\begin{problem*}{20.1}
\Part{a} Calculate the speed of a proton that is accelerated from
rest through a potential difference of $\Delta V = 120\U{V}$.
\Part{b} Calculate the speed of an electron that is accelerated
through the same potential difference.
\end{problem*} % problem 20.1

\begin{solution}
\Part{a}
Conserving energy
\begin{gather}
 E_0 = \frac{1}{2}m_p v^2 = E_1 = e\Delta V \\
 v = \sqrt{\frac{2 e \Delta V}{m_p}}
   = \sqrt{\frac{2 \cdot 1.60\E{-19}\U{C} \cdot 120\U{V}}{1.67\E{-27}\U{kg}}}
   = \ans{152\U{km/s}}
\end{gather}

\Part{b} Replacing $m_p$ with $m_e$
\begin{equation}
v = \sqrt{\frac{2 e \Delta V}{m_e}}
   = \sqrt{\frac{2 \cdot 1.60\E{-19}\U{C} \cdot 120\U{V}}{9.11\E{-31}\U{kg}}}
   = \ans{6.49\U{Mm/s}}
\end{equation}
\end{solution}
