AQA AS quantum physics worked solutions: photons, photoelectric effect, energy levels, fluorescent tubes and de Broglie wavelength, with exam-report guidance.
Constants and assumptions: Use h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C and electron mass mₑ = 9.11 × 10⁻³¹ kg throughout. Thus 1 eV = 1.60 × 10⁻¹⁹ J. Assume non-relativistic electron motion.
Attempt each question before opening the hints. Show equations, substitutions, units and the reason for your method. The mark checklists below are PhysicsUK's own schemes.
Lessons from the examiner reports
These reports identify skills worth practising. The questions below are original PhysicsUK examples, with our own mark schemes.
Define the threshold and distinguish it from the work function: Strong responses addressed the question’s frequency and intensity points coherently. The report identifies confusion between threshold frequency and work function and imprecise descriptions of emission from a metal surface. Frequency is measured in hertz; the work function is an energy. AQA 7407/1, June 2022 — Q04.1; report p. 4
Keep the photoelectric equation in one energy unit: A common error was combining photon energy in joules with electron kinetic energy in electron volts. Convert before subtracting, then give the answer in the requested unit. AQA 7407/1, June 2022 — Q04.2; report p. 5
Follow excitation through to the correct emission transition: The report highlights eV-to-joule errors and answers that stopped after identifying excitation. Stronger answers also chose the transition that produced the requested light and supported it with an energy or wavelength calculation. AQA 7407/1, June 2024 — Q02.1–02.3; report pp. 4–5
Explain what happens in the coating, not just the gas: The report identifies reversed photon-energy/power ratios, wavelength conversion errors, and fluorescent-tube answers that explained the gas but missed the coating’s role in white light. Link photon absorption, excitation, de-excitation and the range of visible photons. AQA 7407/1, June 2025 — Q03.2–03.4; report p. 5
Interpret the threshold and negative graph intercept: The report flags threshold definitions that omitted minimum frequency or emission, prefix-conversion errors, and confusion over the negative energy-axis intercept for different work functions. Use Ek,max = hf − φ: the gradient is h, and a larger φ moves the extrapolated intercept further below zero. AQA 7407/1, June 2025 — Q07.1–07.4; report p. 8
Use the electron energy after the collision: Some candidates used the incoming electron energy when asked for the scattered electron’s wavelength, or applied the photon relation E = hc/λ to that electron. First subtract the excitation energy, then find momentum and use λ = h/p. AQA 7407/1, June 2025 — Q03.1; report p. 5
Connect wavelength to the fixed crystal spacing: The report highlights answers that stated a smaller electron wavelength without explaining the reduced diffraction relative to the graphite spacing. Use increased momentum, shorter de Broglie wavelength and smaller diffraction angles; “less time to spread out” is not the explanation. AQA 7407/1, June 2022 — Q05.1–05.2; report p. 5
Question 1 · Exam-style units and photon counting · 5 marks
Photon energy and photons in a pulse
Specification: 3.2.1.3, 3.2.2.2
A blue laser produces monochromatic light of wavelength 450 nm. Its optical power during a pulse is 2.40 mW and the pulse lasts 5.00 ms. Use the constants at the top of the page.
Calculate the energy of one photon in joules and in electron volts. [3]
Calculate the number of photons in the pulse. [2]
Small hint
The laser power is energy per second for the whole beam; it is not the energy of one photon.
Method hint
Use E = hc/λ, with λ in metres. Divide energy in joules by e to convert to eV. Find pulse energy Pt, then N = Pt/Ephoton.
Complete worked solution and marks
Answer: Photon energy = 4.42 × 10⁻¹⁹ J = 2.76 eV. Number of photons in the pulse = 2.71 × 10¹³.
λ = 450 × 10−9 = 4.50 × 10−7 m; E = hc/λ = (6.63 × 10−34)(3.00 × 108)/(4.50 × 10−7) = 4.42 × 10−19 J.
E in eV = (4.42 × 10−19)/(1.60 × 10−19) = 2.7625 ≈ 2.76 eV.An electron volt is an energy unit, not an electric potential.
P = 2.40 × 10−3 W; t = 5.00 × 10−3 s; Epulse = Pt = 1.20 × 10−5 J.
N = Epulse/Ephoton = (1.20 × 10−5)/(4.42 × 10−19) ≈ 2.71 × 1013 photons.Retain the unrounded single-photon energy in this division.
Check: Multiplying 2.71 × 10¹³ by 4.42 × 10⁻¹⁹ J returns about 1.20 × 10⁻⁵ J. A milliwatt pulse can contain many photons because each photon carries a tiny energy.
Mark checklist · 5 marks
1 mark — Converts wavelength and uses E = hc/λ.
1 mark — Photon energy 4.42 × 10⁻¹⁹ J.
1 mark — Converts to 2.76 eV; accept 2.8 eV with consistent working.
1 mark — Uses the optical power and pulse duration to obtain total pulse energy 1.20 × 10⁻⁵ J.
1 mark — Divides pulse energy by photon energy to obtain 2.71 × 10¹³ photons; carry an earlier numerical photon-energy error forward.
Common mistake: P/E gives photons per second. It does not give the number in a 5.00 ms pulse until you multiply by the duration. For eV conversion, divide joules by 1.60 × 10⁻¹⁹.
Exam technique: State whether you are finding an energy, a count or a count per second. The 2025 report’s photon-power ratio lesson transfers from its ultraviolet lamp to this laser pulse.
A second laser emits at 900 nm with the same optical power and pulse duration. How do the photon energy and number per pulse compare?
Check the transfer answer
The wavelength doubles, so E = hc/λ halves to 2.21 × 10−19 J. Pulse energy is unchanged, so the photon number doubles to 5.43 × 1013. Keeping power fixed does not keep photon arrival rate fixed when wavelength changes.
Question 2 · Exam-style photon explanation · 6 marks
Threshold frequency and changing intensity
Specification: 3.2.2.1
A metal has work function 2.50 eV. Red light supplies photons of energy 1.91 eV; violet light supplies photons of energy 3.55 eV. Use the usual single-photon photoelectric model.
Define threshold frequency for this metal. [2]
Explain why increasing the intensity of the red light cannot produce photoelectric emission in this model. [2]
The violet-light intensity is tripled at the same frequency. State and explain the changes in the number of electrons emitted per second and their maximum kinetic energy. Assume the illuminated area and emission probability per photon stay the same. [2]
Compare the energy of one photon with the work function. More photons at the same frequency do not raise their individual energy. Open the larger diagramSmall hint
Increasing intensity at fixed frequency changes how many photons arrive; it does not change each photon’s energy.
Method hint
Threshold frequency is the minimum frequency needed to emit electrons from the surface. A single photon transfers energy hf to an electron. Compare photon energy with φ, then use Ek,max = hf − φ.
Complete worked solution and marks
Answer: Threshold frequency is the minimum frequency for electron emission from the metal surface. Red light cannot emit electrons in this model. With violet light, tripling intensity triples the emission rate; maximum kinetic energy stays at 1.05 eV.
Threshold frequency is the minimum frequency of incident radiation required for electrons to be emitted from this metal’s surface. At the ideal threshold hf0 = φ; the fastest emitted electrons have zero kinetic energy at that limiting value.
Each red photon has energy 1.91 eV, less than the 2.50 eV needed to escape. In the specified single-photon model an electron receives energy from one photon in an interaction. More red photons still have insufficient individual energy, so there is no photoelectric emission.
Tripling violet-light intensity at fixed frequency triples the photon arrival rate on the same area. With the same emission probability, the number of photoelectrons emitted per second triples. A measured saturated photocurrent would likewise triple if all emitted electrons were collected.
Ek,max = Ephoton − φ = 3.55 − 2.50 = 1.05 eV.Frequency and work function are unchanged, so maximum kinetic energy and stopping-potential magnitude are unchanged.
Check: Threshold frequency and work function are related by hf0 = φ but have different units. Light below threshold cannot be made effective simply by waiting longer or increasing ordinary intensity in this model.
Mark checklist · 6 marks
1 mark — Defines threshold as the minimum frequency, rather than merely a frequency or an energy.
1 mark — Links it to electron emission from the metal surface.
1 mark — Red photon energy is below the work function.
1 mark — Explains that raising intensity changes photon number, not energy per photon, in the single-photon model.
1 mark — Violet emission rate triples because photon arrival rate triples under the stated conditions.
1 mark — Maximum electron kinetic energy remains unchanged because photon frequency/energy and work function remain unchanged; numerical 1.05 eV is a useful check but not required.
Common mistake: Calling threshold frequency an energy confuses f0 with φ. Claiming that three photons automatically share their energies with one electron also contradicts the single-photon model used here.
Exam technique: Include “minimum”, “frequency” and “emitted from the surface” in the threshold definition. Keep frequency and intensity arguments separate. The source was an extended plate-discharge response; this question has a new context and our own point checklist.
A student replaces violet light with higher-frequency light while keeping the photon arrival rate fixed. Above threshold, what happens to emission rate and maximum kinetic energy if the emission probability is unchanged?
Check the transfer answer
The emission rate stays the same under the stated assumption. Maximum kinetic energy increases because hf increases while φ is unchanged. Optical power increases too, since the same number of photons per second now carries more energy each.
Question 3 · Exam-style multi-step calculation · 6 marks
Photoelectron energy, stopping potential and speed
Specification: 3.2.2.1, 3.2.2.2
Light of wavelength 300 nm is incident on a metal of work function 2.50 eV. Electrons are emitted. Use the constants at the top of the page.
Calculate the maximum electron kinetic energy in joules and in electron volts. [3]
Calculate the magnitude of the stopping potential. [1]
Calculate the maximum speed of the emitted electrons. [2]
Small hint
The photon must first supply the work function. Only the remaining energy can become electron kinetic energy.
Method hint
Use Ephoton = hc/λ and Ek,max = Ephoton − φ with both energies in joules. Then eVs = Ek,max and v = √(2Ek,max/me).
φ = 2.50(1.60 × 10−19) = 4.00 × 10−19 J; Ek,max = 6.63 × 10−19 − 4.00 × 10−19 = 2.63 × 10−19 J.Alternatively, photon energy is 4.14375 eV, so subtracting 2.50 eV gives 1.64375 eV.
Ek,max in eV = (2.63 × 10−19)/(1.60 × 10−19) = 1.64375 ≈ 1.64 eV.
At stopping, even the fastest electrons lose all their kinetic energy against the retarding potential: Vs = Ek,max/e = 1.64 V.This is a positive magnitude. The collector must be negative relative to the emitting surface to repel the electrons.
vmax = √(2Ek,max/me) = √[2(2.63 × 10−19)/(9.11 × 10−31)] ≈ 7.60 × 105 m s−1.Other emitted electrons can have smaller kinetic energies because they lose energy before escaping.
The collector is negative relative to the emitter. Vs is the positive stopping-potential magnitude, so eVs = Kmax. This energy sketch illustrates the supplied calculation. Open the larger diagram
Check: The maximum kinetic energy is less than the photon energy. The speed is only about 0.25% of c, so the non-relativistic expression is consistent with the stated model.
Mark checklist · 6 marks
1 mark — Uses wavelength in metres and obtains photon energy 6.63 × 10⁻¹⁹ J or equivalent 4.14 eV.
1 mark — Subtracts the work function in the same units to obtain maximum kinetic energy 2.63 × 10⁻¹⁹ J.
1 mark — Maximum kinetic energy approximately 1.64 eV; carry an earlier numerical energy error forward.
1 mark — Uses eVs = Ek,max to obtain stopping-potential magnitude approximately 1.64 V.
1 mark — Uses Ek,max = ½mev², rearranged for speed.
1 mark — Maximum speed approximately 7.60 × 10⁵ m/s; carry an earlier numerical positive kinetic-energy error forward.
Common mistake: Subtracting 2.50 directly from 6.63 × 10⁻¹⁹ mixes eV and J. Using the whole photon energy for ½mv² omits the work function and overestimates electron speed.
Exam technique: Write the photoelectric energy balance before substituting. An energy of 1.64 eV corresponds to a stopping-potential magnitude of 1.64 V for a single electron, but the energy and voltage are different quantities.
The wavelength is changed to 600 nm. Can photoelectrons be emitted from the same metal? Explain without reporting a negative kinetic energy.
Check the transfer answer
The photon energy is hc/λ = 3.315 × 10−19 J = 2.071875 eV, less than φ = 2.50 eV. No photoelectrons are emitted in the stated model. A negative value from hf − φ signals that the emission condition is not met; it is not an emitted electron’s kinetic energy.
Question 4 · Exam-style gradients and extrapolation · 7 marks
What the photoelectric graph tells you
Specification: 3.2.2.1
A supplied best-fit graph plots maximum photoelectron kinetic energy Ek,max against light frequency f. Its axes use 10−19 J and 1014 Hz. Two points on its solid line are (8.00 × 1014 Hz, 1.28 × 10−19 J) and (11.0 × 1014 Hz, 3.26 × 10−19 J). The dashed part is a mathematical extrapolation below the emission threshold, not negative electron kinetic energy.
Use the gradient to estimate Planck’s constant. [2]
Use the extrapolated energy-axis intercept to find the work function in joules and electron volts. [2]
Determine the threshold frequency from this fitted graph. [1]
A second metal has a greater work function. Describe how its line compares with the first: gradient, energy-axis intercept and threshold frequency. [2]
The negative extension helps determine the work function. No photoelectrons are emitted below the frequency-axis crossing. Open the larger diagramSmall hint
Compare Ek,max = hf − φ with y = mx + b. Include the scale factors on both graph axes.
Method hint
The gradient of this energy-in-joules graph is h. Its extrapolated y-intercept is −φ and its frequency-axis intercept is f0 = φ/h. A new metal changes φ, not h.
Complete worked solution and marks
Answer: Estimated h = 6.60 × 10⁻³⁴ J s; φ = 4.00 × 10⁻¹⁹ J = 2.50 eV; f0 = 6.06 × 10¹⁴ Hz. A larger work function gives a parallel line, a more negative extrapolated energy intercept and a higher threshold frequency.
gradient = (3.26 − 1.28) × 10−19/[(11.0 − 8.00) × 1014] = 6.60 × 10−34 J s.Since Ek,max = hf − φ, this estimates h. Dividing the bare axis numbers gives 0.660; the scale factor is 10−33 J/Hz.
Using a point on the fitted line, b = 1.28 × 10−19 − (6.60 × 10−34)(8.00 × 1014) = −4.00 × 10−19 J.Therefore φ = −b = 4.00 × 10−19 J = 2.50 eV. The work function is positive even though the extrapolated intercept is negative.
f0 = φ/hfit = (4.00 × 10−19)/(6.60 × 10−34) ≈ 6.06 × 1014 Hz.Use the fitted gradient for the graph’s threshold, rather than replacing it with a rounded tabulated constant mid-calculation.
For greater φ, the gradient remains h, so the second line is parallel. The extrapolated energy intercept −φ is more negative, and φ/h is larger, so the frequency-axis crossing shifts to the right. Below each metal’s threshold there are no emitted photoelectrons; the negative extension is an analysis tool.
Check: The fitted h differs from the supplied 6.63 × 10⁻³⁴ J s by about 0.45%, a plausible difference for graph data. J/Hz = J s. A stopping-potential-versus-frequency graph would instead have gradient h/e.
Mark checklist · 7 marks
1 mark — Uses ΔEk/Δf with the scale factors on both axes.
1 mark — Gradient/estimated Planck constant 6.60 × 10⁻³⁴ J s.
1 mark — Identifies intercept = −φ and obtains positive work function 4.00 × 10⁻¹⁹ J.
1 mark — Converts work function to 2.50 eV.
1 mark — Threshold frequency approximately 6.06 × 10¹⁴ Hz using the fitted line.
1 mark — Second-metal line is parallel because h is unchanged.
1 mark — Explains that greater φ makes the extrapolated energy intercept more negative and threshold frequency larger.
Common mistake: Reading a more negative energy intercept as a smaller work function reverses φ = −b. Giving negative kinetic energies for frequencies below threshold also extends the equation beyond the emission condition.
Exam technique: Interpret the plotted quantities, not just the line’s shape. On an energy-in-joules graph the gradient is h; on a stopping-potential graph it is h/e. Label the axes and carry the powers of ten.
If the fitted work function increased to 3.00 eV while hfit stayed the same, find the new threshold frequency and the maximum energy at f = 11.0 × 1014 Hz.
Check the transfer answer
φnew = 4.80 × 10−19 J. f0 = φnew/hfit = 7.27 × 1014 Hz. At the stated frequency, Ek,max = (6.60 × 10−34)(11.0 × 1014) − 4.80 × 10−19 = 2.46 × 10−19 J. Emission still occurs, but the maximum energy is lower.
Question 5 · Exam-style atomic collisions · 6 marks
Excitation, ionisation and energy transfer
Specification: 3.2.2.2, 3.2.2.3
An idealised atom has ground-state energy A = −8.00 eV, excited levels B = −4.80 eV and C = −2.20 eV, and an ionisation level at 0 eV. Only these bound levels are available. Ignore recoil and line broadening.
State the difference between excitation and ionisation. [2]
An incoming electron has kinetic energy 4.50 eV and the atom is initially in A. Identify the only bound excited level it can reach in a collision. If excitation occurs, find the scattered electron’s kinetic energy. [2]
A photon of energy 4.50 eV is incident on an atom in A. Explain why it cannot be absorbed to produce one of the listed bound excited states. [1]
State the minimum energy needed to ionise an atom initially in A. [1]
Energies are relative to the free-electron level at zero. Compare differences between levels with the incoming energy. Open the larger diagramSmall hint
A colliding electron can keep some of its kinetic energy. A photon absorbed in a bound-state transition transfers its whole energy.
Method hint
Calculate each energy gap from A. Excitation leaves the atomic electron bound; ionisation removes it. For an electron collision subtract the excitation gap from incoming kinetic energy. A bound-state photon absorption needs a photon energy equal to an allowed gap.
Complete worked solution and marks
Answer: Excitation moves an atomic electron to a higher bound level; ionisation removes an electron from the atom. The collision can excite A → B, leaving 1.30 eV. A 4.50 eV photon matches neither allowed gap from A. Ionisation from A needs 8.00 eV.
Excitation gives an atomic electron energy to move to a higher permitted bound level. It remains in the atom. Ionisation supplies enough energy to remove an electron, leaving an ion.
ΔEA→B = −4.80 − (−8.00) = 3.20 eV; ΔEA→C = −2.20 − (−8.00) = 5.80 eV.The incoming electron has 4.50 eV, enough for B but not C. If it excites B, Ek,after = 4.50 − 3.20 = 1.30 eV.It transfers exactly the excitation energy and retains the remainder.
A 4.50 eV photon has neither 3.20 eV nor 5.80 eV, so it cannot be absorbed for either listed bound transition from A in this model. It cannot transfer 3.20 eV and keep 1.30 eV while being absorbed. This statement concerns bound-state absorption; it does not rule out all possible scattering processes.
Eionisation,min = 0 − (−8.00) = 8.00 eV.A larger photon energy can ionise the atom, with excess energy appearing as electron kinetic energy when recoil is neglected.
Track two different electrons: the free electron loses kinetic energy, while a bound electron in the atom gains excitation energy. Open the larger diagram
Check: After the excitation collision, 3.20 eV + 1.30 eV = 4.50 eV. A negative bound-level energy is measured relative to the free-electron level; it is not negative kinetic energy.
Mark checklist · 6 marks
1 mark — Excitation moves an atomic electron to a higher allowed bound level without removing it.
1 mark — Ionisation removes an electron from the atom.
1 mark — Identifies B using its 3.20 eV gap; C requires 5.80 eV, more than the incoming energy.
1 mark — Scattered electron retains 4.50 − 3.20 = 1.30 eV.
1 mark — Explains that a 4.50 eV photon matches neither allowed bound-state energy gap from A.
1 mark — Minimum ionisation energy from A is 8.00 eV.
Common mistake: Saying every photon above the first excitation energy is absorbed confuses discrete bound transitions with ionisation. A colliding electron can retain energy, whereas an absorbed photon is removed and transfers all its energy.
Exam technique: Identify the atom’s starting state and calculate an energy difference. Distinguish the incoming free electron from the electron bound in the atom.
A photon of energy 9.20 eV is absorbed by an atom initially in A and ionises it. Neglect recoil. Find the kinetic energy of the removed electron and explain why this does not require a bound-level gap of 9.20 eV.
Check the transfer answer
The photon supplies 8.00 eV to remove the electron and the remainder becomes kinetic energy: Ek = 9.20 − 8.00 = 1.20 eV. The final electron is free, so its kinetic energy is not restricted to the listed bound-state gaps. Above-threshold ionisation differs from a transition between two bound levels.
Question 6 · Exam-style transitions and wavelength · 7 marks
Energy levels, cascades and line spectra
Specification: 3.2.2.3
A different idealised atom has bound levels A = −6.00 eV, B = −3.70 eV and C = −1.60 eV. A sample contains many atoms initially in C. Assume all three downward transitions C → A, C → B and B → A are possible and can occur in the sample. Use 400–700 nm as the visible range.
Find the photon energy for each of these three transitions, in eV. [3]
Calculate the wavelength of the photon from C → B and identify its region of the spectrum. [2]
State the number of distinct photon frequencies that can be emitted and link this to discrete atomic energy levels. [1]
An atom returns to A via B instead of directly from C. Explain why the total emitted photon energy is the same. [1]
This is a different atom from Question 5. Calculate each photon’s energy from a difference, not from a level’s value alone. Open the larger diagramSmall hint
An emitted photon carries the positive difference between the initial higher energy and final lower energy.
Method hint
Use Ephoton = Eupper − Elower. Convert the chosen gap from eV to joules before λ = hc/E. For a cascade, add the two photon energies; for the whole sample count distinct allowed energy differences.
Complete worked solution and marks
Answer: C → A: 4.40 eV; C → B: 2.10 eV; B → A: 2.30 eV. The C → B photon has wavelength 592 nm, in the visible range. Three distinct frequencies are possible; the cascade emits 2.10 + 2.30 = 4.40 eV in total.
For C → B, E = 2.10(1.60 × 10−19) = 3.36 × 10−19 J; λ = (6.63 × 10−34)(3.00 × 108)/(3.36 × 10−19) = 5.91964… × 10−7 m = 592 nm.This is within the stated visible range.
There are three distinct photon frequencies because the three allowed downward energy differences are different and f = ΔE/h. A line spectrum gives evidence for discrete bound energy levels; it is not a continuous range of electron energies. One atom taking a particular path emits either one direct photon or two cascade photons, not all three at once.
The cascade loses 2.10 + 2.30 = 4.40 eV, equal to the direct C → A gap. Energy conservation depends on the same initial and final states, even though the photon number and wavelengths differ. For reference, C → A gives about 283 nm (ultraviolet), while B → A gives about 540 nm (visible).
The labelled arrows show alternative paths. One atom emits one direct photon or two cascade photons; the sample can produce all three distinct spectral lines. Open the larger diagram
Check: The direct 4.40 eV photon has the shortest wavelength because it has the largest energy. Both cascade photons have longer wavelengths than the direct photon.
Mark checklist · 7 marks
1 mark — C → A photon energy 4.40 eV.
1 mark — C → B photon energy 2.10 eV.
1 mark — B → A photon energy 2.30 eV.
1 mark — Uses converted photon energy and hc/E to obtain C → B wavelength approximately 592 nm; carry an earlier numerical positive gap error forward.
1 mark — Identifies this wavelength as visible using the stated range.
1 mark — Three distinct frequencies linked to the three different allowed energy gaps/discrete levels.
1 mark — Cascade energy sum 2.10 + 2.30 = 4.40 eV equals the direct initial-to-final gap.
Common mistake: Using the energy of a level rather than the difference between levels gives the wrong photon energy. Also, the number of levels is not a general formula for the number of spectral lines: count the allowed transitions and any equal gaps.
Exam technique: Follow both the absorption/excitation step and the subsequent emission step. Show a gap and wavelength calculation when asked to identify the emitted light.
A 2.10 eV photon meets an atom in A, or an atom in B. Which of these atoms can be excited by absorbing it in the listed-level model?
Check the transfer answer
The atom in B can absorb it and move to C, because C − B = 2.10 eV. The atom in A cannot use it for either listed upward bound transition: A → B needs 2.30 eV and A → C needs 4.40 eV. The starting state matters.
Question 7 · Exam-style developed explanation · 5 marks
How a fluorescent tube produces white light
Specification: 3.2.2.2, 3.2.2.3
A fluorescent tube contains a low-pressure gas and has a fluorescent coating on its inside surface. Electrons are accelerated through the gas, which produces ultraviolet radiation. The coating produces visible white light.
Explain the energy-transfer sequence from the free electrons to the visible photons. Include what happens in both the gas and the coating, and why the final light contains a range of visible wavelengths. [5]
The cutaway locates the gas and coating. Explain what happens to electrons and photons in each material; the ultraviolet and visible photons are different photons. Open the larger diagramSmall hint
There are two separate excitation processes: collisions in the gas, followed by photon absorption in the coating.
Method hint
Describe free-electron collisions exciting gas atoms, gas de-excitation producing ultraviolet photons, absorption by the coating, and coating de-excitation through smaller energy gaps. A mixture of visible photon energies produces white light.
Complete worked solution and marks
Answer: Electron collisions excite gas atoms; de-excitation produces ultraviolet photons. The coating absorbs ultraviolet photons, becomes excited and emits lower-energy visible photons as it de-excites. Many visible transitions provide the range of colours needed for white light.
Free electrons gain kinetic energy from the electric field. In collisions they transfer energy to bound electrons in gas atoms, moving them to higher allowed energy levels: the gas atoms are excited. Ionisation can also help sustain the discharge, but excitation followed by de-excitation explains the emitted ultraviolet light.
When excited gas atoms return to lower levels, they emit photons with energies equal to the relevant level differences. The transitions in this lamp produce ultraviolet photons.
The coating absorbs ultraviolet photons. Energy from those photons excites electrons in the fluorescent material. The gas and coating are different materials with different energy structures.
The coating then de-excites and emits lower-energy, lower-frequency, longer-wavelength visible photons. The emitting energy differences in the coating are smaller than the gas’s ultraviolet-emitting difference. Some energy can also be transferred to thermal stores.
A mixture of fluorescent materials and many emitting transitions produces a range of visible wavelengths. Their combined emission is perceived as white. Photon number need not be conserved: absorbed ultraviolet energy can be shared among several lower-energy photons, while total energy is conserved.
Check: A single monochromatic visible line would have one colour, rather than white. White light needs a range of visible wavelengths; the coating supplies that range.
Mark checklist · 5 marks
1 mark — Free-electron collisions transfer energy to bound electrons and excite gas atoms.
1 mark — Gas-atom de-excitation emits ultraviolet photons with energy set by level differences.
1 mark — Coating absorbs ultraviolet photons and is excited.
1 mark — Coating de-excitation through smaller emitting energy differences produces lower-energy/longer-wavelength visible photons.
1 mark — Many transitions/materials produce a range of visible wavelengths whose combination gives white light.
Common mistake: Stopping after “electrons collide with gas atoms” leaves the coating unexplained. The ultraviolet light is not simply reflected as white light: it is absorbed and energy is re-emitted in different photons.
Exam technique: Address the coating explicitly and link its smaller emitting energy gaps to visible wavelengths. The 2025 report found that explanations of the gas alone did not answer its coating question.
The coating absorbs 1.00 × 1015 ultraviolet photons per second, each of energy 5.00 eV. It converts 80.0% of that energy into visible photons with mean energy 2.00 eV. Find the visible photon emission rate.
Check the transfer answer
The visible energy per second is 0.800 × (1.00 × 1015) × 5.00 eV. Dividing by 2.00 eV per visible photon gives 2.00 × 1015 visible photons per second. Energy conversion, rather than photon-number conservation, determines this average rate. The same calculation in joules gives the same ratio.
Question 8 · Exam-style de Broglie and diffraction · 9 marks
Electron wavelength before and after a collision
Specification: 3.2.2.2, 3.2.2.4
An electron is accelerated from rest through a potential difference of 6.00 V. It then collides with a stationary atom and excites it through an energy gap of 3.20 eV. Neglect atomic recoil and other energy losses.
Calculate the electron’s initial kinetic energy in joules, its momentum and its de Broglie wavelength. [3]
Calculate its de Broglie wavelength immediately after the excitation collision. [3]
In a separate electron-diffraction experiment with the same graphite target, the accelerating voltage is increased by a factor of four. Explain the changes in wavelength and ring diameter. Use the non-relativistic model and keep the screen geometry fixed. [2]
State which behaviour is supported by electron diffraction, and which behaviour of light is supported by the photoelectric effect. [1]
Small hint
An electron has mass and momentum. Use λ = h/p for it, rather than the photon relation E = hc/λ.
Method hint
Initially Ek = eV = 6.00 eV. For a non-relativistic electron p = √(2meEk). After the collision subtract 3.20 eV before recalculating p and λ. Four times the accelerating voltage gives four times Ek, twice p and half λ.
Complete worked solution and marks
Answer: Initially Ek = 9.60 × 10⁻¹⁹ J, p = 1.32 × 10⁻²⁴ kg m/s and λ = 5.01 × 10⁻¹⁰ m. After excitation, Ek = 2.80 eV and λ = 7.34 × 10⁻¹⁰ m. Four times the accelerating voltage halves wavelength and reduces ring diameter. Electron diffraction supports wave behaviour; the photoelectric effect supports the photon/particle nature of light.
The electric field does work on the electron: Ek,before = eV = (1.60 × 10−19)(6.00) = 9.60 × 10−19 J = 6.00 eV.
Using Ek = p²/(2me), pbefore = √[2(9.11 × 10−31)(9.60 × 10−19)] = 1.32254… × 10−24 ≈ 1.32 × 10−24 kg m s−1.
λbefore = h/p = (6.63 × 10−34)/(1.32254… × 10−24) ≈ 5.01 × 10−10 m.Using p = mev with v ≈ 1.45 × 106 m/s is an equivalent route.
The atom takes 3.20 eV, so Ek,after = 6.00 − 3.20 = 2.80 eV = 4.48 × 10−19 J; pafter = √[2(9.11 × 10−31)(4.48 × 10−19)] = 9.03469… × 10−25 kg m s−1; λafter = h/pafter ≈ 7.34 × 10−10 m.The electron loses energy and momentum, so its associated wavelength increases.
In the separate diffraction experiment, Ek = eV and p = √(2meeV). Four times V means twice the momentum and half the wavelength. That shorter wavelength relative to the unchanged graphite spacing gives smaller diffraction angles, so the rings have smaller diameter. An exact ring-diameter ratio is not needed or implied here.
Electron diffraction supports the wave-like behaviour of electrons. The photoelectric effect supports the particulate/photon nature of electromagnetic radiation, with energy transferred in photon amounts. These are different observations; an electron’s associated wave is not a light wave travelling at c.
Subtract the excitation energy before calculating the outgoing momentum and wavelength. The wave sketches illustrate associated wavelength spacing, not electron paths or light waves. Open the larger diagramThis is the separate diffraction experiment. Four times voltage halves wavelength and reduces ring diameter; the schematic does not assert an exact ring-diameter ratio. Open the larger diagram
Check: The initial speed is about 0.48% of c. Energy falls after excitation, so λ must rise. λafter/λbefore = √(6.00/2.80) ≈ 1.46, which agrees with the computed wavelengths.
Mark checklist · 9 marks
1 mark — Initial electron energy eV = 9.60 × 10⁻¹⁹ J.
1 mark — Finds initial momentum approximately 1.32 × 10⁻²⁴ kg m/s, directly or via speed.
1 mark — Uses λ = h/p to obtain initial wavelength approximately 5.01 × 10⁻¹⁰ m.
1 mark — Subtracts excitation energy to obtain 2.80 eV or 4.48 × 10⁻¹⁹ J after the collision.
1 mark — Uses this reduced energy to find momentum approximately 9.03 × 10⁻²⁵ kg m/s or equivalent speed.
1 mark — Post-collision wavelength approximately 7.34 × 10⁻¹⁰ m; carry a previous numerical positive post-collision energy error forward.
1 mark — Four times voltage gives twice momentum and half de Broglie wavelength.
1 mark — Links shorter wavelength relative to fixed graphite spacing with smaller diffraction angles and ring diameter.
1 mark — Correctly pairs electron diffraction with electron wave behaviour and the photoelectric effect with the photon/particle nature of light.
Common mistake: Using the incoming 6.00 eV for the after-collision wavelength ignores energy transferred to the atom. Using E = hc/λ for the electron treats it as a photon. Its non-relativistic energy–momentum relation is Ek = p²/(2me).
Exam technique: Calculate the energy at the instant asked about, then choose the relation for the actual particle. Explain diffraction through the ratio of wavelength to the fixed target spacing, rather than the time spent in the target.
In another trial the electron starts with 24.0 eV and makes the same 3.20 eV excitation. Find its wavelength after the collision and compare it with its wavelength before the collision.
Check the transfer answer
The remaining energy is 24.0 − 3.20 = 20.8 eV = 3.328 × 10−18 J. Using λ = h/√(2meEk) gives λafter = 2.69 × 10−10 m. Before collision λbefore = 2.51 × 10−10 m. The wavelength increases after energy transfer here too; the fractional change is smaller because the same 3.20 eV is a smaller share of the incoming energy.
Use the simulation to check your predictions
In Photoelectron emission, choose Sodium and set Wavelength to 400 nm. Its model work function is 2.28 eV: predict photon energy about 3.10 eV and maximum electron energy about 0.82 eV. Change Intensity from 40% to 80%; the photocurrent should rise from about 8.4 nA to 16.9 nA while maximum energy stays the same. At 650 nm, predict no emission even at 100% intensity. The Stopping potential tab uses a separate fixed setup: sodium, 300 nm and 60% intensity, rather than the settings from the first tab. Predict a stopping-potential magnitude about 1.85 V. Choose Collector − and increase the voltage magnitude towards 1.85–1.86 V; the current approaches zero. The negative collector potential retards electrons, while the stopping-potential magnitude is reported as positive. Compare brighter light and higher frequency after recording a result. The lab uses more precise SI constants than the rounded values in these questions, so final digits may differ. Stopping-potential measurement and the optional Planck graph are enrichment: AQA AS requires the stopping-potential concept but does not require its experimental determination.
This set covers AQA AS Physics 7407 3.2.2.1–3.2.2.4: the photoelectric effect and stopping potential; excitation, ionisation and the electron volt; discrete energy levels and photon emission; wave–particle duality and de Broglie wavelength. The photon foundation also uses 3.2.1.3. These topics are shared with the first year of AQA A-level 7408; the linked notes cover that shared content and additional particle topics. Photoelectric calculations give maximum electron kinetic energy, not the energy of every emitted electron. Atomic-level models state their assumptions. The graph task interprets supplied data and does not require designing an experiment to measure stopping potential. Details of particular electron-diffraction apparatus, Bragg calculations, relativistic electron motion, pair production and annihilation are outside this set.
These are original PhysicsUK questions, diagrams and mark checklists. The linked examiner lessons are paraphrases of checked AQA reports; applying them to these new contexts is our teaching guidance. The 2022 extended-response source used level descriptors, while this set uses its own point checklists. The graph and energy-level questions have new data. No claim is made that these are official AQA questions or a prediction of their future frequency.
