Many older lifts have a thin beam of light across the doorway. It shines into a small sensor on the other side. Put a hand in the doorway and the closing doors stop and open again. Nothing touched the doors. The only thing that changed was the light reaching the sensor.
Catching the freed electrons
In the open air the photoelectric effect is hard to measure. A freed electron soon hits an air molecule and scatters, and nothing counts it. So the metal is sealed inside a glass tube with the air pumped out. In the vacuum each electron flies in a straight line, with nothing in its way.
The tube holds two electrodes. Light falls on a large curved plate, the emitter E, coated with a metal that frees electrons easily. A rod or plate opposite it, the collector C, gathers the electrons that reach it. The window is often made of quartz, because ordinary glass absorbs ultraviolet light. Such a tube is called a photocell.
A wire joins the collector back to the emitter through a battery and a meter. Without it, every electron that left would leave the emitter a little more positive. The positive emitter would soon pull the next electrons back, and the emission would stop. With the wire, the electrons return to the emitter through the circuit, and the flow is steady.
What the meter measures
An electric current is the charge passing a point each second, \(I = q/t\), measured in amperes (A). In the photocell the charge is carried by electrons, each carrying \(e = 1.6 \times 10^{-19}\,\text{C}\). So a current \(I\) means \(I/e\) electrons reach the collector every second. This current is the photocurrent, and the microammeter reads it in microamperes, \(1\,\mu\text{A} = 10^{-6}\,\text{A}\).
The battery is connected through a potential divider, so its voltage can be varied smoothly and reversed. We write \(V\) for the potential of the collector relative to the emitter. When \(V\) is positive, the field between the electrodes pulls the electrons towards C. This is an accelerating potential. When \(V\) is negative, the field pushes them back towards E. This is a retarding potential.
A photocell is used without the wire that joins the collector back to the emitter. Light keeps falling on the emitter. What happens?
Brighter light, more electrons
Now keep the colour of the light fixed and the collector well positive. Change only the intensity, the power falling on each square metre. The photocurrent rises in proportion. Twice the intensity gives twice the current, and a tenth of the intensity gives a tenth of the current. On a graph, the current against the intensity is a straight line through the origin.
The current counts electrons per second. So a brighter light frees more electrons every second. This fact says nothing yet about how fast each electron moves. The next topic finds a way to measure that separately.
A small lamp stands \(20\,\text{cm}\) from a photocell, and the saturated photocurrent is \(12\,\mu\text{A}\). How many electrons reach the collector each second? The lamp is moved to \(60\,\text{cm}\). What is the current now?
(a) Electrons per second \(= I/e = 12 \times 10^{-6} / 1.6 \times 10^{-19} = 7.5 \times 10^{13}\).
(b) From the chapter on Energy, Power, Intensity and Loudness, a small source spreads its power over a sphere, so the intensity falls as \(1/r^2\). At three times the distance, the intensity is \(1/9\) as large.
(c) The current follows the intensity, so it becomes \(12/9 = 1.33\,\mu\text{A}\).
Values used: \(e = 1.6 \times 10^{-19}\,\text{C}\).
A photocell gives a steady current of \(48\,\mu\text{A}\). How many electrons reach its collector each second? Give the answer in units of \(10^{14}\) electrons per second.
Turning the collector up: saturation
Keep the light fixed and raise \(V\) from zero. The current rises at first, and then it stops rising. However high \(V\) goes, the current stays at the same value, the saturation current. The light frees a fixed number of electrons every second. Once the collector gathers every one of them, a higher \(V\) has nothing more to collect.
Why is the current below saturation at \(V = 0\)? The electrons leave the emitter with a spread of speeds. The slow ones linger near the emitter and form a small cloud of negative charge. The cloud repels the newest slow electrons back into the emitter. A positive collector sweeps the cloud away, and then every electron gets across.
Turning the collector down: the stopping potential
Now reverse the battery, so the collector is negative. An electron crossing towards it is slowed by the field. It arrives only if its kinetic energy is enough to climb the potential difference. The slowest electrons are turned back first, so the current falls gradually. At one value of the retarding potential even the fastest electron just fails, and the current becomes zero.
Try \(V = -0.5\), \(-1.0\) and \(-1.5\,\text{V}\) in the sim. Nine, six and three electrons arrive. Each electron that turns back had less than \(e|V|\) of kinetic energy to spend. The last one to arrive is always the fastest.
The stopping potential \(V_0\) is the least retarding potential of the collector, relative to the emitter, at which the photocurrent becomes zero.
The fastest electron starts with the maximum kinetic energy \(K_{max}\). Crossing the retarding potential \(V_0\), the field takes \(eV_0\) from it. By the work-energy theorem, it stops just at the collector when all its kinetic energy is used up. So \(K_{max} = eV_0\). A voltmeter reading gives the energy of the fastest electron directly.
Ultraviolet light falls on a caesium emitter. The photocurrent becomes zero when the collector is \(1.86\,\text{V}\) below the emitter. Find the maximum kinetic energy of the photoelectrons in eV and in joules, and their greatest speed.
The stopping potential is \(V_0 = 1.86\,\text{V}\).
\(K_{max} = eV_0\), and \(K_{max} = \tfrac12 m_e v_{max}^2\).
\(K_{max} = 1.86\,\text{eV} = 1.86 \times 1.6 \times 10^{-19} = 2.98 \times 10^{-19}\,\text{J}\). Then \(v_{max} = \sqrt{2 \times 2.98 \times 10^{-19} / 9.1 \times 10^{-31}} = 8.1 \times 10^{5}\,\text{m/s}\).
The fastest photoelectron leaves at \(8.1 \times 10^{5}\,\text{m/s}\), about \(800\,\text{km}\) every second. The others are slower.
One more measurement completes the picture. Change the intensity of the light, keeping its colour. The saturation current changes, but the stopping potential does not move at all. A brighter light gives more electrons, but not faster ones.
The stopping potential for a photocell is \(2.5\,\text{V}\). What is the maximum kinetic energy of its photoelectrons? Give the answer in units of \(10^{-19}\,\text{J}\).
The photocurrent has reached its saturation value. The accelerating potential is now doubled. What does the meter read?
A lamp is moved from \(1\,\text{m}\) to \(2\,\text{m}\) from a photocell. What happens to the saturation current and to the stopping potential?
The colour of the light
Now change the frequency \(\nu\) of the light, its colour, and keep everything else fixed. Two facts appear. First, below a certain frequency no electrons come out at all, however bright the light is made. This frequency is the threshold frequency \(\nu_0\). Each metal has its own: caesium works in green light, while copper needs ultraviolet.
Second, above the threshold, the stopping potential grows as the frequency grows. Violet light gives faster electrons than green light from the same metal. Lesson 4 shows that the growth is a straight line.
No waiting
The last fact concerns time. Switch on even a very dim light above the threshold. The first electrons leave within about \(10^{-9}\,\text{s}\), a nanosecond. Dim light gives fewer electrons each second, but the first ones never come late.
The photocell has now given four facts:
- At a fixed colour, the photocurrent is proportional to the intensity.
- The stopping potential, and so \(K_{max}\), does not depend on the intensity.
- Below a threshold frequency there is no emission; above it, \(K_{max}\) grows with the frequency.
- Emission starts at once, with no delay.
The next lesson asks whether the wave picture of light can explain them.
Light below the threshold frequency of a metal falls on it. The intensity is raised a thousand times. What is observed?
With no battery at all, \(V = 0\), a photocell still gives a small current. Why do electrons cross without any accelerating field?
The photoelectrons leave the emitter already moving, with kinetic energies up to \(K_{max}\). With no field to stop them, many of them simply coast across to the collector. The current is below saturation only because slow electrons crowd near the emitter and repel the newest slow ones back.
In 1887 Heinrich Hertz was making the first radio waves with sparks between two metal balls. He noticed that a small receiving spark jumped more easily when ultraviolet light from the main spark fell on it. He reported the effect and left it unexplained. In 1902 Philipp Lenard studied the freed electrons in an evacuated tube, much like Fig 2.1. Using a retarding voltage, he found that brighter light gave more electrons but not faster ones. That result made no sense to anyone until 1905.
The sensor across a lift door, the counter at a shop entrance and the smoke alarm with a light beam all work on one principle. Light falling on a sensor makes a current; a hand or a puff of smoke cuts the light, and the current falls. A circuit notices the change and acts. Early sensors were vacuum photocells like Fig 2.1. Today they are small semiconductor photodiodes, which the chapter on diodes describes.
- Reading \(V_0\) as a negative number of electron volts.\(V_0\) is the size of the retarding potential. \(K_{max} = eV_0\) is positive.
- Thinking a higher accelerating potential always gives more current.Once every freed electron is collected, the current saturates. More \(V\) adds nothing.
- Thinking brighter light gives a larger stopping potential.Intensity changes only the number of electrons per second. \(V_0\) is set by the colour and the metal.
- Using \(1/r\) for the fall of intensity with distance.A small source spreads over a sphere, so the intensity falls as \(1/r^2\).