OxfordAQA International AS and A-level Physics: Electron collisions, photoelectric effect and duality (9630) – Revision Notes

These notes condense sections 3.5.9 to 3.5.11 of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4: collisions of electrons with atoms, the photoelectric effect and wave-particle duality. They are shared AS and A-level content. For full explanations and longer worked examples, read the study guide, then test yourself with the practice questions.

Also useful: the OxfordAQA A-level Physics hub, the printable checklist and the free diagnostics.

Data used in these notes: h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C, electron mass 9.11 × 10⁻³¹ kg.

Key definitions

Term	Meaning
Electronvolt	Energy gained by an electron moved through a p.d. of 1 V; 1 eV = 1.60 × 10⁻¹⁹ J
Excitation	An electron in an atom moves to a higher energy level
Ionisation	An electron is removed from the atom
Ionisation energy	Energy needed to remove an electron from the lowest level
Photon	A packet of electromagnetic energy, E = hf
Threshold frequency f₀	Lowest frequency that releases photoelectrons from a surface
Work function ϕ	Smallest energy that lets an electron escape the metal surface
Stopping potential V(s)	Reverse p.d. that just stops the fastest photoelectrons
de Broglie wavelength	The wavelength λ = h/p associated with a particle of momentum p

Equations

Equation	Use
E = hf = hc/λ	Photon energy
hf = E1 - E2	Photon from a transition between levels
hf = ϕ + Ek(max)	Photoelectric equation
ϕ = hf₀	Work function and threshold frequency
eV(s) = Ek(max)	Stopping potential
λ = h/mv	de Broglie wavelength
Ek = p²/2m	Kinetic energy from momentum
λ(min) = hc/eV	Shortest X-ray wavelength at tube p.d. V

Electron collisions: the essentials

• Energy levels are negative. Zero is ionisation.
• Line spectra are bright lines on a dark background (emission). Each line is one transition. Only certain frequencies appear, so levels are discrete.
• Bigger energy gap means higher frequency and shorter wavelength.
• A free electron can excite an atom if its kinetic energy is at least the gap. It keeps the surplus.
• A photon is absorbed in excitation only if its energy equals a gap exactly.
• Fluorescent tube: electrons accelerated through mercury vapour, collisions excite mercury atoms, de-excitation gives ultraviolet photons, phosphor coating converts these to visible light.

Method: wavelength from a transition

1. Find the gap in eV (higher level minus lower level, so a positive number).
2. Convert to joules: multiply by 1.60 × 10⁻¹⁹.
3. Use λ = hc/ΔE.

Reminder: levels at -3.9 eV and -6.1 eV. Gap = 2.2 eV = 3.52 × 10⁻¹⁹ J. λ = (1.989 × 10⁻²⁵)/(3.52 × 10⁻¹⁹) = 5.65 × 10⁻⁷ m = 565 nm.

X-rays

• X-ray tube: heated filament (thermionic emission), high p.d. across the tube, metal target. Most electron energy becomes heat in the target.
• Continuous spectrum with a minimum wavelength: the electron gives all its energy eV to one photon.
• Line (characteristic) spectrum: sharp peaks whose wavelengths depend on the target metal. They arise from transitions to fill a vacancy in an inner shell.
• Medical use: imaging (bones absorb more than soft tissue) and treating tumours. Photons can ionise tissue, so exposure is minimised.

Method: X-ray tube questions

1. Maximum photon energy = eV (in joules, e × V).
2. Shortest wavelength = hc divided by that energy.
3. Beam power = IV. Electrons per second = I/e.

Reminder: tube p.d. 30 kV. E = 30 × 10³ × 1.60 × 10⁻¹⁹ = 4.8 × 10⁻¹⁵ J, so λ(min) = 1.989 × 10⁻²⁵/4.8 × 10⁻¹⁵ = 4.14 × 10⁻¹¹ m.

Photoelectric effect

Observation	Photon explanation	Wave model fails because
No emission below f₀ at any intensity	One photon gives all its energy to one electron; if hf is less than ϕ, no electron escapes	It predicts any frequency works if bright enough
Emission is immediate	Energy arrives in one photon, with no build-up	It predicts a delay
Ek(max) rises with frequency, not intensity	Surplus energy hf - ϕ goes to kinetic energy	It predicts brighter means faster
Current rises with intensity	More photons per second means more electrons per second	Consistent, but not unique

Method: photoelectric calculations

1. Photon energy = hf or hc/λ.
2. Ek(max) = photon energy - ϕ. Keep all values in J or all in eV.
3. Stopping potential in volts = Ek(max) in eV.
4. Threshold: f₀ = ϕ/h.

Reminder: photon 4.1 eV on a metal with ϕ = 2.3 eV. Ek(max) = 1.8 eV = 2.88 × 10⁻¹⁹ J; V(s) = 1.8 V.

Reminder: a 2.0 mW beam of 520 nm light. Photon energy = 3.83 × 10⁻¹⁹ J, so photons per second = 2.0 × 10⁻³/3.83 × 10⁻¹⁹ = 5.2 × 10¹⁵.

Graph reading: V(s) against f

• Gradient = h/e, so h = gradient × e.
• Intercept on the frequency axis = f₀.
• Intercept on the V(s) axis = -ϕ/e (extrapolated).
• A different metal gives a parallel line shifted along the frequency axis, because the gradient is the same.

Graph reading: Ek(max) against f

• Gradient = h. Intercept on the energy axis = -ϕ. Same-gradient lines for all metals.

Wave-particle duality

• Electron diffraction: particles show wave behaviour.
• Photoelectric effect: electromagnetic waves show particle behaviour.
• λ = h/mv. Larger momentum gives smaller wavelength and less diffraction. Smaller momentum gives larger wavelength and more spreading.
• Diffraction is only noticeable when the wavelength is about the same size as the gap or atomic spacing.
• Large objects have negligible wavelengths.
• Models change as new evidence arrives: light as particles, then waves, then both; the electron as a particle, then also a wave.

Method: electron accelerated through a p.d.

1. Kinetic energy = eV.
2. Momentum p = √(2m × Ek).
3. λ = h/p.

Reminder: electron at 2.5 × 10⁶ m s⁻¹. p = 2.28 × 10⁻²⁴ kg m s⁻¹, λ = 2.91 × 10⁻¹⁰ m. At 5.0 × 10⁶ m s⁻¹, λ = 1.46 × 10⁻¹⁰ m, which is half.

Must-know distinctions

• Excitation vs ionisation: the electron moves up a level vs leaves the atom.
• Electron collision vs photon absorption: at least the gap vs exactly the gap.
• Intensity vs frequency: intensity changes the number of photoelectrons per second. Frequency changes their maximum kinetic energy.
• Work function vs threshold frequency: ϕ is an energy, f₀ is a frequency, and ϕ = hf₀.
• Ek(max) vs typical Ek: some energy is lost inside the metal, so most photoelectrons have less than the maximum.

Quick self-test

1. Convert 7.2 × 10⁻¹⁹ J to eV.
2. Convert 0.85 eV to J.
3. Find the energy of a photon of frequency 6.0 × 10¹⁴ Hz.
4. Find the wavelength of a photon with energy 3.0 eV.
5. A metal has ϕ = 2.5 eV. Find its threshold frequency.
6. Photoelectrons have Ek(max) = 1.4 × 10⁻¹⁹ J. What is the stopping potential?
7. A particle has momentum 4.4 × 10⁻²⁴ kg m s⁻¹. Find its de Broglie wavelength.
8. A beam of electrons passes through a thin crystal. Their speed is doubled. What happens to the spacing of the diffraction rings, and why?
9. Why does increasing the intensity of light above f₀ not change Ek(max)?
10. An atom has a 4.9 eV gap above its ground state. Why can a 6.0 eV electron excite it but a 6.0 eV photon cannot?

Answers

1. 7.2 × 10⁻¹⁹/1.60 × 10⁻¹⁹ = 4.5 eV.
2. 0.85 × 1.60 × 10⁻¹⁹ = 1.36 × 10⁻¹⁹ J.
3. 6.63 × 10⁻³⁴ × 6.0 × 10¹⁴ = 3.98 × 10⁻¹⁹ J.
4. 3.0 eV = 4.8 × 10⁻¹⁹ J, so λ = 1.989 × 10⁻²⁵/4.8 × 10⁻¹⁹ = 4.14 × 10⁻⁷ m, or 414 nm.
5. ϕ = 4.0 × 10⁻¹⁹ J, so f₀ = 4.0 × 10⁻¹⁹/6.63 × 10⁻³⁴ = 6.03 × 10¹⁴ Hz.
6. V(s) = Ek/e = 1.4 × 10⁻¹⁹/1.60 × 10⁻¹⁹ = 0.875 V (0.88 V).
7. λ = 6.63 × 10⁻³⁴/4.4 × 10⁻²⁴ = 1.5 × 10⁻¹⁰ m.
8. Momentum doubles, so λ halves. A shorter wavelength diffracts less, so the rings are closer to the centre.
9. Each photon still has the same energy and releases one electron. More photons release more electrons, but not faster ones.
10. The electron can give up 4.9 eV and keep 1.1 eV. A photon is absorbed whole, and 6.0 eV matches no gap, so it is not absorbed.

Where marks are usually lost

• Leaving an answer in joules when the question asks for electronvolts, or the reverse.
• Writing “the electron absorbs the energy” for a collision without saying it keeps the surplus.
• Saying “brighter light means more energetic electrons”.
• Failing to say that one photon interacts with one electron.
• Using wavelength in nanometres without converting to metres.
• Taking the gradient of a V(s) against f graph as h, forgetting to multiply by e.
• Treating the work function as a frequency.
• Describing diffraction changes without linking momentum, wavelength and gap size.
• Calling the fluorescent tube’s visible light a direct product of mercury excitation, missing the phosphor step.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), version 4.4, for International AS and A-level exams from May/June 2020, published by OxfordAQA.