38.1.3 The Stopping Potential

Photoelectrons leave the cathode with kinetic energy. An electron with energy $E_{elec}$ inside the metal loses energy $Δ E$ as it escapes, so it emerges as a photoelectron with $K = E_{elec} - Δ E$ . The work function energy $E_{0}$ is the minimum energy needed to remove an electron, so the maximum possible kinetic energy of a photoelectron is

K_{max} = E_{elec} - E_{0}

(38.1)

Some photoelectrons reach the anode, creating a measurable current, but many do not. However, as FIGURE 38.5 shows:

A positive anode attracts the photoelectrons. Once all electrons reach the anode, which happens for $Δ V$ greater than about 1 V, a further increase in $Δ V$ does not cause any further increase in the current I. That is why the graph lines become horizontal on the right side of Figure 38.3.
A negative anode repels the electrons. However, photoelectrons leaving the cathode with sufficient kinetic energy can still reach the anode. The current steadily decreases as the anode voltage becomes increasingly negative until, at the stopping potential, all electrons are turned back and the current ceases. This was the behavior observed on the left side of Figure 38.3.

The figure consists of three parts that illustrate the dependence of photoelectric current on anode potential.

FIGURE 38.5 Full Alternative Text

Let the cathode be the point of zero potential energy, as shown in FIGURE 38.6. An electron emitted from the cathode with kinetic energy $K_{i}$ has initial total energy

E_{i} = K_{i} + U_{i} = K_{i} + 0 = K_{i}

When the electron reaches the anode, which is at potential $Δ V$ relative to the cathode, it has potential energy $U = q Δ V = - e Δ V$ and final total energy

E_{f} = K_{f} + U_{f} = K_{f} - e Δ V

From conservation of energy, $E_{f} = E_{i}$ , the electron’s final kinetic energy is

K_{f} = K_{i} + e Δ V

(38.2)

The electron speeds up $(K_{f} > K_{i})$ if $Δ V$ is positive. The electron slows down if $Δ V$ is negative, but it still reaches the anode $(K_{f} > 0)$ if $K_{i}$ is large enough.

The figure illustrates the conservation of energy through an experimental setup.

FIGURE 38.6 Full Alternative Text

An electron with initial kinetic energy $K_{i}$ will stop just as it reaches the anode if the potential difference is $Δ V = - K_{i} / e$ . The potential difference that turns back the very fastest electrons, those with $K = K_{max}$ , and thus stops the current is

Δ V_{stop fastest electrons} = - \frac{K_{max}}{e}

By definition, the potential difference that causes the electron current to cease is $Δ V = - V_{stop}$ , where $V_{stop}$ is the stopping potential. The stopping potential is

V_{stop} = \frac{K_{max}}{e}

(38.3)

Thus the stopping potential tells us the maximum kinetic energy of the photoelectrons.

EXAMPLE 38.1 The classical photoelectric effect

A photoelectric-effect experiment is performed with an aluminum cathode. An electron inside the cathode has a speed of $1.5 \times 10^{6} m / s$ . If the potential difference between the anode and cathode is $- 2.00 V,$ what is the highest possible speed with which this electron could reach the anode?

MODEL Energy is conserved.

SOLVE If the electron escapes with the maximum possible kinetic energy, its kinetic energy at the anode will be given by Equation 38.2 with $Δ V = - 2.00 V$ . The electron’s initial kinetic energy is

\begin{matrix} E_{elec} & = & \frac{1}{2} m v^{2} = \frac{1}{2} (9.11 \times 10^{- 31} kg) {(1.5 \times 10^{6} m / s)}^{2} \\ = & 1.025 \times 10^{- 18} J = 6.41 eV \end{matrix}

Its maximum possible kinetic energy as it leaves the cathode is

K_{i} = K_{max} = E_{elec} - E_{0} = 2.13 eV

where $E_{0} = 4.28 eV$ is the work function of aluminum. Thus the kinetic energy at the anode, given by Equation 38.2, is

K_{f} = K_{i} + e Δ V = 2.13 eV - (e) (2.00 V) = 0.13 eV

Notice that the electron loses 2.00 eV of kinetic energy as it moves through the potential difference of $- 2.00 V,$ so we can compute the final kinetic energy in eV without having to convert to joules. However, we do need joules to find the final speed:

\begin{array}{l} K_{f} & = & \frac{1}{2} m v_{f}^{2} = 0.13 eV = 2.1 \times 10^{- 20} J \\ v_{f} & = & \sqrt{\frac{2 K_{f}}{m}} = 2.1 \times 10^{5} m/s \end{array}

38.1.3 The Stopping Potential

FIGURE 38.5 The photoelectron current depends on the anode potential.

FIGURE 38.6 Energy is conserved.

EXAMPLE 38.1 The classical photoelectric effect