Photoelectric Effect A Level: Equation, Graph and Questions (OCR A)

What the photoelectric effect shows

The photoelectric effect is the release of electrons from a metal surface when electromagnetic radiation shines on it. It matters because it cannot be explained by treating light as a continuous wave, and it gave the first strong evidence that light comes in packets of energy called photons.

You need to know four things for the OCR A course: the experimental observations, why wave theory fails to explain them, how the photon model does explain them with Einstein's equation, and how to use the graph of maximum kinetic energy against frequency. You also need the link to wave-particle duality and the de Broglie wavelength.

This guide is part of the quantum physics topic in Module 4. It also connects to ideas about standing waves in stationary waves, and to the energy-level and nuclear ideas in binding energy and mass defect.

Photons, the electronvolt and the equations

In the photon model, electromagnetic radiation is emitted and absorbed in discrete packets. The energy of one photon depends only on the frequency of the radiation.

Photon energies are tiny in joules, so physicists use the electronvolt (eV). One electronvolt is the kinetic energy gained by an electron accelerated through a p.d. of 1 V, so 1 eV = 1.60 × 10⁻¹⁹ J. To convert from eV to J, multiply by 1.60 × 10⁻¹⁹. To convert from J to eV, divide.

It is helpful to know that hc = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ = 1.99 × 10⁻²⁵ J m. Then E = hc / λ is a single division. A photon of wavelength 450 nm has E = 1.99 × 10⁻²⁵ / 450 × 10⁻⁹ = 4.42 × 10⁻¹⁹ J, which is 2.76 eV.

The gold-leaf electroscope demonstration

1. Clean a zinc plate and attach it to the cap of a gold-leaf electroscope.
2. Charge the electroscope negatively, for example with a rubbed polythene rod. The leaf rises because the leaf and the rod repel each other.
3. Shine ultraviolet light on the zinc. The leaf slowly falls, because electrons are emitted from the zinc and the plate loses its negative charge.
4. Charge the electroscope positively instead. The leaf does not fall, because any emitted electrons are attracted back to the positively charged plate.
5. Place a sheet of glass between the UV lamp and the plate. The leaf stops falling, since glass absorbs the ultraviolet.
6. Shine a bright visible lamp on the plate. The leaf does not fall, even though the lamp delivers a lot of energy overall.

The last two steps are the important ones. Bright visible light has more total energy than a weak UV lamp, yet it does nothing, while the weak UV lamp works. That is the puzzle that classical wave theory cannot solve.

Observations and why wave theory fails

The photon model explains each observation. One photon interacts with one electron, transferring all its energy at once. If hf is less than the work function, no single photon can free an electron, however many photons arrive. If hf is large enough, emission is immediate. Doubling the intensity at a fixed frequency doubles the number of photons arriving per second, so twice as many electrons are emitted per second, each with the same maximum energy.

Einstein's photoelectric equation

Electrons in a metal are held in by attractive forces. The work function φ is the minimum energy needed to remove an electron from the surface of the metal. The threshold frequency f₀ is the lowest frequency of radiation that can release an electron, so φ = hf₀.

Einstein applied conservation of energy to one photon and one electron. The photon gives all its energy hf. Some of that energy, at least φ, is used to escape the metal. The rest becomes kinetic energy.

If hf < φ there are no photoelectrons. If hf = φ, the electron barely escapes, and KEmax = 0, so the frequency is the threshold frequency. For higher frequencies, KEmax = hf − φ = h(f − f₀).

Stopping potential. You can measure KEmax by applying a retarding p.d. to a photocell until the photocurrent just falls to zero. At that stopping potential Vs, the work done on the electron, eVs, equals KEmax. So eVs = KEmax, and a stopping potential of 0.67 V corresponds to KEmax = 0.67 eV.

The KEmax against frequency graph

Rearrange Einstein's equation into the form y = mx + c: KEmax = hf − φ. A graph of KEmax against f is therefore a straight line.

• Gradient = h. It is the same for every metal, so lines for different metals are parallel.
• x-intercept = f₀. This is the threshold frequency, where KEmax = 0.
• y-intercept = −φ. This is found by extending the line back to f = 0, although no emission is possible below f₀.

If a question gives a graph of stopping potential Vs against f, then eVs = hf − φ gives Vs = (h/e)f − φ/e. The gradient is h/e, so you multiply by e to find h. Read the axis labels carefully: if the vertical axis is in eV, the gradient has units of eV Hz⁻¹ and you must convert to J before quoting h.

Wave-particle duality and the de Broglie wavelength

The photoelectric effect shows light behaving as particles. Yet light also diffracts and interferes, which only waves do. So light has a dual nature, behaving as a wave in some experiments and as particles in others.

In 1924 Louis de Broglie suggested that particles such as electrons can show wave behaviour too. The wavelength depends on the particle's momentum p = mv.

Electron diffraction is the evidence. When electrons are accelerated through a p.d. and directed at a thin sheet of graphite, the atoms in the graphite act like a diffraction grating. The screen shows a pattern of concentric rings. Diffraction is a wave property, so the electrons are behaving as waves. The spacing between carbon atoms is about 10⁻¹⁰ m, which is why the electron wavelength needs to be of a similar size for the effect to show.

If the accelerating p.d. is increased, the electrons gain momentum, the wavelength becomes shorter, and the diffraction rings move closer to the centre. This agrees with the de Broglie equation, and it is a common exam question.

Worked examples

Common mistakes

• Not converting eV to J before using h and f in the same equation, or forgetting that 1 eV = 1.60 × 10⁻¹⁹ J.
• Mixing up wavelength and frequency, or forgetting to convert nm to m (1 nm = 10⁻⁹ m).
• Reading the work function from the wrong intercept. It is the negative of the y-intercept, and the x-intercept is f₀.
• Saying that the gradient of the graph is the work function. The gradient is h.
• Describing the threshold as a threshold intensity. It is a threshold frequency.
• Saying that electron diffraction shows electrons are "only" waves. It shows wave behaviour, and other experiments show particle behaviour.

How OCR A examines this topic

Photoelectric effect questions mix short calculations with explanation. A typical question gives a wavelength or frequency and a work function, then asks for KEmax or the stopping potential, and follows with an explanation of what happens when the intensity or frequency changes.

• Calculations: photon energy in J and eV, KEmax, threshold frequency, work function, number of photons per second.
• Graph questions: read h, f₀ or φ from a KEmax-f or stopping potential graph, or sketch how the line changes for a different metal.
• Explanations: use the photon model to account for the threshold frequency, instant emission and the independence of KEmax from intensity.
• Duality: the de Broglie equation, electron diffraction and how the pattern changes when the electron speed changes.
• Synoptic links: energy conservation, stationary waves and standing electron waves in atoms, and nuclear energy in binding energy and mass defect.

Practice questions

1. Calculate the energy, in J and in eV, of a photon of frequency 7.5 × 10¹⁴ Hz.2 marks

   Show the answer

   Answer: 4.97 × 10⁻¹⁹ J and 3.1 eV

   E = hf = 6.63 × 10⁻³⁴ × 7.5 × 10¹⁴ = 4.97 × 10⁻¹⁹ J. In eV: 4.97 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 3.1 eV.

2. A red laser emits light of wavelength 633 nm with a power of 1.0 mW. Calculate the number of photons emitted each second.3 marks

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   Answer: 3.2 × 10¹⁵ s⁻¹

   E = hc / λ = 1.989 × 10⁻²⁵ / 633 × 10⁻⁹ = 3.14 × 10⁻¹⁹ J. N = P / E = 1.0 × 10⁻³ / 3.14 × 10⁻¹⁹ = 3.2 × 10¹⁵ s⁻¹.

3. A metal has a work function of 4.3 eV. (a) Calculate KEmax and the stopping potential when light of wavelength 250 nm is used. (b) Show that light of wavelength 500 nm cannot release electrons.5 marks

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   Answer: (a) KEmax = 0.67 eV (1.1 × 10⁻¹⁹ J) and Vs = 0.67 V; (b) the 500 nm photon energy is 2.5 eV, below 4.3 eV

   (a) E = 1.989 × 10⁻²⁵ / 250 × 10⁻⁹ = 7.96 × 10⁻¹⁹ J = 4.97 eV. KEmax = 4.97 − 4.3 = 0.67 eV = 1.08 × 10⁻¹⁹ J. eVs = KEmax, so Vs = 0.67 V. (b) E = 1.989 × 10⁻²⁵ / 500 × 10⁻⁹ = 3.98 × 10⁻¹⁹ J = 2.5 eV, which is less than 4.3 eV.

4. A metal has a work function of 2.3 eV. Calculate the longest wavelength of light that can cause photoemission.3 marks

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   Answer: 5.4 × 10⁻⁷ m, or 540 nm

   φ = 2.3 × 1.60 × 10⁻¹⁹ = 3.68 × 10⁻¹⁹ J. λ = hc / φ = 1.989 × 10⁻²⁵ / 3.68 × 10⁻¹⁹ = 5.40 × 10⁻⁷ m.

5. Light above the threshold frequency shines on a metal. Describe and explain what happens to (a) the number of electrons emitted per second, and (b) their maximum kinetic energy, when the intensity of the light is doubled with the frequency unchanged.4 marks

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   Answer: (a) The number per second doubles. (b) KEmax is unchanged.

   Doubling the intensity doubles the number of photons arriving per second, so twice as many electrons are released per second. Each photon still has the same energy hf, and KEmax = hf − φ is unchanged.

6. An electron travels at 2.0 × 10⁶ m s⁻¹. Calculate its de Broglie wavelength.2 marks

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   Answer: 3.6 × 10⁻¹⁰ m

   p = mv = 9.11 × 10⁻³¹ × 2.0 × 10⁶ = 1.82 × 10⁻²⁴ kg m s⁻¹. λ = h / p = 6.63 × 10⁻³⁴ / 1.82 × 10⁻²⁴ = 3.6 × 10⁻¹⁰ m.

Next steps

For another wave idea that shows quantisation, read stationary waves. For more on energy at the nuclear scale, see binding energy and mass defect. To test yourself on photon calculations and graphs, practise this topic with step-by-step marking, free to start. The rest of the course is on the Module 4 hub.

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