Saturation current and stopping potential

Keep the frequency ν and intensity fixed and make A more and more positive. The photocurrent grows, then levels off: at a large enough accelerating potential every electron emitted by C reaches A. This largest current is the saturation current, and raising the potential further does not increase it.

Now make A negative with respect to C. The electrons are pushed back, and only the more energetic ones reach A, so the current falls quickly.

At one sharply defined negative potential of A the current becomes zero. This smallest retarding potential that stops the photocurrent, for a given frequency, is the cut-off or stopping potential V₀.

The photoelectrons do not all leave with the same energy. The stopping potential is just enough to turn back the fastest of them, so the maximum kinetic energy is Kmax = eV₀ (Eq. 11.1).

Repeat with the same frequency at higher intensities I₂ and I₃ (I₃ > I₂ > I₁). The saturation current rises in proportion to the intensity, but the stopping potential stays exactly the same (Fig. 11.3).

So, for light of a given frequency, the stopping potential, and with it Kmax, is independent of the intensity. Kmax depends on the light source and the emitter material, not on how bright the light is.

Exercise 11.3: a cut-off voltage of 1.5 V means Kmax = 1.5 eV = 1.5 × 1.6 × 10⁻¹⁹ J = 2.4 × 10⁻¹⁹ J.