Answer
3.32 eV
Work Step by Step
Use equation 27–5b to calculate the work function.
$$ W_o =hf-KE_{max} =\frac{hc}{\lambda}-KE_{max}$$
The stopping potential of 1.64 volts tells us that the electrons have a maximum kinetic energy of 1.64 eV. In other words, the maximum KE is the product of stopping voltage and the magnitude of the electron’s charge.
In this problem, the value of hc is useful. Use the result stated in Problem 29, $hc=1240\;eV \cdot nm$.
$$ W_o=\frac{1240\;eV \cdot nm }{250nm}-1.64eV=3.32eV$$