electrostatic attraction, we can write that the two forces are equal; so we obtain:

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

And re-arranging the equation for v we find:

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