The photoelectric effect and its laws

A door that stops when you walk in

A lift door stops closing when you step into its beam. A small device on one side watches a light that shines across the doorway. When you block the light, a signal changes. That device is a photocell. It works because light can knock electrons out of a metal.

The quick version of this idea is in the course Modern Physics: Light, Atoms, and Quanta. Here we look at the experiment itself, with its graphs. Then we see why it broke the wave picture of light.

First sightings

• In 1887 Heinrich Hertz was studying radio waves. He saw that sparks jumped across a gap more easily when ultraviolet light fell on the metal knobs.
• In 1888 Wilhelm Hallwachs lit a clean zinc plate with ultraviolet light. A zinc plate with a negative charge lost its charge. A plate with a positive charge did not.
• So light made negative charge leave the metal. In 1899 and 1900, J. J. Thomson and Philipp Lenard measured the charge and mass of these particles. They are electrons.

The electrons that leave are called photoelectrons. The whole effect is the photoelectric effect.

The photocell

The air is pumped out of the glass tube, so the electrons do not hit air molecules on their way. Ordinary glass stops ultraviolet light, so the window is made of quartz. Light of a chosen frequency $\nu$ and a chosen intensity falls on the emitter C.

The electrons that leave C go to the collector A when A is positive. They are pushed back when A is negative. The battery has a key that reverses it, so we can set the collector voltage on either side of zero.

What we change, and what we see

We have three controls. We change one at a time.

1. The intensity (frequency and voltage fixed). The photocurrent is in proportion to the intensity. Twice the intensity gives twice the current.
2. The collector voltage (frequency and intensity fixed). As A becomes more positive, the current rises and then levels off. This is the saturation current: every electron that leaves C now reaches A. At zero voltage a current still flows, since the electrons leave with kinetic energy. When A is made negative, the current falls. At the stopping potential $V_0$, with A at $-V_0$, it is zero.
3. The frequency (intensity fixed). The stopping potential rises as the frequency rises. Below a certain frequency, the threshold frequency $\nu_0$, there is no current for any intensity.

The blue and orange curves have the same stopping potential. Only their saturation currents differ. The green curve, for a higher frequency, has a larger stopping potential.

Move the frequency below $7.0 \times 10^{14}$ Hz. The curve disappears. Now go back above it and raise the intensity. Which readout changes, and which stays the same? The saturation current grows. The stopping potential does not move.

Stopping potential and energy

When A is at $-V_0$, even the fastest electron stops just before it reaches A. The work done against the field is $eV_0$. So

Here $e = 1.6 \times 10^{-19}$ C is the charge of an electron. A stopping potential of $1.5$ V means that the fastest electrons have $K_{\max} = 1.5$ eV.

The laws of photoelectric emission

Many measurements, by Lenard and others, gave four laws.

• For a metal and a frequency above the threshold, the saturation current is in proportion to the intensity.
• The stopping potential, and so $K_{\max}$, does not depend on the intensity. It rises in a straight line with the frequency.
• Below the threshold frequency no electron leaves, however intense the light.
• The emission starts at once, in less than $10^{-9}$ s, even for very faint light.

Why the wave picture fails

In 1900 light was a wave: the electromagnetic wave of Maxwell and Hertz. The wave picture makes three predictions. All three fail.

1. A bigger wave gives faster electrons. The energy of a wave grows with its intensity. A stronger wave should drive the electrons out with more energy. In the experiment, $K_{\max}$ stays the same.
2. No threshold. A strong wave of any frequency should free electrons if we wait long enough. In the experiment, below $\nu_0$ nothing leaves.
3. A delay. A faint wave spreads its energy over the whole plate. One electron would need time to collect the energy to leave. In the experiment, there is no delay.

A worked example. A lamp gives an intensity of $0.10\ \text{W/m}^2$ on a metal with a work function of $3.0$ eV. Suppose one atom collects light from an area of $1.0 \times 10^{-19}\ \text{m}^2$. How long would a wave take to give one electron $3.0$ eV?

1. Power taken in by one atom. $P = IA = 0.10\ \text{W/m}^2 \times 1.0 \times 10^{-19}\ \text{m}^2 = 1.0 \times 10^{-20}\ \text{W}$.
2. The energy in joule. $3.0\ \text{eV} \times 1.6 \times 10^{-19}\ \text{J/eV} = 4.8 \times 10^{-19}\ \text{J}$.
3. The time. $t = \dfrac{4.8 \times 10^{-19}\ \text{J}}{1.0 \times 10^{-20}\ \text{W}} = 48\ \text{s}$.
4. Check. Nearly a minute. The measured delay is less than $10^{-9}$ s, so the wave picture is wrong by a factor of about $5 \times 10^{10}$.

In 1905 Albert Einstein found a way out. It is the next lesson.

Practise with new numbers

Try it in Python

The program lists the stopping potential and the saturation current of a photocell with a lithium cathode, at five frequencies. The saturation current follows the intensity. The stopping potential follows the frequency.

Common mistakes

• Thinking a brighter light gives faster electrons. A brighter light gives more electrons. Their largest kinetic energy stays the same.
• Thinking that a very intense light of low frequency will work in the end. Below the threshold frequency no electron leaves, however intense the light.
• Reading the stopping potential as a positive voltage on the collector. To stop the electrons, the collector is negative, at $-V_0$.
• Saying the current is zero at zero voltage. The electrons leave with kinetic energy, so some reach the collector.

Remember these

• Photocurrent (saturation) is in proportion to the intensity.
• $K_{\max} = eV_0$, and $V_0$ rises in a straight line with the frequency.
• No emission below the threshold frequency $\nu_0$, and no delay above it.
• Waves cannot explain these three facts.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A lamp shines on a photocell. It is made twice as bright, with the same colour. What happens to (a) the saturation current and (b) the stopping potential?

2. Why is the window of a photocell for ultraviolet light made of quartz and not of ordinary glass?

3. A photocell has a current when the collector is at $-1.0$ V. There is no current when it is at $-1.5$ V. What can you say about the stopping potential?

4. Light of a frequency below the threshold frequency shines on a metal for an hour, at very high intensity. How many photoelectrons leave?