Answer
428 nm
Work Step by Step
At the minimum frequency, the ejected electrons have no kinetic energy. The energy of a photon equals the work function, according to equation 27–5a.
$$KE=hf_{min}-W_o=0$$
$$f_{min}=\frac{W_o}{h}$$
$$\lambda_{max}=\frac{c}{f_{min}}=\frac{hc}{W_o}$$
In this problem, the value of hc is useful. Use the result stated in Problem 29, $hc=1240\;eV \cdot nm$.
$$\lambda_{max}=\frac{1240\;eV \cdot nm }{2.90eV}=428nm$$
