Effect of potential difference on photoelectric current

Keeping the frequency and intensity of the incident light fixed, the anode's potential relative to the cathode is now varied and the resulting photocurrent recorded. Starting from a positive potential and increasing it further, the photocurrent rises and eventually flattens out at a maximum, constant value called the saturation current, reached once every single photoelectron emitted from the cathode is being collected at the anode -- increasing the potential further cannot increase the current any more, because there are simply no more electrons left to collect.

Making the anode potential negative instead (a retarding potential, since it now pushes emitted electrons back toward the cathode rather than pulling them across) does not make the current drop to zero immediately, because different photoelectrons leave the cathode with different kinetic energies, and the most energetic ones can still fight their way across against a modest retarding field. As the retarding potential is made more and more negative, progressively more photoelectrons are turned back, and the current steadily falls, until at one particular negative potential, $-V_0$, even the most energetic photoelectrons are stopped and the current drops all the way to zero. This particular value $V_0$ is called the stopping potential or cut-off potential: the minimum retarding potential that is just sufficient to stop the most energetic photoelectrons and reduce the photocurrent to zero.

At the stopping potential, the work done against the retarding field, $eV_0$, exactly equals the initial kinetic energy of the fastest photoelectron, so

where $v_{max}$ is the maximum speed among the emitted photoelectrons. Rearranging gives the maximum speed directly, $v_{max}=\sqrt{2eV_0/m}$, and substituting the known electron charge and mass gives the numerical form $v_{max}=5.93\times10^{5}\sqrt{V_0}\ \text{m s}^{-1}$ (7.2), and equation (7.1) rearranged for the energy in electron-volts gives simply $K_{max}(\text{in eV})=V_0$ (7.3), since $K_{max}=eV_0$ means the numerical value of $K_{max}$ in eV always equals the numerical value of $V_0$ in volts. …