Special theory of relativity > Motion of relativistic particles in electric and magnetic fields > Problem 14.4.8

Statement

$14.4.8$ How fast does an electron move around a heavy nucleus with charge $ez$ in a
circular orbit of radius $R$?

Solution

We can solve this problem using Newton's second law for the radial direction of the motion.
In this case the radial acceleration is:

\begin{equation}
a_r = \frac{v^2}{R}
\end{equation}

The force acting on the electron is the electric force:

\begin{equation}
F_e = \frac{(ez)e}{4 \pi \varepsilon_0 R^2}
\end{equation}

This force points in the radial direction, like the acceleration of the electron.

Using Newton's second law:

\begin{equation}
F_e = m_e a_r \rightarrow \frac{m_e v^2}{R} = \frac{(ez)e}{4 \pi \varepsilon_0 R^2}
\end{equation}

Then:

\begin{equation}
v = \sqrt{\frac{e^2 z}{4 \pi \varepsilon_0 m_e R}}
\end{equation}

Answer

\begin{equation}
v = \sqrt{\frac{e^2 z}{4 \pi \varepsilon_0 m_e R}}
\end{equation}