The shape of the lesson
By the end of the lesson, learners can
- Hunt for the threshold: Pick a metal, then lower the wavelength until electrons just begin to escape.
- Four ideas that carry the marks: The photoelectric effect is mostly about telling intensity and frequency apart.
- Photon: E = hf = hc/λ, momentum p = E/c = h/λ.
- Photoelectric: hf = Φ + KE_max; threshold f₀ = Φ/h; intensity sets rate, frequency sets KE_max.
- de Broglie: λ = h/p; for an electron through voltage V, p = √(2m₋eV).
- Energy levels: hf = E₁ − E₂ gives the wavelength of each spectral line.
One picture every idea returns to
One photon frees at most one electron, and only if the photon's energy exceeds the work function Φ. Below the threshold frequency f₀ = Φ/h there is no emission at any intensity. Above it, Einstein's equation gives the maximum kinetic energy: hf = Φ + KE_max. Intensity sets the rate of emission; frequency sets the maximum energy.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, The experiment that broke wave theory.. Shine light on a metal and electrons can fly off. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Hunt for the threshold | Pick a metal, then lower the wavelength until electrons just begin to escape. Project tab 2 of the lab (Photoelectric effect) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Four ideas that carry the marks | The photoelectric effect is mostly about telling intensity and frequency apart. | Cooperative Concept Mapping |
| 30 to 43 min | Check and plenary | The quick-check questions from the student lesson, answered on mini whiteboards one at a time; re-teach the one most get wrong, then a random call on the last. | Individual, then whole class |
| 43 to 45 min | Exit and homework | Set the worksheet below (5 questions, 46 marks) from the topic practice set. Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Four ideas that carry the marks, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Wave theory predicts that any frequency should eventually free electrons if you wait or brighten the light. | The experiment shows the opposite: there is a sharp threshold, emission is instantaneous, and KE_max rises with frequency, not intensity. These are the points that "explain why the photoelectric effect is evidence for the particle nature of light". |
The photoelectric effect: topic worksheet
5 questions, 46 marks. Written in the style of the papers; no past paper question is reproduced. These questions cover the whole topic, so sibling lessons share them.
The work function of sodium is 2.3 eV.
(a) State what is meant by the work function. [1]
(b) Express the work function in joules. [1]
(c) Calculate the threshold frequency for sodium. [2]
(d) Calculate the corresponding threshold wavelength. [2]
(e) State and explain what happens when light of wavelength 600 nm shines on sodium. [2]
Ultraviolet light of wavelength 400 nm is shone on a metal of work function 2.0 eV.
(a) Calculate the energy of each photon, in electronvolts. [2]
(b) Calculate the maximum kinetic energy of the photoelectrons, in electronvolts and in joules. [2]
(c) Calculate the maximum speed of the photoelectrons. [3]
(d) State the stopping voltage that would just prevent these electrons from reaching a collector. [1]
(e) State the effect on the maximum kinetic energy of doubling the intensity. [1]
Fig. 5.1 shows how the maximum kinetic energy of photoelectrons varies with the frequency of the incident light for a particular metal.
(a) State what the gradient of the line represents. [1]
(b) State what is found at the point marked f₀. [1]
(c) State what the intercept marked −Φ represents. [1]
(d) The line crosses the frequency axis at f₀ = 5.0 × 10¹⁴ Hz. Calculate the work function of the metal, in joules and in electronvolts. [3]
(e) A second metal has a larger work function. State how its line would differ from the one shown. [2]
A hydrogen atom emits a photon when its electron falls from the n = 4 level (−0.85 eV) to the n = 1 level (−13.6 eV). The photon then strikes a metal of work function 4.5 eV.
(a) Calculate the energy of the emitted photon, in eV. [1]
(b) Calculate the wavelength of the photon and state the region of the spectrum. [3]
(c) Calculate the maximum kinetic energy of the photoelectron released from the metal, in eV. [2]
(d) A photon from the n = 3 to n = 2 transition (1.89 eV) instead strikes the same metal. Explain why no photoelectron is emitted. [2]
(e) State what determines the number of photoelectrons emitted per second by the metal. [1]
Ultraviolet light of wavelength 254 nm falls on a zinc surface of work function 4.3 eV. This question follows the energy from photon to photoelectron.
(a) Calculate the energy of each photon, in eV. [2]
(b) Calculate the maximum kinetic energy of the photoelectrons, in eV and in joules. [2]
(c) Calculate the momentum of one of these photoelectrons. [2]
(d) Calculate the de Broglie wavelength of such a photoelectron. [2]
(e) Calculate the momentum of one incident photon, and comment on how it compares with the photoelectron's momentum. [3]
Answer key and marking notes
- (a)The minimum energy needed to remove an electron from the surface of the metal ✓
- (b)Φ = 2.3 × 1.60 × 10⁻¹⁹ = 3.7 × 10⁻¹⁹ J ✓
- (c)f₀ = Φ/h = 3.7 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ ✓ = 5.6 × 10¹⁴ Hz ✓
- (d)λ₀ = c/f₀ = 3.00 × 10⁸ / 5.6 × 10¹⁴ ✓ = 540 nm ✓
- (e)600 nm is longer than 540 nm, so f < f₀ ✓; the photon energy is below Φ, so no electrons are emitted ✓
- (a)E = 1240/400 ✓ = 3.10 eV ✓
- (b)KE_max = hf − Φ = 3.10 − 2.0 = 1.10 eV ✓ = 1.10 × 1.60 × 10⁻¹⁹ = 1.76 × 10⁻¹⁹ J ✓
- (c)½mv² = 1.76 × 10⁻¹⁹; v = √(2 × 1.76 × 10⁻¹⁹ / 9.11 × 10⁻³¹) ✓✓ = 6.2 × 10⁵ m s⁻¹ ✓
- (d)Vₛ = KE_max/e = 1.10 V ✓
- (e)No effect ✓ (intensity changes only the rate, not KE_max)
- (a)The Planck constant h ✓ (KE_max = hf − Φ, gradient = h)
- (b)The threshold frequency: below it, no photoelectrons are emitted ✓
- (c)The (negative of the) work function Φ of the metal ✓
- (d)Φ = hf₀ = 6.63 × 10⁻³⁴ × 5.0 × 10¹⁴ ✓ = 3.3 × 10⁻¹⁹ J ✓ = 2.1 eV ✓
- (e)Same gradient (parallel line) ✓, but a larger threshold frequency and a more negative intercept ✓
- (a)ΔE = −0.85 − (−13.6) = 12.75 eV ✓
- (b)λ = 1240/12.75 ✓ = 97 nm ✓; ultraviolet ✓
- (c)KE_max = 12.75 − 4.5 = 8.25 eV ✓✓
- (d)1.89 eV is less than the work function 4.5 eV ✓; a single photon cannot supply enough energy to free an electron, so none is emitted ✓
- (e)The intensity (number of photons arriving per second) ✓
- (a)E = 1240/254 ✓ = 4.88 eV ✓
- (b)KE_max = 4.88 − 4.3 = 0.58 eV ✓ = 0.58 × 1.60 × 10⁻¹⁹ = 9.3 × 10⁻²⁰ J ✓
- (c)p = √(2m₋KE) = √(2 × 9.11 × 10⁻³¹ × 9.3 × 10⁻²⁰) ✓ = 4.1 × 10⁻²⁵ N s ✓
- (d)λ = h/p = 6.63 × 10⁻³⁴ / 4.1 × 10⁻²⁵ ✓ = 1.6 × 10⁻⁹ m (1.6 nm) ✓
- (e)pₙₘₒₜₒₙ = h/λ = 6.63 × 10⁻³⁴ / 254 × 10⁻⁹ = 2.6 × 10⁻²⁷ N s ✓; this is far smaller than the photoelectron's momentum (4.1 × 10⁻²⁵ N s) ✓, about 160 times smaller ✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
When light below the threshold frequency shines on a metal, increasing the intensity:
- Astill produces no photoelectrons
- Bproduces faster photoelectrons
- Clowers the threshold frequency
- Dproduces more photoelectrons
The maximum kinetic energy of the photoelectrons depends on:
- Athe frequency of the light and the metal used
- Bthe intensity of the light only
- Cthe time the light is applied
- Dthe area of the metal
Einstein's photoelectric equation is:
- Ahf = Φ + KE_max
- Bhf = Φ − KE_max
- Chf = Φ × KE_max
- DKE_max = Φ + hf
For light above the threshold frequency, increasing the intensity increases:
- Athe number of photoelectrons emitted per second
- Bthe maximum kinetic energy of the electrons
- Cthe threshold frequency
- Dthe work function
Quiz key
Below the threshold, each photon carries less than the work function, so no single photon can free an electron. Adding more such photons changes nothing.
KE_max = hf − Φ. It rises with frequency and falls with the work function of the metal, but is independent of intensity.
The photon energy hf is shared: Φ is used to escape the surface, and the rest becomes the maximum kinetic energy. So hf = Φ + KE_max.
Intensity is the rate of arrival of photons, so it sets the rate of photoelectrons (the photocurrent). It does not change KE_max, which depends on frequency.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.