Answer
a. 1.7 eV.
b. 3.0 eV.
Work Step by Step
a. The photons with the lowest frequency and the longest wavelength have the smallest energy. The work function of the metal must be less than or equal to that energy for those photons to eject electrons. Set the work function equal to the energy of a 750-nm photon.
Use the result stated in Problem 29, $hc=1240\;eV \cdot nm$.
$$W_0=hf_{min}=\frac{hc}{\lambda_{max}}=\frac{1240\;eV \cdot nm }{750nm}=1.7eV$$
b. For work functions greater than a threshold value, the photomultiplier only works for incident uv wavelengths. Set the threshold work function equal to the energy of a 410-nm photon.
$$W_0=hf_{max}=\frac{hc}{\lambda_{min}}=\frac{1240\;eV \cdot nm }{410nm}=3.0eV$$
