Wave and Particle Nature of Light

Lessons

Senior 5 · Physics

Wave and Particle Nature of Light

5 lessons

Lesson 3 of 5

Factors Affecting Photoelectric Emission and Its Applications

25 min

Brighter or bluer?

You can change a beam of light in two independent ways: make it brighter, or change its colour. In the last lesson you saw that the photon model predicts quite different effects for each. This lesson tests those predictions on the full current–voltage characteristic of a photocell, lists every factor that controls photoemission, and then looks at devices that put the effect to work, from an astronomer's single-photon detector to the solar panels spreading across Rwanda.

The current–voltage characteristic

With the apparatus of the last lesson, the light is kept fixed and the collector potential $V$ is varied from negative to positive while the photocurrent $I$ is recorded. The graph shows three beams.

Large positive $V$ : every photoelectron that leaves the emitter is collected, so the current reaches a steady saturation current $I_{s}$ . Raising $V$ further cannot increase it, because no more electrons are being emitted.
Small or negative $V$ : slower electrons are turned back, so the current falls.
$V = - V_{s}$ : even the fastest electrons are stopped, and the current is zero.

Compare the curves. A and B have the same frequency, so the same stopping potential (0.8 V), but B has twice the photon rate and twice the saturation current. A and C have the same photon rate and so the same saturation current, but C has the higher frequency, so its electrons are faster and its stopping potential (1.6 V) is larger.

The saturation current is fixed by the rate at which electrons are emitted. Not every photon frees an electron: many are reflected, or give their energy to electrons that never reach the surface. The fraction that do is the quantum efficiency $\eta$ . If $N$ photons arrive per second,

I_s = \eta N e = \eta \frac{P\lambda}{hc} e

where $P$ is the light power reaching the emitter.

Definition

The saturation current $I_{s}$ is the largest photocurrent a photocell gives for a particular beam, reached when every emitted electron is collected. The quantum efficiency $\eta$ of a surface is the number of photoelectrons emitted per photon that falls on it. Example: a photocathode with $\eta = 0.20$ emits one electron for every five photons.

The factors that affect photoelectric emission

Factor	What it changes	What it does not change
Frequency $f$ of the radiation	Whether emission happens at all ( $f \ge f_0$ ); $E_{k,\max}$ and $V_{s}$ , which rise linearly with $f$	The photocurrent, for a fixed photon rate
Intensity (photon rate)	The number of electrons per second, so the photocurrent and $I_{s}$ , in direct proportion	$E_{k,\max}$ and $V_{s}$
The metal (its work function $\phi$ )	$f_{0}$ ; and, at a given $f$ , $E_{k,\max} = hf - \phi$	The gradient $\frac{h}{e}$ of the $V_{s}$ – $f$ line
State of the surface	Oxide or dirt layers change $\phi$ and lower $\eta$	Planck's constant
Collector potential $V$	The fraction of emitted electrons collected, up to saturation	The number emitted

For a small lamp, the intensity at the cell falls as the inverse square of the distance, so moving the lamp twice as far away cuts the photocurrent to a quarter. The stopping potential does not change at all.

Worked example 1: the saturation current

Light of wavelength 420 nm and power 3.0 mW falls on a photocathode of quantum efficiency 0.20. Find the saturation current.

Solution

Photons per second: $N = \frac{P\lambda}{hc} = \frac{(3.0 \times 10^{-3})(420 \times 10^{-9})}{(6.63 \times 10^{-34})(3.00 \times 10^{8})} = 6.3 \times 10^{15}\,\mathrm{s^{-1}}$ .

Electrons per second: $\eta N = 0.20 \times 6.33 \times 10^{15} = 1.27 \times 10^{15}\,\mathrm{s^{-1}}$ .

$I_s = (1.27 \times 10^{15})(1.60 \times 10^{-19}) = 2.0 \times 10^{-4}\,\mathrm{A} = 0.20\,\mathrm{mA}$ .

Worked example 2: reading the characteristic

For curve C above, the emitter has work function 2.1 eV. (a) Find the maximum kinetic energy of the photoelectrons. (b) Find the wavelength of the light. (c) The lamp producing beam A is moved from 20 cm to 40 cm from the cell. Describe the new curve.

Solution

(a) $E_{k,\max} = eV_s = 1.6$ eV.

(b) $hf = \phi + E_{k,\max} = 2.1 + 1.6 = 3.7$ eV, so $\lambda = \frac{1240\,\mathrm{eV\,nm}}{3.7\,\mathrm{eV}} = 335$ nm, in the near ultraviolet.

(c) Doubling the distance from a small lamp quarters the intensity, so the saturation current falls to $\frac{1}{4}$ of 1 unit, 0.25 units. The frequency is unchanged, so the curve still meets the axis at $- 0.8$ V.

Applications

Light-beam switches and counters

A beam of light falls on a photocell; when an object or person breaks the beam, the current drops and a relay switches. This is the principle of automatic doors, burglar alarms, the safety beam that stops a lift door closing on you, and the counters that tally bottles on a factory conveyor. The first such sensors were vacuum photocells. Most modern ones are semiconductor photodiodes, which use the internal photoelectric effect: the photon frees an electron inside the material instead of out of it, but the rule "one photon, one electron, and only if $h f$ is large enough" is the same.

The photomultiplier tube

A photomultiplier detects light so faint that individual photons can be counted. A photon frees one electron from the photocathode. The electron is accelerated to the first dynode, where it knocks out several secondary electrons; these are accelerated to the next dynode, and so on.

If each dynode releases $\delta$ electrons for each one that strikes it, and there are $n$ dynodes, the gain is $G = \delta^n$ . Photomultipliers are used in astronomical photometers, in scintillation counters that detect nuclear radiation, and in medical scanners.

Worked example 3: the gain of a photomultiplier

A photomultiplier has 10 dynodes, each releasing 4 secondary electrons per incident electron. (a) Find its gain. (b) Find the charge reaching the anode for one photoelectron. (c) Faint starlight delivers 1000 photons per second to its photocathode, whose quantum efficiency is 0.25. Find the anode current.

Solution

(a) $G = 4^{10} = 1.05 \times 10^{6}$ .

(b) $Q = Ge = (1.05 \times 10^{6})(1.60 \times 10^{-19}) = 1.7 \times 10^{-13}$ C: a pulse large enough for electronics to count.

(c) Photoelectrons per second $= 0.25 \times 1000 = 250$ . Using a gain of $1.0 \times 10^{6}$ , $I = (250)(1.0 \times 10^{6})(1.60 \times 10^{-19}) = 4 \times 10^{-11}$ A, which a sensitive meter can measure, although the starlight alone (photons of about 2.5 eV) delivers only about $4 imes 10^{-16}$ W.

Image sensors and solar cells

The camera in a phone is an array of millions of tiny silicon photodiodes (pixels). During an exposure each pixel collects the electrons freed by the photons that strike it; the charge, proportional to the number of photons, is read out and turned into a number. A solar (photovoltaic) cell also uses the internal photoelectric effect: photons free electrons in silicon, and the built-in field of a p–n junction (studied in Senior 6) drives them round a circuit, so light energy becomes electrical energy directly. Silicon responds to photons above about 1.1 eV, that is, wavelengths shorter than about 1100 nm, which covers the visible spectrum and the near infrared.

Warning

"Brighter light makes faster electrons." It does not. Intensity changes how many electrons come out, not how fast. If a question changes only the intensity, the stopping potential is unchanged; if it changes only the frequency at a fixed photon rate, the saturation current is unchanged.

Tip

Read the graph from both ends. On an $I$ – $V$ characteristic, the left end (where $I = 0$ ) tells you about the photon energy; the right end (the plateau) tells you about the number of photons.

How do solar panels work? by Richard Komp (TED-Ed). Watch for how a photon frees an electron in silicon and how the junction makes the electrons flow as a current.

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Practice

Explain why the photocurrent reaches a saturation value as the collector potential is increased.
A beam delivers $5.0 \times 10^{15}$ photons per second to a surface of quantum efficiency 0.10. Find the saturation current.
Sketch, on one set of axes, the $I$ – $V$ characteristics for (a) a beam, (b) the same beam with three times the intensity. Label the stopping potential.
A small lamp 0.30 m from a photocell gives a saturation current of 9.0 μA. Find the saturation current when the lamp is 0.90 m away, and state what happens to the stopping potential.
A photomultiplier with 8 dynodes has a gain of $3.9 \times 10^{5}$ . Find the number of secondary electrons released per electron at each dynode.
Explain why a photovoltaic cell made of silicon can use infrared light of wavelength 1000 nm, while a vacuum photocell with a sodium emitter cannot.
The door sensor of a shop in Huye uses a vacuum photocell lit by a yellow-green LED of wavelength 560 nm. Which of the emitters caesium, sodium and zinc would work? Justify your choice.

Show answersHide answers

At high enough $V$ every emitted photoelectron reaches the collector. The current is then limited by the rate of emission, which depends only on the photon rate and the quantum efficiency, so raising $V$ further cannot increase it.
$I_s = \eta N e = (0.10)(5.0 \times 10^{15})(1.60 \times 10^{-19}) = 8.0 \times 10^{-5}$ A $= 80\,\mu\mathrm{A}$ .
Both curves meet the $V$ -axis at the same point, $- V_{s}$ ; curve (b) has a plateau three times as high.
The distance is tripled, so the intensity falls to $\frac{1}{9}$ : $I_s = 1.0\,\mu\mathrm{A}$ . The stopping potential is unchanged.
$\delta = (3.9 \times 10^{5})^{\frac{1}{8}} = 5.0$ .
A 1000 nm photon has energy $\frac{1240}{1000} = 1.24$ eV. This is more than silicon's 1.1 eV, so it can free an electron inside the silicon, but far less than sodium's work function (2.28 eV), so it cannot free an electron from the sodium surface.
Photon energy $= \frac{1240}{560} = 2.21$ eV. Only caesium ( $\phi \approx 2.1$ eV) works: it gives photoelectrons with up to about 0.1 eV. Sodium (2.28 eV) and zinc (4.3 eV) have work functions above 2.21 eV, so no electrons are emitted however bright the LED.

Key points

On the $I$ – $V$ curve, $V_{s}$ (left end) depends on frequency; $I_{s}$ (plateau) depends on photon rate: $I_s = \eta N e$ .
Emission is controlled by frequency (and $f_{0}$ ), intensity, the metal's work function, the state of the surface and the collector potential.
Photocells and photodiodes switch and count; photomultipliers (gain $\delta^n$ ) count single photons; camera sensors and solar cells use the internal photoelectric effect.

Sources and credits

Photo: RMA 1P21 photomultiplier tube by Ladd Observatory, Wikimedia Commons, licensed under CC BY-SA 4.0.
Map: Rwanda PVOUT photovoltaic power potential map, Global Solar Atlas 2.0, The World Bank, ESMAP and Solargis, Wikimedia Commons, licensed under CC BY 4.0.
Video: How do solar panels work? - Richard Komp, TED-Ed, embedded from YouTube.

Written by Ganzaa. The characteristic graph, the photomultiplier diagram, the worked examples and the practice exercises are by Ganzaa. This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Brighter or bluer?

2The current–voltage characteristic

3The factors that affect photoelectric emission

4Applications

Light-beam switches and counters

The photomultiplier tube

Image sensors and solar cells

Brighter or bluer?

The current–voltage characteristic

The factors that affect photoelectric emission

Applications