1 \begin{problem*}{28.10} % photoelectric effect
2 Electrons are ejected from a metallic surface with speeds ranging up
3 to $4.60\E{5}\U{m/s}$ when light with a wavelength of $625\U{nm}$ is
4 used. \Part{a} What is the work function of the surface? \Part{b}
5 What is the cutoff frequency for this surface?
12 hf &= \frac{hc}{\lambda} = \phi + K_\text{max}
13 = \phi + \frac{1}{2} m_e v_\text{max}^2 \\
14 \phi &= \frac{hc}{\lambda} - \frac{1}{2} m_e v_\text{max}^2
15 = \frac{6.63\E{-34}\U{J$\cdot$s}\cdot3.00\E{8}\U{m/s}}{625\E{-9}\U{m}}
16 - \frac{1}{2}\cdot 9.11\E{-31}\U{kg}\cdot(4.60\E{5}\U{m/s})^2
17 = \ans{2.21\E{-19}\U{J}} = \ans{1.38\U{eV}}
21 Light at the cutoff frequency only barely supplies enough energy to
22 overcome the work function.
24 hf_\text{cut} &= \phi \\
25 f_\text{cut} &= \frac{\phi}{h} = \ans{334\U{THz}} \\