\begin{problem*}{20.20}
In 1911, Ernest Rutherford and his assistants Hans Geiger and Ernest
Mardsen conducted an experiment in which they scattered alpha
particles from thin sheets of gold.  An alpha particle, having a
charge of $q_\alpha = +2e$ and a mass of $m = 6.64\E{-27}\U{kg}$ is a
product of certain radioactive decays.  The results of the experiment
lead Rutherford to the idea that most of the mass of an atom is in a
very small nucleus, whith electrons in orbit around it, in his
planetary model of the atom.  Assume that an alpha particle, initially
very far from a gold nucleus, is fired with a velocity $v =
2.00\E{7}\U{m/s}$ directly toward the nucleus (charge $Q = +79e$).
How close does the alpha particle get to the nucleus before turning
around?  Asume that the gold nucleus remains stationary.
\end{problem*} % problem 20.20

\begin{solution}
Let $r$ be the distance between the alpha particle and the gold
nucleus.  Conserving energy between the initial point at $r=\infty$
where the energy is all kinetic
\begin{equation}
 E_0 = \frac{1}{2}m v_0^2
\end{equation}
And the point of closest approach where the energy is all electric potential
\begin{equation}
 E_1 = k_e \frac{(2e)(79e)}{r}
\end{equation}
We have
\begin{align}
 E_0 = \frac{1}{2}mv^2 &= E_1 = k_e \frac{158 e^2}{r} \\
 r &= \frac{2 \cdot 158 \cdot k_e e^2}{m v^2}
    = \frac{316 \cdot 8.99\E{9}\U{N m$^2$/C$^2$} \cdot (1.60\E{-19}\U{C})^2}{6.64\E{-27}\U{kg} \cdot (2.00\E{7}\U{m/s})^2}
    = \ans{2.74\E{-14}\U{m}}
\end{align}
Which is significantly less than the $r_e \sim 10^{-10}\U{m}$ radius
of the gold atom.
\end{solution}
