\begin{problem*}{11.36}
For a hydrogen atom in its ground state, use the Bohr model to
compute \Part{a} the orbital speed of the electron, \Part{b} the
kinetic energy of the electron, and \Part{c} the electric potential
energy of the atom.
\end{problem*} % problem 11.36

\begin{solution}
\Part{a}
The Bohr radius is given by
\begin{align}
  r_n &= \frac{n^2\hbar^2}{mke^2} = n^2 r_1 \\
  r_1 &= 0.529\E{-10}\U{m}
\end{align}
where $\hbar=h/(2\pi)=1.055\E{-34}\U{J$\cdot$s}$ and
$k=9.00\E{9}\U{Nm$^2$/C$^2$}$.  So the velocity is
\begin{align}
  v_n &= \frac{n\hbar}{mr_n} \\
  v_1 &= \frac{\hbar}{mr_1} = \ans{2.19\U{Mm/s}} \;.
\end{align}
\begin{equation}
\end{equation}

\Part{b}
The kinetic energy is then
\begin{equation}
  K = \frac{1}{2} m v^2 = \ans{13.6\U{eV}} \;.
\end{equation}

\Part{c}
The electric potential energy is
\begin{equation}
  U = -k\frac{e^2}{r} = -27.2\U{eV} \;.
\end{equation}
\end{solution}
