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+### Statement
+
+$14.4.8.$ [Insert the problem statement]
+
+### Solution
+
+\documentclass[12pt,a4paper]{article}
+\usepackage[english]{babel}
+\usepackage{float}
+\usepackage{wrapfig}
+\usepackage{lmodern}
+\usepackage[T1]{fontenc}
+\usepackage[utf8]{inputenc}
+\usepackage{microtype}
+\usepackage{graphicx}
+\usepackage{booktabs}
+\usepackage{amsmath,amssymb}
+\usepackage{hyperref}
+\usepackage{csquotes}
+\usepackage{geometry}
+\usepackage{subcaption}
+\usepackage{tikz}
+\usepackage{array}
+\usepackage{pgfplots}
+\usepackage{wrapfig}
+\usepackage{subcaption}
+
+\begin{document}
+
+$14.4.8$ How fast does an electron move around a heavy nucleus with charge $ez$ in a
+circular orbit of radius $R$?
+
+We can solve this problem using Newton's second law for the radial direction of the motion.
+In this case the 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}
+
+\end{document}
+
+#### Answer
+
+[Insert a concise answer or boxed result]