Photons & the Photoelectric Effect
The Ray & Wave Optics topic ended with strong evidence that light is a wave: two beams of light can cancel each other in interference, and light spreads round edges in diffraction. This topic goes inside the atom, in three parts: light that arrives in packets, the energy levels of atoms, and the nucleus, with its decays and the energy stored inside it. This first page starts with an experiment the wave picture could not explain. Light falling on a clean metal surface knocks electrons out of it, but only light of a high enough frequency does so; light of a lower frequency frees none, however bright it is made. Explaining that took a new idea. Light carries its energy in separate packets, called photons, and the frequency sets the size of each packet.
The syllabus lists this material in its Modern Physics section, in three lines: "Photoelectric effect", "Atomic structure" and "Fundamentals of nuclear physics". None of the four 2025–26 papers (December 2025 to April 2026) has asked a single question on it so far, so every example and question on this page is new, written in the style of the real papers. The papers give their constants inside each question, as they do with g, and so does this page. The topic draws on several earlier ones (the electric field, momentum and optics), and if time before the exam is short, those earlier topics matter more.
Energy in packets: the photon
In 1900 Max Planck was trying to explain the light given out by hot objects: why a heated iron bar glows dull red, then orange, then white as it gets hotter. Every theory that let the energy take any value failed. Planck found that the measurements fit exactly if the energy given out or taken in at a frequency ν can change only in whole steps of one fixed size, E = hν. Each step is called a quantum, and h is a new constant of nature, the Planck constant:
h = 6.6 × 10⁻³⁴ J·s.
In 1905 Albert Einstein went further. Light itself, he proposed, travels as separate packets of energy, later named photons, and each photon of light of frequency ν carries energy hν. Light is an electromagnetic wave with speed c = λν, so the energy can also be written with the wavelength: E = hν = hc/λ. A higher frequency, which means a shorter wavelength, gives a bigger packet. Blue light comes in bigger packets than red, and ultraviolet in bigger ones still.
The constant is tiny, so each packet is tiny. A photon of green light carries about 4 × 10⁻¹⁹ J, as the example below shows, and even a weak lamp sends out billions of billions of photons every second. That is why light looks smooth. It is also why brightness and colour are separate things. For light of one frequency every photon is the same size, so a brighter beam carries more photons each second, not bigger ones. A beam of power P delivers energy P each second, so it carries N = P/(hν) photons per second.
You have met h before. January 2026, Q1 asked for the base unit of mass, which since 2019 has been defined by fixing h at exactly 6.62607015 × 10⁻³⁴ J·s; the Quantities & Units topic explains how.
Green light has a wavelength of 500 nm. The energy of one photon of this light is ( ) (h = 6.6 × 10⁻³⁴ J·s, c = 3.0 × 10⁸ m/s)
A. 4.0 × 10⁻²⁸ J B. 1.3 × 10⁻²⁷ J C. 1.1 × 10⁻⁴⁸ J D. 4.0 × 10⁻¹⁹ J
Step 1, the wavelength in metres: 1 nm = 10⁻⁹ m, so λ = 500 × 10⁻⁹ m = 5.0 × 10⁻⁷ m.
Step 2, the energy: E = hc/λ = (6.6 × 10⁻³⁴ × 3.0 × 10⁸) ÷ (5.0 × 10⁻⁷) = (1.98 × 10⁻²⁵) ÷ (5.0 × 10⁻⁷) ≈ 4.0 × 10⁻¹⁹ J. The answer is D.
A leaves the wavelength as 500, in nanometres: 1.98 × 10⁻²⁵ ÷ 500 ≈ 4.0 × 10⁻²⁸ J, a billion times too small. B, h/λ, leaves out c, and the units give it away: J·s ÷ m is kg·m/s, not joules. (h/λ is in fact the photon's momentum, which comes up at the end of this page.) C turns the frequency upside down, using ν = λ/c instead of c/λ.
A green laser pointer of power 1.0 mW gives out 1.0 × 10⁻³ J every second, so it sends out N = 1.0 × 10⁻³ ÷ 4.0 × 10⁻¹⁹ = 2.5 × 10¹⁵ photons per second.
A smaller unit: the electronvolt
Photon energies of a few times 10⁻¹⁹ J are awkward to write, so atomic physics uses a smaller unit. The Potential & Potential Difference page showed that when a charge q crosses a potential difference U, the field does work qU on it. For an electron or a proton, q = e = 1.6 × 10⁻¹⁹ C, so crossing 1 V changes its energy by 1.6 × 10⁻¹⁹ C × 1 V = 1.6 × 10⁻¹⁹ J. That amount of energy is the electronvolt, eV, which the Charged Particles in Electric Fields page met briefly:
1 eV = 1.6 × 10⁻¹⁹ J.
To turn joules into electronvolts, divide by 1.6 × 10⁻¹⁹; to turn electronvolts into joules, multiply. The green photon above carries 4.0 × 10⁻¹⁹ ÷ 1.6 × 10⁻¹⁹ = 2.5 eV. The same sum at the two ends of the visible range gives about 1.8 eV for red light of 700 nm and 3.1 eV for violet light of 400 nm. Photons of visible light carry a few electronvolts each, and ultraviolet photons more; the next page measures the energy levels of atoms in the same unit. The electronvolt has one more advantage, used later on this page. A particle of charge e that crosses U volts gains or loses exactly U electronvolts, so the number of volts and the number of electronvolts are the same.
Light that knocks out electrons
The effect is easy to show. Fix a clean zinc plate to an electroscope and charge it negatively, so that the leaves spread apart. Shine ultraviolet light from an arc lamp on the zinc and the leaves fall together at once: the light is knocking electrons out of the plate, and the plate is losing its negative charge. Put a sheet of ordinary glass in the beam, which stops ultraviolet but lets visible light through, and the leaves stay apart, however bright the lamp. Charge the plate positively instead and the ultraviolet makes no difference, because any electrons it frees are pulled straight back. Electrons knocked out of a metal by light are called photoelectrons, and the effect is the photoelectric effect. Heinrich Hertz first noticed it in 1887, while experimenting with sparks.
To measure the effect, the metal is sealed inside an evacuated glass tube, a phototube, as in the figure. Light falls on the metal plate, the cathode K, and the photoelectrons cross the vacuum to a second electrode, the anode A. They return to K round the outside circuit, so a sensitive meter in the circuit shows a current, the photocurrent. A voltage between A and K can help the electrons across or, with the supply turned round, hold them back.
Experiments with phototubes gave four results.
The wave picture predicts something different every time. A wave delivers its energy smoothly over the whole surface, and a brighter wave delivers it faster. Then light of any frequency should free electrons if it is bright enough or shines for long enough, because the electrons would soak up energy until they had enough to leave. Dim light should need a noticeable wait before the first electron came out. And brighter light should shake the electrons harder and throw them out faster. None of this happens.
Einstein's equation
Einstein's photons explain all four results with one rule: a photon gives all of its energy, hν, to a single electron, at once, or it gives none. The electron then has to get out. The electrons in a metal are held in by the attraction of the positive ions, and leaving takes work. The least work, needed by the most loosely held electrons at the surface, is the work function W₀ of the metal. It depends only on the metal: about 2.3 eV for sodium and about 4.5 eV for tungsten. Whatever is left of the photon's energy after the electron has used W₀ to escape becomes kinetic energy. By the conservation of energy, for the fastest electrons,
hν = W₀ + Ek, so Ek = hν − W₀.
Here Ek is the largest kinetic energy, the one the fastest electrons have. Electrons from deeper inside the metal lose more energy on the way out and leave more slowly, which is why there is a range of speeds; some books write the largest value as Ek,max or ½mvmax².
The four results now follow one by one.
A threshold. An electron can leave only if hν is at least W₀. The lowest frequency that works is given by hν₀ = W₀, so ν₀ = W₀/h. Since λ = c/ν, the longest wavelength that works is λ₀ = hc/W₀. Below ν₀ each photon is too small, and an electron practically never collects the energy of two photons. Brighter light only sends more photons that are each too small, so no electrons leave, however bright the light or however long it shines.
No delay. One photon is enough to free one electron, so nothing has to build up.
Energy set by frequency. Ek = hν − W₀ grows with ν. Brighter light of the same frequency has photons of the same size, so it leaves Ek unchanged.
Number set by intensity. Each photon frees at most one electron. Not every photon frees one, but the number freed each second stays in proportion to the number arriving, and a brighter beam brings more photons. Einstein received the Nobel Prize in 1921 for this explanation.
Light of frequency 8.0 × 10¹⁴ Hz falls on a sodium surface. The work function of sodium is 2.3 eV. The maximum kinetic energy of the photoelectrons is ( ) (h = 6.6 × 10⁻³⁴ J·s, 1 eV = 1.6 × 10⁻¹⁹ J)
A. 1.0 eV B. 3.3 eV C. 5.6 eV D. 2.3 eV
Step 1, the photon energy: hν = 6.6 × 10⁻³⁴ × 8.0 × 10¹⁴ = 5.28 × 10⁻¹⁹ J, and 5.28 × 10⁻¹⁹ ÷ 1.6 × 10⁻¹⁹ = 3.3 eV.
Step 2, the equation: Ek = hν − W₀ = 3.3 − 2.3 = 1.0 eV, which is 1.6 × 10⁻¹⁹ J. The answer is A.
B is the photon energy: it forgets the energy spent getting out. C adds the work function instead of subtracting it. D is the work function itself, the energy the electron spends leaving, not the energy it leaves with.
The same numbers give sodium's threshold: ν₀ = W₀/h = (2.3 × 1.6 × 10⁻¹⁹) ÷ (6.6 × 10⁻³⁴) ≈ 5.6 × 10¹⁴ Hz, and λ₀ = c/ν₀ ≈ 5.4 × 10⁻⁷ m, about 540 nm. Green, blue and violet light free electrons from sodium; yellow, orange and red light do not, however bright. Not even the yellow light of a sodium street lamp, at 589 nm, can do it.
The stopping voltage
Ek can be measured with the phototube itself. Turn the supply round, so that the anode A is at a lower potential than the cathode K, by a voltage U. Now the field between them pushes the electrons back towards K. As the Potential & Potential Difference page showed, the field does work −eU on an electron that crosses from K to A against it, so the electron arrives with eU less kinetic energy than it left with, if it arrives at all. Electrons that left K with less than eU are turned back before they reach A.
Raise U from zero and the slower electrons drop out first, so the photocurrent falls. At one voltage even the fastest electrons are just stopped, and the current falls to zero. This is the stopping voltage Uc, and at that point the field has taken away all of the largest kinetic energy:
eUc = Ek = hν − W₀.
In electronvolts this needs no arithmetic: a stopping voltage of 1.0 V means Ek = 1.0 eV. The sodium surface in the worked example above, with Ek = 1.0 eV, has a stopping voltage of 1.0 V.
Ultraviolet light of frequency 1.6 × 10¹⁵ Hz falls on a tungsten cathode. The work function of tungsten is 4.5 eV. The stopping voltage is ( ) (h = 6.6 × 10⁻³⁴ J·s, e = 1.6 × 10⁻¹⁹ C)
A. 6.6 V B. 4.5 V C. 3.4 × 10⁻¹⁹ V D. 2.1 V
Step 1, the photon energy in eV: hν = 6.6 × 10⁻³⁴ × 1.6 × 10¹⁵ = 1.056 × 10⁻¹⁸ J, and 1.056 × 10⁻¹⁸ ÷ 1.6 × 10⁻¹⁹ = 6.6 eV.
Step 2, the largest kinetic energy: Ek = 6.6 − 4.5 = 2.1 eV.
Step 3, the voltage: eUc = Ek, so Uc = 2.1 V; the number of electronvolts is the number of volts. The answer is D.
A is the photon energy read as a voltage, with the work function forgotten. B is the work function read as a voltage. C is Ek in joules, 2.1 × 1.6 × 10⁻¹⁹ ≈ 3.4 × 10⁻¹⁹ J, written in volts: the last step, Uc = Ek/e, has been left out.
Brighter or bluer: the photocurrent graph
Sweep the voltage between A and K from reverse to forward and record the photocurrent I at each voltage U. The graph has three parts, each with a reason.
On the far left, at the stopping voltage −Uc, the current is zero: the reverse field turns back even the fastest electrons. Moving right, the reverse voltage weakens, more electrons get through, and the current rises. At U = 0 the current is not zero, because the electrons leave K with kinetic energy and some reach A unaided. With a forward voltage the field helps them across, and once it is large enough every electron that leaves K reaches A. The current then stops rising. This level is the saturation current. If n electrons leave K each second, the saturation current is I = ne, so it measures how many electrons the light frees each second.
The figure shows the same phototube lit three ways. Bright and dim green light have the same frequency, so their photons are the same size, the electrons leave with the same Ek, and the two curves reach zero at the same stopping voltage. The bright light frees more electrons each second, so its saturation current is higher: twice the intensity, twice the saturation current. Blue light has a higher frequency than green, so its photons are bigger, its electrons leave faster, and its curve reaches zero at a larger reverse voltage. The height of its plateau depends only on how bright the blue beam is.
The figure above shows the photocurrent I against the voltage U for one phototube lit by bright green, dim green and blue light. Which statement is correct?
A. The bright green light has the highest frequency, because it gives the largest current.
B. The photoelectrons freed by the blue light have the largest maximum kinetic energy.
C. The photoelectrons freed by the dim green light are slower than those freed by the bright green light.
D. The blue light is the brightest, because its curve reaches zero at the largest reverse voltage.
Read the stopping points first. The two green curves reach zero at the same reverse voltage, so their electrons have the same largest kinetic energy. The blue curve reaches zero at a larger reverse voltage, and eUc = Ek, so Ek is largest for blue. The answer is B.
A reads the height of a curve as its frequency. The saturation current counts electrons, and two curves with the same stopping point have the same frequency. C: the same stopping voltage means the same Ek; dim light gives fewer electrons, not slower ones. D reads the stopping voltage as brightness, but it measures the photon energy, which is set by the frequency.
The Ek–ν graph
Einstein's equation, Ek = hν − W₀, has the form y = kx + b from the maths course. Plotted against ν, Ek is a straight line with slope h and intercept −W₀ on the Ek-axis. It crosses the ν-axis where Ek = 0, at ν₀. Only the part to the right of ν₀ can be measured, because below ν₀ there are no photoelectrons; the part below the axis is drawn dashed, as an extension of the line, to show where it meets the Ek-axis.
Two things follow. First, the slope is h for every metal, so the lines for different metals are parallel. A metal with a larger work function has its line moved to the right, with a larger ν₀ and a lower intercept, −W₀. Second, the line does not pass through the origin, so Ek is not proportional to ν. For metal 1 in the figure below (W₀ = 3.3 × 10⁻¹⁹ J), light of 1.0 × 10¹⁵ Hz gives Ek = 6.6 × 10⁻¹⁹ − 3.3 × 10⁻¹⁹ = 3.3 × 10⁻¹⁹ J. Light of twice that frequency gives 13.2 × 10⁻¹⁹ − 3.3 × 10⁻¹⁹ = 9.9 × 10⁻¹⁹ J, three times as much. Doubling the frequency doubles the photon energy, but W₀ is taken away only once, so Ek more than doubles.
Robert Millikan, who doubted Einstein's idea, measured stopping voltages at several frequencies and published the results in 1916. His points lay on a straight line, and its slope gave h, matching the value Planck had found from glowing objects. A graph of Uc against ν is the same line divided by e: its slope is h/e and its intercept on the Uc-axis is −W₀/e.
For a certain metal, the graph of the maximum kinetic energy Ek of the photoelectrons against the frequency ν of the light is a straight line. It crosses the ν-axis at 5.0 × 10¹⁴ Hz, and at 1.5 × 10¹⁵ Hz it gives Ek = 6.6 × 10⁻¹⁹ J. The work function of the metal is ( )
A. 6.6 × 10⁻¹⁹ J B. 9.9 × 10⁻¹⁹ J C. 3.3 × 10⁻¹⁹ J D. 5.0 × 10¹⁴ J
Step 1, the slope: from 5.0 × 10¹⁴ Hz to 1.5 × 10¹⁵ Hz the line rises 6.6 × 10⁻¹⁹ J over 1.0 × 10¹⁵ Hz, a slope of 6.6 × 10⁻³⁴ J·s. That is h, as it must be.
Step 2, the work function: W₀ = hν₀ = 6.6 × 10⁻³⁴ × 5.0 × 10¹⁴ = 3.3 × 10⁻¹⁹ J. Extending the line back to ν = 0 gives the same number, as the intercept −3.3 × 10⁻¹⁹ J. The answer is C. This is metal 1 in the figure.
A is the Ek read from the graph, not the work function. B is the photon energy at 1.5 × 10¹⁵ Hz, hν = 9.9 × 10⁻¹⁹ J, the energy before the work function has been taken away. D is the threshold frequency with its unit changed; a frequency becomes an energy only when it is multiplied by h.
Wave and particle
So is light a wave or a stream of particles? The Interference & Diffraction of Light page showed fringes from two slits, light spreading behind a narrow slit, the bright Poisson spot and polarization, all things that only a wave can do. The photoelectric effect shows light handing over its energy in whole packets of hν, one electron at a time, which only particles do. Both sets of results are real, so light is both, and neither picture alone is complete. As light travels and spreads it behaves as a wave, and the wave tells where the photons are likely to arrive; when it gives up or takes in energy, it does so one photon at a time. This is wave–particle duality. The shorter the wavelength, the more the particle side shows: X-rays and γ-rays behave much like particles, while radio waves behave as waves.
Photons carry momentum as well as energy, p = h/λ, as Arthur Compton found in 1923 when X-rays bounced off electrons like small balls. In 1924 Louis de Broglie turned the idea round. If a wave can behave as particles, perhaps a particle with momentum p behaves as a wave, with wavelength
λ = h/p = h/(mv).
These matter waves were found in 1927, when a beam of electrons fired through a thin metal foil made a pattern of rings, just as X-rays do. For everyday objects the wavelength is far too small to notice: a 0.1 kg ball at 10 m/s has p = 1 kg·m/s and λ = 6.6 × 10⁻³⁴ m, far smaller than any gap it could pass through. An electron's mass is so small that its wavelength can match the spacing of atoms, about 10⁻¹⁰ m, as the next example shows. For a particle given a kinetic energy Ek, the momentum comes from p = √(2mEk), from the Momentum & Impulse topic.
An electron, starting from rest, is accelerated through a potential difference of 150 V. Its de Broglie wavelength is about ( ) (h = 6.6 × 10⁻³⁴ J·s, electron mass m = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C)
A. 1.0 × 10⁻¹⁰ m B. 8.3 × 10⁻⁹ m C. 1.4 × 10⁻¹⁰ m D. 2.8 × 10⁻¹⁷ m
Step 1, the kinetic energy: the field does work eU = 1.6 × 10⁻¹⁹ × 150 = 2.4 × 10⁻¹⁷ J, as on the Charged Particles in Electric Fields page.
Step 2, the momentum: p = √(2mEk) = √(2 × 9.1 × 10⁻³¹ × 2.4 × 10⁻¹⁷) = √(4.4 × 10⁻⁴⁷) ≈ 6.6 × 10⁻²⁴ kg·m/s.
Step 3, the wavelength: λ = h/p = 6.6 × 10⁻³⁴ ÷ 6.6 × 10⁻²⁴ = 1.0 × 10⁻¹⁰ m. The answer is A. That is about the spacing of atoms in a crystal, which is why crystals diffract electron beams, and why electron microscopes show far finer detail than light microscopes: visible light has wavelengths thousands of times longer.
B treats the electron as a photon of the same energy: λ = hc/E = 1.98 × 10⁻²⁵ ÷ 2.4 × 10⁻¹⁷ ≈ 8.3 × 10⁻⁹ m. hc/E holds only for light; for a particle with mass, use λ = h/p. C drops the 2 from p = √(2mEk), which makes p too small by a factor of √2 and λ too large. D divides h by the energy instead of the momentum.
Your turn! · Practice questions
Things to remember
| Concept | What to know |
|---|---|
| Photon | E = hν = hc/λ, with h = 6.6 × 10⁻³⁴ J·s. A shorter wavelength means a bigger photon. A brighter beam of the same light has more photons, N = P/(hν) per second, not bigger ones. |
| Electronvolt | 1 eV = 1.6 × 10⁻¹⁹ J, the energy a charge e gains across 1 V. Visible photons carry about 1.8–3.1 eV. Convert to joules before using E = hν. |
| The four results | A threshold frequency for each metal; no delay; the largest kinetic energy set by the frequency; the number of electrons per second set by the intensity. The wave picture fails on all four. |
| Einstein's equation | Ek = hν − W₀. One photon gives all its energy to one electron. The work function W₀ depends only on the metal. |
| Threshold | ν₀ = W₀/h and λ₀ = hc/W₀. Below ν₀ (above λ₀) no electrons come out, however bright the light or however long it shines. |
| Stopping voltage | eUc = Ek = hν − W₀. Uc in volts equals Ek in eV. Set by the frequency and the metal, not by the intensity. |
| Photocurrent graph | Brighter light of the same colour: a higher saturation current, the same stopping voltage. A higher frequency: the stopping voltage moves further out. The current is not zero at U = 0. |
| Ek–ν graph | A straight line of slope h, the same for every metal, crossing the ν-axis at ν₀, with Ek-intercept −W₀. Doubling ν more than doubles Ek. |
| Wave–particle duality | Interference, diffraction and polarization show waves; the photoelectric effect shows particles. Light is both. |
| de Broglie wavelength | λ = h/p = h/(mv); for a given kinetic energy, p = √(2mEk). More momentum, shorter wavelength. An electron accelerated through 150 V: about 10⁻¹⁰ m. |
| Common slips | λ left in nm; eV and J mixed; W₀ forgotten or added; Ek taken as proportional to ν; brighter light taken to mean faster electrons; hc/E used for a particle with mass. |