Answer
a. 2.3 eV.
b. 0.85 volts.
Work Step by Step
a. At the threshold wavelength, the KE of the photoelectrons is zero, because they barely make it out of the material. The work function equals the energy of the incoming photon.
Use equation 27–4, E = hf, to find the energy of a photon. For electromagnetic radiation, we also know that $f=c/\lambda$.
$$W_o=hf-KE_{max}=hf-0=\frac{hc}{\lambda}=\frac{1240 eV\cdot nm}{550nm}=2.3eV$$
b. The stopping voltage is the voltage that stops the fastest ejected electrons. The change in potential energy equals the maximum kinetic energy. Use equation 27–5b to calculate the maximum kinetic energy.
$$KE_{max}=hf-W_o=\frac{hc}{\lambda}-W_o=\frac{1240 eV\cdot nm}{400nm}-2.255eV=0.845eV$$
The stopping voltage for a 0.845eV electron is 0.845 volts $\approx 0.85 V$.
