Answer:
[tex]v=\sqrt{\frac{kZe^2}{mr}}[/tex]
Explanation:
The electrostatic attraction between the nucleus and the electron is given by:
[tex]F=k\frac{(e)(Ze)}{r^2}=k\frac{Ze^2}{r^2}[/tex] (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:
[tex]F=m\frac{v^2}{r}[/tex] (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
[tex]k\frac{Ze^2}{r^2}=m\frac{v^2}{r}[/tex]
And solving for v, we find an expression for the speed of the electron:
[tex]v=\sqrt{\frac{kZe^2}{mr}}[/tex]
