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Notes · PhysicsUK · A-Levels

Particles and radiation

This chapter opens the sub-atomic world: the particles that make up the atom, the antiparticles that mirror them, and the exchange particles that carry the four fundamental forces. It classifies matter into hadrons and leptons, builds baryons and mesons from quarks, and applies the conservation laws that decide which interactions can happen. It closes with the quantum evidence that light and matter are both wave and particle - the photoelectric effect, discrete atomic energy levels and the de Broglie wavelength.

6 sections·~18 min reading time·3 competencies·Level Foundation 1 · Standard 2 · Advanced 3

T·0222 / 13
Exam profile
AO1 · Recall the constituents of the atom, the particle classification, the fundamental forces and exchange particles, and the photoelectric and energy-level equationsAO2 · Calculate specific charge, photon energy, threshold frequency, stopping potential and de Broglie wavelength, and apply the conservation laws to interactionsAO3 · Interpret beta-decay energy spectra as evidence for the neutrino and electron diffraction as evidence for wave-particle duality
Operators:statedescribeexplaincalculatededuceshow that

basic level

AS-Level requires the constituents of the atom, radioactive decay equations, the particle zoo and its conservation laws, the photoelectric effect, atomic energy levels and the de Broglie relation.

higher level

The full A-Level uses the same content synoptically - conservation laws underpin nuclear physics later in the course, and the quantum ideas here return in fields, capacitance and nuclear energy.

Depth

Reading depth: In depth

Text

Text size: Standard

Contents · 6 sections▾
  1. Particles and radiation
    • 01The nuclear atom, specific charge and isotopes○
    • 02Stability, radioactive decay and the neutrino◐
    • 03Antiparticles, photons, annihilation and pair production◐
    • 04Classification, quarks and conservation laws●
    • 05The photoelectric effect and atomic energy levels●
    • 06Wave-particle duality and the de Broglie wavelength●
§ 01

The nuclear atom, specific charge and isotopes#

●○○FoundationLPAQA 7408 3.2.1.1LPDfE GCE Physics - constituents of the atom

Key points

An atom is a tiny, dense, positively charged nucleus of protons and neutrons (together the nucleons) surrounded by orbiting electrons. On the SI scale a proton and a neutron each have a mass of about 1.67×10−27 kg1.67\times10^{-27}\,\text{kg}1.67×10−27kg, while an electron is about 183618361836 times lighter at 9.11×10−31 kg9.11\times10^{-31}\,\text{kg}9.11×10−31kg. The proton carries charge +e+e+e and the electron −e-e−e, where the elementary charge e=1.60×10−19 Ce = 1.60\times10^{-19}\,\text{C}e=1.60×10−19C; the neutron is uncharged. A neutral atom has equal numbers of protons and electrons.
A nuclide is specified by two numbers written as ZAX^{A}_{Z}\text{X}ZA​X. The proton (atomic) number ZZZ is the number of protons; it fixes the element. The nucleon (mass) number AAA is the total number of protons and neutrons, so the neutron number is N=A−ZN = A - ZN=A−Z. Sodium-23, 1123Na^{23}_{11}\text{Na}1123​Na, therefore has 111111 protons, 111111 electrons and 121212 neutrons.
The specific charge of a particle is its charge divided by its mass, in C kg−1\text{C}\,\text{kg}^{-1}Ckg−1; it is largest for the electron because that particle has the greatest charge-to-mass ratio. For a nucleus or ion, add the charges of the protons (and subtract any extra or missing electrons) and divide by the total mass of the nucleons. Specific charge is the quantity a mass spectrometer effectively measures and a favourite short calculation in this topic.
Isotopes are nuclei of the same element (same ZZZ) with different numbers of neutrons and therefore different AAA. Because chemistry and the electron structure depend only on ZZZ, isotopes are chemically identical but differ in mass and in nuclear stability. Carbon exists as stable 12C^{12}\text{C}12C and 13C^{13}\text{C}13C and as radioactive 14C^{14}\text{C}14C, the last of which underlies radiocarbon dating.
specific charge=Qm\text{specific charge} = \dfrac{Q}{m}specific charge=mQ​

Specific charge

Charge in coulombs divided by mass in kilograms, measured in C kg−1\text{C}\,\text{kg}^{-1}Ckg−1.

N=A−ZN = A - ZN=A−Z

Neutron number

The nucleon number minus the proton number.

Worked example

Specific charge of a nucleus

Find the specific charge of a 24He^{4}_{2}\text{He}24​He nucleus (an alpha particle). Take each nucleon as 1.67×10−27 kg1.67\times10^{-27}\,\text{kg}1.67×10−27kg and e=1.60×10−19 Ce = 1.60\times10^{-19}\,\text{C}e=1.60×10−19C.

  1. 01Charge

    Two protons give Q=2e=2×1.60×10−19=3.20×10−19 CQ = 2e = 2 \times 1.60\times10^{-19} = 3.20\times10^{-19}\,\text{C}Q=2e=2×1.60×10−19=3.20×10−19C.

  2. 02Mass

    Four nucleons give m=4×1.67×10−27=6.68×10−27 kgm = 4 \times 1.67\times10^{-27} = 6.68\times10^{-27}\,\text{kg}m=4×1.67×10−27=6.68×10−27kg.

  3. 03Divide

    Specific charge =Q/m= Q/m=Q/m.

    3.20×10−196.68×10−27=4.79×107 C kg−1\dfrac{3.20\times10^{-19}}{6.68\times10^{-27}} = 4.79\times10^{7}\,\text{C}\,\text{kg}^{-1}6.68×10−273.20×10−19​=4.79×107Ckg−1

Result: The alpha particle has a specific charge of 4.8×107 C kg−14.8\times10^{7}\,\text{C}\,\text{kg}^{-1}4.8×107Ckg−1.

Exam focus

  • State the relative charges and masses of the proton, neutron and electron and give the numbers of each in a stated nuclide or ion.
  • Calculate the specific charge of a nucleus, ion or particle in C kg−1\text{C}\,\text{kg}^{-1}Ckg−1.

Typical mistakes

  • Changing the nucleon number when an atom becomes an ion - forming an ion changes only the electron count, never the number of protons or neutrons.
  • Using the atomic mass in grams or unified mass units instead of kilograms when finding a specific charge.

Active revision

Calculate the specific charge of a 37Li+^{7}_{3}\text{Li}^{+}37​Li+ ion, taking the mass of a nucleon as 1.67×10−27 kg1.67\times10^{-27}\,\text{kg}1.67×10−27kg and e=1.60×10−19 Ce = 1.60\times10^{-19}\,\text{C}e=1.60×10−19C.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Physics 7408 specification (AQA)

§ 02

Stability, radioactive decay and the neutrino#

●●○StandardLPAQA 7408 3.2.1.2LPDfE GCE Physics - radioactive decay

The continuous energy spectrum of beta particles

Beta-minus energy spectrumLine chart: Number of beta particles by Kinetic energy / MeV00.20.40.60.8100.20.40.60.81Number of beta particlesKinetic energy / MeV
Fig. 1Beta particles are emitted with all energies up to a sharp maximum. The missing energy is carried by the antineutrino - the evidence for its existence.

Key points

The protons in a nucleus repel one another electrostatically, yet nuclei hold together, so a stronger, shorter-range attraction must act between nucleons: the strong nuclear force. It is attractive from about 3 fm3\,\text{fm}3fm down to about 0.5 fm0.5\,\text{fm}0.5fm, becoming strongly repulsive at separations smaller than that (which stops the nucleus collapsing), and it is negligible beyond a few femtometres. Because it acts only over such short ranges, very large nuclei cannot be fully bound by it and become unstable.
An unstable nucleus reaches a more stable configuration by radioactive decay. Alpha decay emits a helium nucleus 24α^{4}_{2}\alpha24​α, reducing ZZZ by 2 and AAA by 4; it occurs in heavy nuclei such as uranium. In beta-minus decay a neutron changes into a proton, emitting an electron −10β^{0}_{-1}\beta−10​β and an antineutrino: ZZZ rises by 1 while AAA is unchanged. Every decay equation must balance both nucleon number and proton number (charge).
Beta decay posed a crisis: the emitted electrons carry a continuous range of energies up to a maximum, not the single fixed energy that a two-body decay demands. If only a proton and an electron were produced, energy and momentum could not both be conserved. Pauli proposed a third, almost undetectable particle - the (anti)neutrino - to carry the missing energy and momentum. The continuous beta spectrum, with its sharp maximum energy, is the direct experimental evidence for the neutrino.
The neutrino is a lepton: it is electrically neutral, has a very small mass and interacts only through the weak interaction, so it passes through matter almost unimpeded. Writing the beta-minus equation in full, ZAX→ Z+1AY+ −10β+νˉe^{A}_{Z}\text{X} \rightarrow\, ^{A}_{Z+1}\text{Y} + \,^{0}_{-1}\beta + \bar{\nu}_eZA​X→Z+1A​Y+−10​β+νˉe​, shows the antineutrino explicitly; a genuine understanding of it will be completed by the quark picture later in the chapter.
ZAX→ Z−2A−4Y+ 24α^{A}_{Z}\text{X} \rightarrow\, ^{A-4}_{Z-2}\text{Y} + \,^{4}_{2}\alphaZA​X→Z−2A−4​Y+24​α

Alpha decay

The nucleus loses two protons and two neutrons.

ZAX→ Z+1AY+ −10β+νˉe^{A}_{Z}\text{X} \rightarrow\, ^{A}_{Z+1}\text{Y} + \,^{0}_{-1}\beta + \bar{\nu}_eZA​X→Z+1A​Y+−10​β+νˉe​

Beta-minus decay

A neutron becomes a proton, emitting an electron and an electron antineutrino.

Worked example

Balancing a decay chain

Thorium-234 decays by beta-minus emission to protactinium (Pa). Write the balanced nuclear equation.

  1. 01Conserve nucleon number

    Beta-minus emission does not change AAA, so the daughter still has A=234A = 234A=234.

  2. 02Conserve proton number

    The electron has Z=−1Z = -1Z=−1, so the daughter's proton number rises by 1: 90→9190 \rightarrow 9190→91 (protactinium).

  3. 03Write the equation

    Include the electron antineutrino.

    90234Th→ 91234Pa+ −10β+νˉe^{234}_{90}\text{Th} \rightarrow\, ^{234}_{91}\text{Pa} + \,^{0}_{-1}\beta + \bar{\nu}_e90234​Th→91234​Pa+−10​β+νˉe​

Result: 90234Th→ 91234Pa+ −10β+νˉe^{234}_{90}\text{Th} \rightarrow\, ^{234}_{91}\text{Pa} + \,^{0}_{-1}\beta + \bar{\nu}_e90234​Th→91234​Pa+−10​β+νˉe​, with nucleon and proton numbers balanced.

Exam focus

  • Write balanced nuclear equations for alpha and beta-minus decay, conserving nucleon and proton number.
  • Explain how the continuous energy spectrum of beta particles is evidence for the existence of the (anti)neutrino.

Typical mistakes

  • Leaving the antineutrino out of a beta-minus decay equation, or omitting the electron's proton number of −1-1−1.
  • Describing the strong force as attractive at all separations - it becomes repulsive below about 0.5 fm0.5\,\text{fm}0.5fm.

Active revision

Uranium-238 decays by alpha emission to thorium, which then decays by beta-minus emission. Write both balanced nuclear equations, including the antineutrino.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

§ 03

Antiparticles, photons, annihilation and pair production#

●●○StandardLPAQA 7408 3.2.1.3LPAQA 7408 3.2.1.4LPDfE GCE Physics - antiparticles and photons

Key points

Every particle has a corresponding antiparticle with the same mass and rest energy but opposite charge (and opposite baryon/lepton number). The electron's antiparticle is the positron +10e^{0}_{+1}\text{e}+10​e; the proton's is the antiproton; the neutrino's is the antineutrino. A particle and its antiparticle can be created together or destroyed together, but never singly, because charge and other quantum numbers must be conserved.
Electromagnetic radiation is quantised into photons, each carrying energy E=hf=hc/λE = hf = hc/\lambdaE=hf=hc/λ, where h=6.63×10−34 J sh = 6.63\times10^{-34}\,\text{J}\,\text{s}h=6.63×10−34Js is the Planck constant. Photon and particle energies at this scale are conveniently measured in electronvolts: 1 eV=1.60×10−19 J1\,\text{eV} = 1.60\times10^{-19}\,\text{J}1eV=1.60×10−19J is the energy gained by an electron accelerated through one volt. Rest energy is given by Einstein's mass-energy relation E0=mc2E_0 = m c^2E0​=mc2, so an electron's rest energy is 0.511 MeV0.511\,\text{MeV}0.511MeV.
In annihilation a particle meets its antiparticle and their entire mass is converted to electromagnetic energy. An electron and a positron at rest annihilate to produce two photons (two, so that momentum is conserved), each of energy equal to the rest energy of one particle, 0.511 MeV0.511\,\text{MeV}0.511MeV. This is the physics behind the PET (positron emission tomography) medical scanner.
Pair production is the reverse: a single photon of sufficient energy, passing near a nucleus, converts into a particle-antiparticle pair. The photon must carry at least the total rest energy of the pair - for an electron-positron pair, at least 2×0.511=1.02 MeV2 \times 0.511 = 1.02\,\text{MeV}2×0.511=1.02MeV; any excess becomes kinetic energy. A nearby nucleus is needed to conserve momentum. Both processes are direct, everyday demonstrations that mass and energy are interconvertible.
E=hf=hcλE = hf = \dfrac{hc}{\lambda}E=hf=λhc​

Photon energy

The energy of a photon of frequency fff or wavelength λ\lambdaλ.

E0=mc2E_0 = m c^2E0​=mc2

Rest energy

The energy equivalent of a particle's rest mass; for an electron 0.511 MeV0.511\,\text{MeV}0.511MeV.

hfmin⁡=2mc2hf_{\min} = 2 m c^2hfmin​=2mc2

Threshold for pair production

The photon must supply at least the total rest energy of the particle-antiparticle pair.

Worked example

Minimum energy for pair production

Find the minimum photon energy in joules, and its frequency, needed to create an electron-positron pair. Rest energy of each is 0.511 MeV0.511\,\text{MeV}0.511MeV; h=6.63×10−34 J sh = 6.63\times10^{-34}\,\text{J}\,\text{s}h=6.63×10−34Js, 1 eV=1.60×10−19 J1\,\text{eV} = 1.60\times10^{-19}\,\text{J}1eV=1.60×10−19J.

  1. 01Total rest energy

    Two particles: Emin⁡=2×0.511=1.022 MeV=1.022×106 eVE_{\min} = 2 \times 0.511 = 1.022\,\text{MeV} = 1.022\times10^{6}\,\text{eV}Emin​=2×0.511=1.022MeV=1.022×106eV.

  2. 02Convert to joules

    Emin⁡=1.022×106×1.60×10−19=1.64×10−13 JE_{\min} = 1.022\times10^{6} \times 1.60\times10^{-19} = 1.64\times10^{-13}\,\text{J}Emin​=1.022×106×1.60×10−19=1.64×10−13J.

  3. 03Find the frequency

    Rearrange E=hfE = hfE=hf to f=E/hf = E/hf=E/h.

    f=1.64×10−136.63×10−34=2.47×1020 Hzf = \dfrac{1.64\times10^{-13}}{6.63\times10^{-34}} = 2.47\times10^{20}\,\text{Hz}f=6.63×10−341.64×10−13​=2.47×1020Hz

Result: The photon needs at least 1.64×10−13 J1.64\times10^{-13}\,\text{J}1.64×10−13J, a frequency of 2.5×1020 Hz2.5\times10^{20}\,\text{Hz}2.5×1020Hz (a gamma-ray photon).

Exam focus

  • Calculate photon energy from frequency or wavelength and convert between joules, electronvolts and MeV\text{MeV}MeV.
  • Find the minimum photon energy (or frequency) for pair production, and the photon energy released in annihilation.

Typical mistakes

  • Producing a single photon in annihilation - two are required so that momentum is conserved when the particles are initially at rest.
  • Forgetting that pair production needs at least twice the rest energy of the particle, and that a nucleus must be present.

Active revision

Calculate the minimum frequency of a photon that can produce an electron-positron pair, given the electron rest energy is 0.511 MeV0.511\,\text{MeV}0.511MeV and h=6.63×10−34 J sh = 6.63\times10^{-34}\,\text{J}\,\text{s}h=6.63×10−34Js.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

§ 04

Classification, quarks and conservation laws#

●●●AdvancedLPAQA 7408 3.2.1.5LPAQA 7408 3.2.1.6LPAQA 7408 3.2.1.7LPDfE GCE Physics - particle classification

Classification of particles

The particle familiesGraph, Particles → Hadrons, Particles → Leptons, Hadrons → Baryons (qqq), Hadrons → Mesons (q-qbar), Baryons (qqq) → proton uud, Baryons (qqq) → neutron udd, Mesons (q-qbar) → pion, Mesons (q-qbar) → kaon, Leptons → electron, Leptons → neutrinoParticlesHadronsLeptonsBaryons (qqq)Mesons (q-qbar)proton uudneutron uddpionkaonelectronneutrino
Fig. 2Matter divides into hadrons (which feel the strong force and are built from quarks) and leptons (fundamental). Hadrons split into baryons (three quarks) and mesons (quark-antiquark).

Key points

Matter particles divide into two great families. Hadrons feel the strong nuclear force and are made of quarks; leptons do not feel the strong force and are, as far as we know, fundamental. Hadrons split again into baryons (three quarks, e.g. the proton and neutron) and mesons (a quark and an antiquark, e.g. pions and kaons). Leptons include the electron, the muon and their neutrinos, together with their antiparticles.
Four fundamental forces act through the exchange of virtual particles. The electromagnetic force is carried by the virtual photon; the weak interaction (responsible for beta decay and any change of quark or lepton type) by the W and Z bosons; the strong force between nucleons by pions; and, between quarks, by gluons. Feynman diagrams represent these interactions, with the exchange particle passing between the two interacting particles.
Every interaction must conserve certain quantities. Charge, baryon number and lepton number are always conserved. Baryon number is +13+\tfrac{1}{3}+31​ per quark (so +1+1+1 for a baryon, −1-1−1 for an antibaryon, 000 for a meson); lepton number is +1+1+1 for a lepton and −1-1−1 for an antilepton, counted separately for the electron and muon families. Strangeness is conserved in strong and electromagnetic interactions but can change by 000 or ±1\pm 1±1 in weak interactions - which is exactly why strange particles are produced quickly (strongly) but decay slowly (weakly).
The up, down and strange quarks carry charges +23e+\tfrac{2}{3}e+32​e, −13e-\tfrac{1}{3}e−31​e and −13e-\tfrac{1}{3}e−31​e, with the strange quark also carrying strangeness −1-1−1. A proton is uud\text{uud}uud (charge +23+23−13=+1+\tfrac{2}{3}+\tfrac{2}{3}-\tfrac{1}{3} = +1+32​+32​−31​=+1) and a neutron is udd\text{udd}udd (charge 000). Beta-minus decay is, at the quark level, a down quark turning into an up quark: d→u+ −10β+νˉe\text{d} \rightarrow \text{u} + \,^{0}_{-1}\beta + \bar{\nu}_ed→u+−10​β+νˉe​, mediated by a W−W^-W− boson - the deepest explanation of the process met earlier.
proton=uud,neutron=udd\text{proton} = \text{uud}, \qquad \text{neutron} = \text{udd}proton=uud,neutron=udd

Quark composition of nucleons

The proton is two up and one down quark; the neutron one up and two down.

d→u+ −10β+νˉe\text{d} \rightarrow \text{u} + \,^{0}_{-1}\beta + \bar{\nu}_ed→u+−10​β+νˉe​

Beta-minus decay at the quark level

A down quark becomes an up quark via the weak interaction (a W−W^-W− boson).

Worked example

Testing an interaction with conservation laws

Can the strong interaction K−+p→K++K0+Ω−\text{K}^{-} + \text{p} \rightarrow \text{K}^{+} + \text{K}^{0} + \Omega^{-}K−+p→K++K0+Ω− proceed? Given: charge, baryon number and strangeness must be conserved. K−\text{K}^{-}K− has S = -1, K+\text{K}^{+}K+ and K0\text{K}^{0}K0 have S = +1, Ω−\Omega^{-}Ω− has S = -3.

  1. 01Charge

    Left: −1+1=0-1 + 1 = 0−1+1=0. Right: +1+0+(−1)=0+1 + 0 + (-1) = 0+1+0+(−1)=0. Charge is conserved.

  2. 02Baryon number

    Left: mesons 000, proton +1+1+1, total +1+1+1. Right: two mesons 000, Ω−\Omega^{-}Ω− is a baryon +1+1+1, total +1+1+1. Conserved.

  3. 03Strangeness

    Left: −1+0=−1-1 + 0 = -1−1+0=−1. Right: (+1)+(+1)+(−3)=−1(+1) + (+1) + (-3) = -1(+1)+(+1)+(−3)=−1. Conserved.

    Sleft=−1=SrightS_{\text{left}} = -1 = S_{\text{right}}Sleft​=−1=Sright​

Result: Charge, baryon number and strangeness are all conserved, so this strong interaction is allowed.

Exam focus

  • Classify a named particle as a baryon, meson or lepton and give the quark composition of the proton, neutron, pions and kaons.
  • Apply conservation of charge, baryon number, lepton number and strangeness to decide whether a proposed interaction can occur.

Typical mistakes

  • Counting lepton number across families - the electron and muon lepton numbers are conserved separately.
  • Assuming strangeness is always conserved - it may change by ±1\pm 1±1 in weak interactions, which is how strange particles decay.

Active revision

Test whether the interaction p+π−→K++Σ−\text{p} + \pi^{-} \rightarrow \text{K}^{+} + \Sigma^{-}p+π−→K++Σ− can proceed by checking charge, baryon number and strangeness.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

§ 05

The photoelectric effect and atomic energy levels#

●●●AdvancedLPAQA 7408 3.2.2.1LPAQA 7408 3.2.2.2LPAQA 7408 3.2.2.3LPDfE GCE Physics - photoelectric effect

Maximum kinetic energy against frequency

Photoelectric line: Ek(max) vs frequencyGraph of Ek(max) = hf - phi, roots at x = 4.831, y-intercept at y = -2, increasing, on the interval x from 0 to 1224681012−2−1123f0 (threshold)−phiEk = 0Ek(max) = hf -phiEk(max) / eVFrequency f / 1014 Hz
Fig. 3Above the threshold frequency f0f_0f0​ the maximum kinetic energy rises linearly with frequency; the gradient is the Planck constant and the y-intercept is minus the work function.

Key points

When light of high enough frequency strikes a metal surface, electrons are emitted at once. Three features defeat the wave theory of light: emission occurs only above a threshold frequency f0f_0f0​ however bright the light; above f0f_0f0​ electrons are emitted the instant the light arrives, with no delay; and the maximum kinetic energy of the emitted electrons depends on the frequency, not the intensity. Einstein explained all three by treating light as photons, each delivering energy hfhfhf to a single electron in a one-to-one interaction.
Each metal has a work function ϕ\phiϕ, the minimum energy needed to release an electron from its surface. If a photon supplies hfhfhf, the electron uses ϕ\phiϕ to escape and keeps the rest as kinetic energy, giving Einstein's photoelectric equation hf=ϕ+Ek,max⁡hf = \phi + E_{k,\max}hf=ϕ+Ek,max​. Emission is impossible when hf<ϕhf < \phihf<ϕ, so the threshold frequency is f0=ϕ/hf_0 = \phi/hf0​=ϕ/h. A graph of Ek,max⁡E_{k,\max}Ek,max​ against fff is a straight line of gradient hhh, with x-intercept f0f_0f0​ and y-intercept −ϕ-\phi−ϕ - a beautiful confirmation of the photon model.
The maximum kinetic energy can be measured with a stopping potential VsV_sVs​: the reverse voltage just large enough to stop the fastest electrons reaching the collector, so that eVs=Ek,max⁡eV_s = E_{k,\max}eVs​=Ek,max​. Increasing the intensity at fixed frequency releases more electrons per second (a larger current) but does not increase their maximum energy - decisive evidence against the wave picture.
Electrons in an atom occupy discrete energy levels, quoted as negative energies because the electron is bound (zero is taken at ionisation). An electron can be excited to a higher level by absorbing exactly the right energy, from a photon or a colliding electron; it later falls back, emitting a photon whose energy equals the difference between the levels, hf=E2−E1hf = E_2 - E_1hf=E2​−E1​. This is why hot gases emit line spectra, and it is the principle of the fluorescent tube, in which ultraviolet photons from excited mercury atoms are re-emitted as visible light by a phosphor coating.
hf=ϕ+Ek,max⁡hf = \phi + E_{k,\max}hf=ϕ+Ek,max​

Einstein's photoelectric equation

Photon energy equals the work function plus the maximum kinetic energy of the electron.

f0=ϕhf_0 = \dfrac{\phi}{h}f0​=hϕ​

Threshold frequency

The lowest frequency that can release an electron.

hf=E2−E1hf = E_2 - E_1hf=E2​−E1​

Photon from a transition

The photon energy equals the difference between the two energy levels.

Atomic energy levels and emission

Energy levels of a hydrogen-like atomenergy level diagram, 4 levels, 3 transitions, Data: n=1 (ground), n=2, n=3, n=4Energy / eVn = 1 (ground)n = 2n = 3n = 4n = 3 to n = 2, 1.89 eVn = 2 to n = 1, 10.2 eVabsorption
Fig. 4Discrete energy levels (eV). An electron excited to a higher level falls back, emitting a photon whose energy equals the level difference.
Worked example

Photoelectric maximum kinetic energy

Light of wavelength 250 nm250\,\text{nm}250nm strikes a metal of work function 3.0 eV3.0\,\text{eV}3.0eV. Find the maximum kinetic energy of the emitted electrons in electronvolts. Take h=6.63×10−34 J sh = 6.63\times10^{-34}\,\text{J}\,\text{s}h=6.63×10−34Js, c=3.00×108 m s−1c = 3.00\times10^{8}\,\text{m}\,\text{s}^{-1}c=3.00×108ms−1, 1 eV=1.60×10−19 J1\,\text{eV} = 1.60\times10^{-19}\,\text{J}1eV=1.60×10−19J.

  1. 01Photon energy

    E=hc/λ=(6.63×10−34×3.00×108)/(250×10−9)=7.96×10−19 JE = hc/\lambda = (6.63\times10^{-34} \times 3.00\times10^{8})/(250\times10^{-9}) = 7.96\times10^{-19}\,\text{J}E=hc/λ=(6.63×10−34×3.00×108)/(250×10−9)=7.96×10−19J.

  2. 02Convert to eV

    7.96×10−19/1.60×10−19=4.97 eV7.96\times10^{-19}/1.60\times10^{-19} = 4.97\,\text{eV}7.96×10−19/1.60×10−19=4.97eV.

  3. 03Apply the photoelectric equation

    Ek,max⁡=hf−ϕ=4.97−3.0E_{k,\max} = hf - \phi = 4.97 - 3.0Ek,max​=hf−ϕ=4.97−3.0.

    Ek,max⁡=4.97−3.0=1.97 eVE_{k,\max} = 4.97 - 3.0 = 1.97\,\text{eV}Ek,max​=4.97−3.0=1.97eV

Result: The fastest electrons leave with about 2.0 eV2.0\,\text{eV}2.0eV (equivalently 3.2×10−19 J3.2\times10^{-19}\,\text{J}3.2×10−19J).

Exam focus

  • Use hf=ϕ+Ek,max⁡hf = \phi + E_{k,\max}hf=ϕ+Ek,max​ to find work function, threshold frequency or maximum kinetic energy, and interpret an Ek,max⁡E_{k,\max}Ek,max​-against-fff graph (gradient =h= h=h).
  • Calculate the wavelength of a photon emitted or absorbed in a transition between two stated energy levels.

Typical mistakes

  • Thinking a brighter source gives faster photoelectrons - intensity changes the number of electrons, not their maximum kinetic energy.
  • Mixing joules and electronvolts: energy-level differences in eV must be converted before using E=hfE = hfE=hf with hhh in J s\text{J}\,\text{s}Js.

Active revision

A metal has work function 2.0 eV2.0\,\text{eV}2.0eV. Light of wavelength 400 nm400\,\text{nm}400nm falls on it. Calculate the maximum kinetic energy of the emitted electrons in joules.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA) · GCE AS and A level subject content for the sciences (Department for Education)

§ 06

Wave-particle duality and the de Broglie wavelength#

●●●AdvancedLPAQA 7408 3.2.2.4LPDfE GCE Physics - wave-particle duality

de Broglie wavelength against momentum

lambda = h / pGraph of lambda = h/p, decreasing, on the interval x from 0.3 to 30.511.522.530.511.522.5p = 1 -> 0.66 nmlambda = h/pWavelength / nmMomentum p / 10−24 kg m s−1
Fig. 5The de Broglie wavelength is inversely proportional to momentum: doubling the momentum halves the wavelength. Here the electron momentum is in units of 10−24 kg m s−110^{-24}\,\text{kg}\,\text{m}\,\text{s}^{-1}10−24kgms−1.

Key points

The photoelectric effect shows light behaving as particles, yet interference and diffraction show it behaving as a wave: light has a dual nature and shows whichever behaviour the experiment probes. In 1924 de Broglie proposed that this duality is universal - that particles of matter also have an associated wavelength, given by λ=h/p\lambda = h/pλ=h/p, where p=mvp = mvp=mv is the momentum. The more massive or faster the particle, the shorter its wavelength.
The prediction was confirmed by electron diffraction: a beam of electrons passed through a thin crystal or graphite film produces rings, exactly as X-rays of the same wavelength would. Because the ring spacing depends on the de Broglie wavelength, increasing the accelerating voltage raises the electrons' momentum, shortens their wavelength and shrinks the rings - a direct, quantitative demonstration that matter diffracts.
The wavelengths involved are astonishingly small. An electron accelerated through a few hundred volts has a wavelength of the order of 10−10 m10^{-10}\,\text{m}10−10m, comparable to atomic spacings, which is why crystals act as diffraction gratings for electrons and why the electron microscope - using these tiny wavelengths - resolves far finer detail than any light microscope. Everyday objects have utterly negligible wavelengths, which is why we never see a cricket ball diffract.
Duality reshapes how we picture the quantum world: an electron is neither a tiny billiard ball nor a spread-out wave, but an entity whose wave nature governs where it is likely to be found and whose particle nature governs how it is detected. The two aspects are complementary - a single experiment reveals one or the other, never both at once.
λ=hp=hmv\lambda = \dfrac{h}{p} = \dfrac{h}{mv}λ=ph​=mvh​

de Broglie wavelength

The wavelength associated with a particle of momentum ppp.

p=2mEkp = \sqrt{2 m E_k}p=2mEk​​

Momentum from kinetic energy

For an electron accelerated through a pd VVV, Ek=eVE_k = eVEk​=eV.

Worked example

de Broglie wavelength of an accelerated electron

An electron is accelerated from rest through a potential difference of 100 V100\,\text{V}100V. Find its de Broglie wavelength. Take me=9.11×10−31 kgm_e = 9.11\times10^{-31}\,\text{kg}me​=9.11×10−31kg, e=1.60×10−19 Ce = 1.60\times10^{-19}\,\text{C}e=1.60×10−19C, h=6.63×10−34 J sh = 6.63\times10^{-34}\,\text{J}\,\text{s}h=6.63×10−34Js.

  1. 01Kinetic energy gained

    Ek=eV=1.60×10−19×100=1.60×10−17 JE_k = eV = 1.60\times10^{-19} \times 100 = 1.60\times10^{-17}\,\text{J}Ek​=eV=1.60×10−19×100=1.60×10−17J.

  2. 02Momentum

    p=2mEk=2×9.11×10−31×1.60×10−17p = \sqrt{2 m E_k} = \sqrt{2 \times 9.11\times10^{-31} \times 1.60\times10^{-17}}p=2mEk​​=2×9.11×10−31×1.60×10−17​.

    p=2.915×10−47=5.40×10−24 kg m s−1p = \sqrt{2.915\times10^{-47}} = 5.40\times10^{-24}\,\text{kg}\,\text{m}\,\text{s}^{-1}p=2.915×10−47​=5.40×10−24kgms−1
  3. 03Wavelength

    λ=h/p=6.63×10−34/5.40×10−24=1.23×10−10 m\lambda = h/p = 6.63\times10^{-34}/5.40\times10^{-24} = 1.23\times10^{-10}\,\text{m}λ=h/p=6.63×10−34/5.40×10−24=1.23×10−10m.

Result: The electron has a de Broglie wavelength of 1.23×10−10 m1.23\times10^{-10}\,\text{m}1.23×10−10m, comparable to atomic spacings - hence electron diffraction.

Exam focus

  • Calculate the de Broglie wavelength of a particle from its momentum, or from an accelerating voltage via Ek=eVE_k = eVEk​=eV and p=2mEkp = \sqrt{2mE_k}p=2mEk​​.
  • Explain how electron diffraction provides evidence for the wave nature of matter and how changing the accelerating voltage changes the pattern.

Typical mistakes

  • Using energy in place of momentum in λ=h/p\lambda = h/pλ=h/p, or forgetting that p=2mEkp = \sqrt{2 m E_k}p=2mEk​​ links kinetic energy to momentum.
  • Thinking a particle is a wave and a particle simultaneously - it shows one behaviour or the other depending on the experiment.

Active revision

An electron is accelerated from rest through 200 V200\,\text{V}200V. Calculate its de Broglie wavelength. Take me=9.11×10−31 kgm_e = 9.11\times10^{-31}\,\text{kg}me​=9.11×10−31kg, e=1.60×10−19 Ce = 1.60\times10^{-19}\,\text{C}e=1.60×10−19C, h=6.63×10−34 J sh = 6.63\times10^{-34}\,\text{J}\,\text{s}h=6.63×10−34Js.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Contents

Section -- / 06

    • 01The nuclear atom, specific charge and isotopes○
    • 02Stability, radioactive decay and the neutrino◐
    • 03Antiparticles, photons, annihilation and pair production◐
    • 04Classification, quarks and conservation laws●
    • 05The photoelectric effect and atomic energy levels●
    • 06Wave-particle duality and the de Broglie wavelength●

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References & sources

Sources

Department for Education

  • GCE AS and A level subject content for the sciences

AQA

  • AQA A-level Physics 7408 specification

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