The photoelectric effect — OCR A-Level Physics
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The photoelectric effect explained
This row is a guided-reading instruction for section 4.5.2, so it carries no assessed content of its own.
Read the full explanation
Read the specification statements that follow it as a connected account of the photoelectric effect. First identify the effect itself: electrons are emitted from a metal surface when electromagnetic radiation above a threshold frequency is incident on it. Then study the simple demonstration, such as a clean zinc plate on a gold-leaf electroscope, and note what is observed when ultraviolet radiation is used and when it is not. Link the observations to photon energy, work function and the idea that one photon interacts with one electron. Use the reading to build a clear causal chain from incident photon to emitted electron, and check that you can describe the apparatus, the observation and the explanation without mixing them up.
(i) photoelectric effect, including a simple experiment to demonstrate this effect
The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency is incident on it. In a simple demonstration, a clean zinc plate is fixed to the cap of a gold-leaf electroscope. When the plate is negatively charged, the gold leaf diverges; shining ultraviolet radiation on the plate causes the leaf to fall because electrons are emitted and the charge is neutralised. If the plate is positively charged, the leaf stays diverged because emitted electrons are attracted back. Radiation below the threshold frequency produces no emission, however intense it is, which supports the photon model rather than a wave model. The emitted electrons have a maximum kinetic energy given by hf = φ + Ek(max).
(ii) demonstration of the photoelectric effect using, e.g. gold-leaf electroscope and zinc plate
A gold-leaf electroscope with a clean zinc plate on its cap can demonstrate the photoelectric effect. The zinc plate is cleaned with emery paper to remove the oxide layer, then charged negatively; the gold leaf diverges because like charges repel. When ultraviolet radiation is shone on the zinc plate, electrons are emitted from the surface, the negative charge is reduced and the leaf falls. If the plate is charged positively, the leaf remains diverged because emitted electrons are attracted back to the positive plate. If visible light is used instead of ultraviolet, no emission occurs because its frequency is below the threshold frequency for zinc. This shows that emission depends on frequency, not on intensity alone.
(b) a one-to-one interaction between a photon and a surface electron
The photoelectric effect is explained by treating light as a stream of photons, each carrying energy E = hf, where h is the Planck constant and f the frequency. Emission occurs when a single photon is absorbed by a single electron at or near the metal surface in a one-to-one interaction. All the photon energy is transferred to that electron; it is not shared among many electrons and does not accumulate over time. The electron uses energy equal to the work function φ to escape the surface, and any surplus becomes kinetic energy. If hf is below φ, no emission occurs however intense the beam, because intensity means more photons, not more energetic ones. A useful check is to compare hf with φ for one photon and one electron.
(c) Einstein’s photoelectric equation hf = φ + KEmax
Einstein’s photoelectric equation is hf = φ + KEmax, where hf is the energy of one absorbed photon, φ is the work function of the metal surface and KEmax is the maximum kinetic energy of an emitted photoelectron. The equation is a statement of energy conservation for a single one-to-one photon-electron interaction: the photon energy is used partly to escape the surface and the remainder becomes kinetic energy. Rearranged, KEmax = hf − φ, so KEmax rises linearly with frequency and is zero at the threshold frequency f₀ = φ/h. Because φ is a property of the metal, different metals give different intercepts on a KEmax against f graph, while the gradient is h. Intensity does not appear in the equation, so it cannot change KEmax.
(d) work function; threshold frequency
The work function φ is the minimum energy required to remove an electron from the surface of a particular metal; it is a property of the metal and is often quoted in electronvolts, where 1 eV = 1.60 × 10⁻¹⁹ J. The threshold frequency f₀ is the lowest frequency of incident radiation that can cause photoelectric emission from that surface. It follows from Einstein’s equation with KEmax = 0, giving hf₀ = φ, so f₀ = φ/h. Radiation below f₀ cannot release electrons no matter how intense it is, because each photon then carries less than φ. The corresponding threshold wavelength is λ₀ = c/f₀ = hc/φ. Work function and threshold frequency are linked by φ = hf₀, so a larger work function means a higher threshold frequency.
(e) the idea that the maximum kinetic energy of the photoelectrons is independent of the intensity of the incident radiation
In the photon model, intensity is the energy arriving per unit area per unit time, so a more intense beam of fixed frequency delivers more photons per second. Each photon still carries the same energy hf, so each emitted electron gains the same maximum kinetic energy KEmax = hf − φ. Increasing intensity therefore increases the number of photoelectrons emitted per second, which raises the photocurrent, but it does not change KEmax. To change KEmax you must change the frequency of the radiation or use a metal with a different work function. This independence of KEmax from intensity is a key piece of evidence for the photon model and cannot be explained by a continuous wave picture.
(f) the idea that rate of emission of photoelectrons above the threshold frequency is directly proportional to the intensity of the incident radiation.
Above the threshold frequency, each incident photon has enough energy to release one photoelectron. Doubling the intensity of monochromatic radiation doubles the number of photons arriving per second, so the rate of emission of photoelectrons doubles. The photoelectric current is therefore directly proportional to intensity, provided the frequency is above the threshold and the metal surface is unchanged. Below the threshold frequency, increasing intensity does not cause emission because no single photon has sufficient energy. A useful check is to compare currents for two intensities at the same frequency: if the current doubles when intensity doubles, the proportionality is confirmed. This relationship is separate from the maximum kinetic energy of the emitted electrons, which depends on frequency, not intensity.
Your focus
- Identify the key ideas in section 4.5.2 and organise them into a coherent account.
- Describe the simple demonstration of the photoelectric effect and the observations it produces.
- Explain the observations using photon energy, threshold frequency and work function.
Show all 24 objectives
- Describe the photoelectric effect and a simple experiment that demonstrates it.
- Apply hf = φ + Ek(max) to calculate maximum kinetic energy or work function.
- Explain how the observations support the photon model of electromagnetic radiation.
- Describe how a gold-leaf electroscope and zinc plate demonstrate the photoelectric effect.
- Explain the observations for negative and positive charging and for different frequencies.
- Relate the demonstration to the threshold frequency and the photon model.
- Describe the photoelectric interaction as a one-to-one event between a single photon and a single surface electron.
- Use E = hf to compare the energy of one photon with the energy needed to release an electron.
- Explain why intensity affects the number of emitted electrons rather than the energy of each one.
- State and apply Einstein’s photoelectric equation hf = φ + KEmax to photoelectric emission.
- Rearrange the equation to find KEmax, φ or the threshold frequency from given data.
- Interpret the gradient and intercept of a graph of maximum kinetic energy against frequency.
- Define work function and threshold frequency and state the relationship φ = hf₀.
- Calculate threshold frequency or threshold wavelength from a given work function.
- Explain why radiation below the threshold frequency produces no photoelectric emission.
- Explain why the maximum kinetic energy of photoelectrons is independent of the intensity of the incident radiation.
- Relate changes in intensity to changes in the number of photoelectrons emitted per second and the photocurrent.
- Use the photon model to justify the observed independence of KEmax from intensity.
- Describe the relationship between the rate of emission of photoelectrons and the intensity of incident radiation above the threshold frequency.
- Explain why the photoelectric current is proportional to intensity at constant frequency above the threshold.
- Distinguish between the effects of intensity and frequency on photoelectric emission.
The photoelectric effect exam tips
Marking Points
- The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation is incident on it.
- Emission occurs only when the frequency of the radiation is at or above a threshold frequency that depends on the metal.
- The intensity of the radiation affects the number of electrons emitted per second, not the maximum kinetic energy of each electron.
- The maximum kinetic energy of an emitted electron is given by hf = φ + Ek(max), where φ is the work function.
- In the zinc-plate demonstration, ultraviolet radiation discharges a negatively charged electroscope but not a positively charged one.
- The observations support the photon model, in which one photon interacts with one electron.
- The zinc plate must be clean, for example rubbed with emery paper, so that the oxide layer does not prevent electron emission.
- The electroscope is charged negatively and the gold leaf diverges because the leaf and stem carry like charges.
- Ultraviolet radiation incident on the zinc plate causes electrons to be emitted, reducing the negative charge so the leaf falls.
- A positively charged electroscope does not discharge because emitted electrons are attracted back to the positive plate.
- Visible light does not discharge the negatively charged electroscope because its frequency is below the threshold frequency for zinc.
- The demonstration shows that emission depends on the frequency of the radiation, supporting the photon model.
- A photon is a quantum of electromagnetic radiation carrying energy E = hf, where h is the Planck constant and f is the frequency.
- In the photoelectric effect one photon interacts with one surface electron, so the interaction is one-to-one rather than spread over many electrons.
- The whole photon energy hf is transferred to that single electron; energy is not shared or accumulated over time.
- Emission requires hf to be at least the work function φ; surplus energy appears as kinetic energy of the emitted electron.
- Increasing intensity increases the number of photons per second, so it increases the rate of emission, not the energy delivered per photon.
- hf is the energy of one photon, with h the Planck constant and f the frequency of the incident radiation.
- φ is the work function, the minimum energy needed to remove an electron from the metal surface.
- KEmax is the maximum kinetic energy of an emitted photoelectron, corresponding to electrons that lose no energy in escaping.
- The equation expresses conservation of energy for a single photon absorbed by a single electron.
- Rearranging gives KEmax = hf − φ, so KEmax increases with frequency and is zero at the threshold frequency f₀ = φ/h.
- A graph of KEmax against f has gradient h and intercept −φ on the KEmax axis.
- The work function φ is the minimum energy needed to remove an electron from the surface of a metal.
- φ is a property of the metal surface and may be expressed in joules or electronvolts, with 1 eV = 1.60 × 10⁻¹⁹ J.
- The threshold frequency f₀ is the minimum frequency of incident radiation that produces photoelectric emission from that surface.
- At the threshold, KEmax = 0, so hf₀ = φ and therefore f₀ = φ/h.
- The corresponding threshold wavelength is λ₀ = c/f₀ = hc/φ.
- Radiation below the threshold frequency produces no emission at any intensity, because each photon carries less than the work function.
- Intensity is the energy transferred per unit area per unit time, so for fixed frequency it is proportional to the number of photons arriving per second.
- Each photon has energy hf, so increasing intensity at fixed frequency does not increase the energy carried by an individual photon.
- KEmax = hf − φ depends only on the photon frequency and the work function of the metal, so it is independent of intensity.
- Increasing intensity increases the number of photoelectrons emitted per second and hence the photocurrent.
- Changing KEmax requires a change in frequency or a change in the metal used, not a change in intensity.
- This independence supports the photon model over a continuous wave model of light.
- Above the threshold frequency, the rate of emission of photoelectrons is directly proportional to the intensity of the incident radiation.
- Intensity is the energy per unit area per unit time; for monochromatic light it is proportional to the number of photons incident per second.
- Each absorbed photon can release at most one photoelectron, so doubling the photon arrival rate doubles the emission rate.
- The photoelectric current is proportional to the rate of emission of photoelectrons, so current is proportional to intensity at constant frequency above threshold.
- Below the threshold frequency, increasing intensity does not produce photoelectrons because individual photons lack sufficient energy.
Examiner Tips
- 💡Read the whole of section 4.5.2 before attempting questions, so the demonstration and the theory are linked.
- 💡Write a short summary in your own words that names the apparatus, the observation and the photon explanation.
- 💡Check that you can distinguish threshold frequency, work function and maximum kinetic energy of emitted electrons.
- 💡Use the reading to prepare definitions and equations, then test yourself by describing the zinc-plate demonstration from memory.
- 💡State the threshold condition clearly: emission requires f ≥ f₀, where f₀ is the threshold frequency.
- 💡Use hf = φ + Ek(max) and identify each term before substituting values.
- 💡When describing the demonstration, say what happens to the gold leaf and explain why in terms of emitted electrons.
- 💡Compare photon and wave explanations explicitly when the question asks why the wave model fails.
- 💡Describe the charging step, the cleaning step and the observation in a logical sequence.
- 💡Name the radiation used, ultraviolet, and state that its frequency exceeds the threshold frequency for zinc.
- 💡Explain the role of the electroscope: it detects the loss of charge when electrons leave the plate.
- 💡Use the positive-charge control observation to strengthen the explanation of electron emission.
- 💡State explicitly that one photon interacts with one electron, then use E = hf to justify whether emission is possible.
- 💡When a question mentions intensity, translate it into photons per second and link that to emission rate rather than to electron energy.
- 💡Check the frequency against the threshold before discussing kinetic energy, so your reasoning follows the one-to-one condition.
- 💡Quote the equation in the form hf = φ + KEmax before substituting values, so the energy balance is clear.
- 💡Convert frequencies and work functions into consistent SI units, using joules or electronvolts throughout, before calculating KEmax.
- 💡When reading a KEmax against f graph, identify the gradient as h and the intercept on the KEmax axis as −φ.
- 💡Convert between joules and electronvolts carefully, using 1 eV = 1.60 × 10⁻¹⁹ J, and keep units consistent in calculations.
- 💡Use f₀ = φ/h or λ₀ = hc/φ to move between work function, threshold frequency and threshold wavelength.
- 💡State that below the threshold frequency no emission occurs at any intensity, and justify this using the one-photon-one-electron model.
- 💡Separate the two effects clearly: intensity controls the number of electrons per second, while frequency controls the energy of each electron.
- 💡Use KEmax = hf − φ to justify why KEmax is unchanged when only the intensity is altered.
- 💡When describing evidence for photons, contrast the constant KEmax with the wave prediction that brighter light should give more energetic electrons.
- 💡State clearly that the proportionality is between rate of emission and intensity, and that it applies only above the threshold frequency.
- 💡When explaining, link intensity to the number of photons per second and then to the number of photoelectrons emitted per second.
- 💡Use the term 'rate of emission' rather than vague phrases such as 'amount of electrons' to show precise understanding.
Common Mistakes
- Treating this row as a question to answer; it is a reading instruction, so the learner should use it to organise study of the following statements.
- Reading the photoelectric effect as a gradual heating effect; the emission depends on photon frequency, not on intensity alone.
- Confusing the work function with the photon energy; the work function is the minimum energy needed to release an electron from the metal surface.
- Assuming any colour of light will cause emission; radiation below the threshold frequency produces no emission however intense it is.
- Believing that brighter light of any colour will eventually cause emission; below the threshold frequency no emission occurs regardless of intensity.
- Thinking that increasing intensity increases the maximum kinetic energy of emitted electrons; it increases the number of electrons emitted per second instead.
- Confusing the work function with the threshold frequency; the work function is an energy and the threshold frequency is the minimum frequency.
- Describing the electroscope observation without linking it to electron emission and charge neutralisation.
- Forgetting to clean the zinc plate; an oxide layer can prevent emission and make the demonstration fail.
- Charging the electroscope positively and expecting the leaf to fall; a positive charge attracts emitted electrons back.
- Using a bright visible lamp and expecting discharge; intensity does not compensate for frequency below the threshold.
- Saying the leaf falls because the plate is heated; the effect is due to electron emission, not temperature.
- Thinking that many photons can combine their energies to release one electron. Correction: the interaction is one-to-one, so a single photon must supply at least the work function.
- Believing that a very intense low-frequency beam will eventually cause emission. Correction: if hf is less than φ, no single photon has enough energy, so no emission occurs at any intensity.
- Assuming the electron gradually absorbs energy from the wave until it escapes. Correction: absorption is a single instantaneous photon-electron event, not a gradual build-up.
- Writing the equation with the work function subtracted from kinetic energy, for example KEmax = φ − hf. Correction: the photon energy supplies the work function and the remainder is kinetic energy, so hf = φ + KEmax.
- Treating φ as the energy of the photon. Correction: φ is a property of the metal surface, while hf depends on the radiation used.
- Using the equation with intensity in place of frequency. Correction: intensity controls the number of photons per second, not the energy of each photon.
- Confusing the work function with the threshold frequency. Correction: φ is an energy in joules or electronvolts, while f₀ is a frequency in hertz, related by φ = hf₀.
- Thinking that a brighter beam below the threshold will cause emission. Correction: intensity adds more photons, but each photon still has energy below φ, so no electrons are released.
- Treating the work function as the same for all metals. Correction: φ depends on the metal, so different surfaces have different threshold frequencies.
- Believing that a more intense beam gives photoelectrons more kinetic energy. Correction: intensity increases the number of photons per second, so it increases the emission rate, while KEmax stays the same.
- Thinking that intensity and frequency both affect KEmax equally. Correction: KEmax depends on frequency and work function through KEmax = hf − φ, and is independent of intensity.
- Assuming a brighter beam can raise the maximum kinetic energy above hf − φ. Correction: no electron can gain more than one photon’s energy minus the work function in a one-to-one interaction.
- Thinking that increasing intensity increases the maximum kinetic energy of the photoelectrons. Correction: intensity changes the number of photoelectrons per second, while maximum kinetic energy depends on the frequency of the radiation.
- Believing that any intensity will cause emission if it is large enough. Correction: emission requires the frequency to be above the threshold frequency; below threshold, no emission occurs regardless of intensity.
- Confusing intensity with frequency. Correction: intensity relates to the number of photons per second per unit area, whereas frequency determines the energy of each photon.
- Assuming the proportionality holds below the threshold frequency. Correction: the direct proportionality applies only above the threshold frequency.