\begin{problem*}{28.10} % photoelectric effect
Electrons are ejected from a metallic surface with speeds ranging up
to $4.60\E{5}\U{m/s}$ when light with a wavelength of $625\U{nm}$ is
used.  \Part{a} What is the work function of the surface?  \Part{b}
What is the cutoff frequency for this surface?
\end{problem*}

\begin{solution}
\Part{a}
Conserving energy
\begin{align}
  hf &= \frac{hc}{\lambda} = \phi + K_\text{max}
    = \phi + \frac{1}{2} m_e v_\text{max}^2 \\
  \phi &= \frac{hc}{\lambda} - \frac{1}{2} m_e v_\text{max}^2
    = \frac{6.63\E{-34}\U{J$\cdot$s}\cdot3.00\E{8}\U{m/s}}{625\E{-9}\U{m}}
      - \frac{1}{2}\cdot 9.11\E{-31}\U{kg}\cdot(4.60\E{5}\U{m/s})^2
    = \ans{2.21\E{-19}\U{J}} = \ans{1.38\U{eV}}
\end{align}

\Part{b}
Light at the cutoff frequency only barely supplies enough energy to
overcome the work function.
\begin{align}
  hf_\text{cut} &= \phi \\
  f_\text{cut} &= \frac{\phi}{h} = \ans{334\U{THz}} \\
\end{align}
\end{solution}
