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

This guide covers sections 3.5.9 (collisions of electrons with atoms), 3.5.10 (photoelectric effect) and 3.5.11 (wave-particle duality) of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4, for exams from May/June 2020 onwards. Sections 3.1 to 3.5 are shared International AS and A-level content, so everything here applies to both.

Use it with the OxfordAQA A-level Physics hub and the printable checklist. The companion pages are the revision notes and the practice questions. If you are unsure where you stand, take a free diagnostic first.

What you must be able to do

Section	Outcome
3.5.9	Explain ionisation and excitation, including in the fluorescent tube; use the electronvolt and convert between eV and J
3.5.9	Use line spectra as evidence for discrete energy levels; use hf = E1 - E2
3.5.9	Describe characteristic and line spectra for X-rays, the basic X-ray tube and medical uses
3.5.10	Use the photon model, the Planck constant and E = hf = hc/λ
3.5.10	Explain threshold frequency, work function and stopping potential; use hf = ϕ + Ek(max)
3.5.11	Explain why electron diffraction and the photoelectric effect show duality; use λ = h/mv
3.5.11	Explain how and why diffraction changes with momentum; appreciate how ideas about matter change over time

The specification says the experimental determination of stopping potential is not required, and that details of particular methods of particle diffraction are not expected. Energy levels may be given in eV or J.

Values used below are supplied data: 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.

3.5.9 Collisions of electrons with atoms

The electronvolt

One electronvolt is the energy an electron gains when it is moved through a potential difference of one volt. Since W = QV, 1 eV = 1.60 × 10⁻¹⁹ J.

To go from eV to J, multiply by 1.60 × 10⁻¹⁹. To go from J to eV, divide. Example: 4.8 eV = 4.8 × 1.60 × 10⁻¹⁹ = 7.68 × 10⁻¹⁹ J, and 2.4 × 10⁻¹⁹ J = 1.5 eV.

Ionisation and excitation

• Ionisation: an electron is removed from the atom, leaving a positive ion. The energy needed from the lowest level is the ionisation energy.
• Excitation: an electron in the atom moves to a higher energy level. The atom stays neutral.

An atom’s energy levels are negative, with zero at ionisation. A level of -9.0 eV means 9.0 eV is needed to free that electron.

A free electron colliding with an atom can excite it if its kinetic energy is at least the gap to a higher level. It gives up exactly that gap and keeps the rest. A photon is different: it is absorbed only if its energy equals a gap exactly, or exceeds the ionisation energy (the surplus becomes kinetic energy of the freed electron).

Fluorescent tube

A fluorescent tube holds low-pressure mercury vapour.

1. A high voltage accelerates free electrons along the tube.
2. Fast electrons collide with mercury atoms and excite them. Some collisions ionise atoms, which releases more electrons and keeps the discharge going.
3. Excited atoms de-excite and emit photons, mostly in the ultraviolet.
4. The phosphor coating on the inside of the glass absorbs the ultraviolet photons and re-emits lower-energy photons as visible light.

Line spectra and discrete levels

A hot, low-pressure gas emits only certain wavelengths, seen as bright lines. Each line is a photon from one transition between two levels:

hf = E1 - E2

where E1 is the higher level. Only certain energy differences exist, so only certain frequencies appear. This is evidence that energy levels are discrete.

Worked example: a hypothetical atom

Atom X has levels (in eV): n = 1, -9.0; n = 2, -4.2; n = 3, -2.4; n = 4, -1.1.

(a) Ionisation energy from n = 1: 9.0 eV.

(b) Longest wavelength among these levels comes from the smallest gap, 4 to 3: 1.3 eV = 2.08 × 10⁻¹⁹ J. λ = hc/E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(2.08 × 10⁻¹⁹) = 9.56 × 10⁻⁷ m, or 956 nm.

(c) Transition 3 to 2: ΔE = 1.8 eV = 2.88 × 10⁻¹⁹ J, so f = ΔE/h = 4.34 × 10¹⁴ Hz and λ = 691 nm.

(d) An electron with 5.0 eV collides with an atom in the ground state. The 1 to 2 gap is 4.8 eV, so it can excite the atom to n = 2 and leave with 0.2 eV. A photon of 5.0 eV could not do this, because 5.0 is not a gap.

X-rays

An X-ray tube is an evacuated glass tube.

• A heated filament releases electrons by thermionic emission.
• A high p.d. (tens of kilovolts) accelerates them towards a metal target (the anode).
• When they hit, most of their energy becomes thermal energy in the target, so it needs cooling. A small fraction becomes X-ray photons.

The X-ray spectrum has two features. A continuous spectrum comes from electrons slowing in the target. Its maximum photon energy is eV, when one electron gives all its energy to one photon, so there is a minimum wavelength λ(min) = hc/eV. A line (characteristic) spectrum appears as sharp peaks at wavelengths fixed by the target metal. An incoming electron ejects an inner-shell electron from a target atom, an outer electron drops into the gap, and the energy difference is emitted as an X-ray photon. Changing the metal changes the lines.

Worked example. An X-ray tube runs at 45 kV with a beam current of 12 mA.

• Maximum photon energy = 45 keV = 45 × 10³ × 1.60 × 10⁻¹⁹ = 7.2 × 10⁻¹⁵ J.
• λ(min) = hc/E = (1.989 × 10⁻²⁵)/(7.2 × 10⁻¹⁵) = 2.76 × 10⁻¹¹ m.
• Power delivered by the beam = IV = 12 × 10⁻³ × 45 × 10³ = 540 W.
• Electrons per second = I/e = 12 × 10⁻³/1.60 × 10⁻¹⁹ = 7.5 × 10¹⁶ s⁻¹.

Medical use. X-rays pass through soft tissue more easily than bone, so an image on a detector shows bones as lighter shadows. X-rays are also used to treat cancer by damaging cells in a tumour. Because X-ray photons can ionise tissue, exposure is kept as low as possible.

3.5.10 Photoelectric effect

The photon model

Electromagnetic radiation travels as packets of energy called photons. Each photon has energy

E = hf = hc/λ

where h is the Planck constant, 6.63 × 10⁻³⁴ J s. Higher frequency means more energy per photon. Intensity (at a given frequency) is the number of photons arriving per second per unit area.

What is observed

When ultraviolet light hits a clean metal surface, electrons can be emitted (photoelectrons).

1. Below a threshold frequency f₀ there is no emission, however bright the light and however long you wait.
2. Above f₀, emission starts at once.
3. The maximum kinetic energy of the photoelectrons depends on frequency, not on intensity.
4. Increasing the intensity (same frequency, above f₀) increases the number of photoelectrons per second, so the photocurrent rises.

The wave model predicts that a bright enough beam would eventually free electrons at any frequency, after a delay while energy builds up, and that brighter light would give faster electrons. Observation contradicts all of this.

The photon explanation

One photon interacts with one electron, giving it all its energy. The work function ϕ is the smallest energy that lets an electron escape from the metal surface. If hf < ϕ, no electron can escape, whatever the number of photons. So the threshold frequency is f₀ = ϕ/h. Above it, the surplus appears as kinetic energy:

hf = ϕ + Ek(max)

Ek(max) is the maximum kinetic energy, for electrons that lose nothing on the way out. Others have less.

The stopping potential V(s) is the reverse p.d. that just stops the fastest photoelectrons reaching the collector: eV(s) = Ek(max).

Worked example: calculation

A metal has work function 3.2 eV. Light of wavelength 240 nm falls on it.

• Photon energy = hc/λ = (1.989 × 10⁻²⁵)/(240 × 10⁻⁹) = 8.29 × 10⁻¹⁹ J = 5.18 eV.
• ϕ = 3.2 × 1.60 × 10⁻¹⁹ = 5.12 × 10⁻¹⁹ J.
• Ek(max) = 8.29 × 10⁻¹⁹ - 5.12 × 10⁻¹⁹ = 3.17 × 10⁻¹⁹ J = 1.98 eV.
• Stopping potential = 1.98 V.
• Threshold frequency f₀ = ϕ/h = 5.12 × 10⁻¹⁹/6.63 × 10⁻³⁴ = 7.72 × 10¹⁴ Hz, which is a wavelength of 388 nm.
• Light of 450 nm gives no emission: it is above the threshold wavelength.

Worked example: interpreting data

Rearranging gives V(s) = (h/e)f - ϕ/e. A graph of V(s) against f is a straight line with gradient h/e and a frequency-axis intercept f₀.

f / 10¹⁴ Hz	6.0	7.0	8.0	9.0	10.0
V(s) / V	0.39	0.80	1.22	1.63	2.04

A best-fit line passes close to (6.0, 0.39) and (10.0, 2.04).

• Gradient = (2.04 - 0.39)/((10.0 - 6.0) × 10¹⁴) = 4.13 × 10⁻¹⁵ V s.
• h = gradient × e = 4.13 × 10⁻¹⁵ × 1.60 × 10⁻¹⁹ = 6.60 × 10⁻³⁴ J s.
• The line crosses V(s) = 0 at f₀ = 6.0 × 10¹⁴ - 0.39/4.13 × 10⁻¹⁵ = 5.06 × 10¹⁴ Hz.
• ϕ = hf₀ = 6.60 × 10⁻³⁴ × 5.06 × 10¹⁴ = 3.34 × 10⁻¹⁹ J, about 2.1 eV.

3.5.11 Wave-particle duality

Evidence both ways

• Electron diffraction suggests particles have wave properties. A beam of electrons passed through a thin crystalline sample or a narrow gap produces a diffraction pattern, which only waves do.
• The photoelectric effect suggests electromagnetic waves have a particulate nature, since energy is transferred in discrete photons.

de Broglie wavelength

λ = h/mv, where mv is the momentum.

Worked example. An electron moves at 4.0 × 10⁶ m s⁻¹. Momentum = 9.11 × 10⁻³¹ × 4.0 × 10⁶ = 3.64 × 10⁻²⁴ kg m s⁻¹, so λ = 6.63 × 10⁻³⁴/3.64 × 10⁻²⁴ = 1.82 × 10⁻¹⁰ m. That is comparable to the spacing between atoms, so a crystal diffracts it.

Compare a proton at the same speed (mass 1.67 × 10⁻²⁷ kg): λ = 9.93 × 10⁻¹⁴ m, much smaller because its momentum is about 1800 times greater. A 0.16 kg ball at 30 m s⁻¹ has λ = 1.4 × 10⁻³⁴ m. No gap is that small, so you never see it diffract.

Momentum and diffraction

Diffraction is noticeable when the wavelength is about the size of the gap or the atomic spacing. If momentum increases, λ = h/p decreases, so the pattern spreads less and the angles to the maxima and minima get smaller. Halving the momentum doubles the wavelength and widens the pattern.

Changing ideas about matter

Light was described as a stream of particles, then shown to behave as a wave through interference experiments, and then needed photons to explain the photoelectric effect. Electrons were found to be particles, yet de Broglie proposed in 1924 that they had a wavelength, and diffraction experiments in the 1920s supported this. Models are kept while they explain the evidence and are revised when they do not.

Common errors

• Using 1.60 × 10⁻¹⁹ the wrong way round when converting eV and J.
• Saying brighter light gives faster photoelectrons. It gives more of them.
• Forgetting that hf = ϕ + Ek(max) needs consistent units: all joules, or all eV.
• Saying an electron needs exactly a gap’s energy to excite an atom. A photon does; an electron needs at least that much.
• Using the electron’s kinetic energy as its momentum in λ = h/mv.

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.