Answer
$KE_{max}=0.92eV$
$v=5.7\times10^5m/s$
Work Step by Step
In this problem, the value of hc is useful. Use the result stated in Problem 29, $hc=1240\;eV \cdot nm$.
Use equation 27–5b to calculate the maximum kinetic energy.
$$KE_{max}=hf-W_o=\frac{hc}{\lambda}-W_o$$
$$KE_{max}=\frac{1240\;eV \cdot nm }{365nm}-2.48eV=0.92eV$$
To calculate the electron speed, set this energy equal to $\frac{1}{2}mv^2$ and solve for the speed v of $5.7\times10^5m/s$.
The energy is much less than the electron's rest energy, so the classical expression for KE is sufficient.