Answer
0.62 eV
Work Step by Step
In this problem, the value of hc is useful. Use the result stated in Problem 29, $hc=1240\;eV \cdot nm$.
The most energetic photon of visible light has the highest frequency and the shortest wavelength. Use a wavelength of 400 nm.
Use equation 27–5b to calculate the maximum kinetic energy.
$$KE_{max}=hf-W_o=\frac{hc}{\lambda}-W_o$$
$$KE_{max}=\frac{1240\;eV \cdot nm }{400nm}-2.48eV=0.62eV$$
