Answer
3.46 eV
Work Step by Step
Use equation 27–5b to calculate the work function.
$$ W_o =hf-KE_{max} =\frac{hc}{\lambda}-KE_{max}$$
In this problem, the value of hc is useful. Use the result stated in Problem 29, $hc=1240\;eV \cdot nm$.
$$ =\frac{1240\;eV \cdot nm }{255nm}-1.40eV=3.46eV$$
