What are the fundamental forces involved in particle interactions?

In the context of Edexcel International A Levels Physics, the fundamental forces involved in particle interactions are the gravitational force, electromagnetic force, strong nuclear force, and weak nuclear force. These forces govern how particles interact with each other, with each force having its own range and strength. Understanding these forces is crucial for analyzing particle interactions and conservation laws.

How does the conservation of energy apply to particle interactions?

In particle interactions, the conservation of energy principle states that the total energy before the interaction must equal the total energy after the interaction. This includes kinetic energy, potential energy, and rest mass energy. For Edexcel International A Levels students, it's important to recognize that energy conservation is a fundamental concept that applies universally, ensuring that energy is neither created nor destroyed during particle interactions.

What is the role of conservation of momentum in particle interactions?

The conservation of momentum is a key principle in particle interactions, stating that the total momentum of a closed system remains constant if no external forces act on it. For students studying Further Mechanics, Fields and Particles in Edexcel International A Levels, this means that the vector sum of the momenta of all particles before an interaction must equal the vector sum after the interaction. This principle helps in predicting the outcomes of collisions and other interactions.

Can you explain the concept of conservation of charge in particle interactions?

Conservation of charge is a fundamental principle stating that the total electric charge in an isolated system remains constant over time. In the context of particle interactions, this means that the sum of charges before and after an interaction must be equal. For Edexcel International A Levels Physics students, understanding this concept is crucial as it applies to all particle interactions, ensuring that charge is neither created nor destroyed.

What is the significance of conservation laws in understanding particle interactions?

Conservation laws, including those of energy, momentum, and charge, are vital for understanding particle interactions as they provide constraints that must be satisfied in any physical process. For Edexcel International A Levels students, these laws are essential tools for analyzing and predicting the outcomes of interactions in Further Mechanics, Fields and Particles. They help simplify complex problems by reducing the number of unknowns and ensuring that fundamental principles are upheld.

Why is "Getting the baryon-number or lepton-number bookkeeping wrong when checking an interaction" a common mistake in Particle Interactions and Conservation?

Why it happens: Students forget that mesons have baryon number 0, that antiparticles carry the OPPOSITE sign (e.g. an antineutrino has L = -1), or they mix up which particles are leptons. How to avoid it: Tabulate B and L explicitly for every particle: baryons B = +1, antibaryons -1, mesons and leptons B = 0; leptons L = +1, antileptons L = -1, hadrons L = 0. Sum each column before and after and compare. Source: Edexcel IAL Physics Unit 4 examiner reports

Why is "Saying annihilation produces a single photon (ignoring momentum conservation)" a common mistake in Particle Interactions and Conservation?

Why it happens: Students focus only on energy (E = 2mc^2) and forget that momentum must also be conserved. How to avoid it: Annihilation of a particle–antiparticle pair at rest produces **two** photons travelling in **opposite directions**, so the total momentum stays zero. A lone photon always carries momentum, so it cannot be the sole product. Source: Edexcel IAL Physics Unit 4 examiner reports

Why is "Using $mc^2$ instead of $2mc^2$ for the minimum photon energy in pair production" a common mistake in Particle Interactions and Conservation?

Why it happens: Students calculate the rest energy of only one particle, forgetting that a pair (particle AND antiparticle) is created. How to avoid it: Pair production creates **two** particles of equal mass, so the minimum photon energy is E_min = 2mc^2 (e.g. 1.022\ MeV for an electron–positron pair), not mc^2. Source: Edexcel IAL Physics Unit 4 examiner reports

Why is "Misremembering quark charges or the proton/neutron quark composition" a common mistake in Particle Interactions and Conservation?

Why it happens: The fractional charges (+23, -13) and which of uud/udd is the proton are easy to swap. How to avoid it: Learn that up = +23e, down = -13e; the **proton is uud** (charge +1) and the **neutron is udd** (charge 0). Check by adding the quark charges — a proton MUST come out to +1. Source: Edexcel IAL Physics Unit 4 examiner reports

Why is "Unit slips in $\Delta E = c^2\Delta m$ — leaving mass in u, or forgetting to square $c$" a common mistake in Particle Interactions and Conservation?

Why it happens: Masses are often quoted in u or MeV/c^2, and c is large so squaring is easy to overlook. How to avoid it: Convert m to **kilograms** first (1\ u = 1.66 × 10^-27\ kg), and always use c^2 = 9.00 × 10^16. To convert the resulting joules to MeV, divide by 1.60 × 10^-13 (NOT 1.60 × 10^-19). Source: Edexcel IAL Physics Unit 4 examiner reports

Why is "Emitting a neutrino instead of an antineutrino in beta-minus decay" a common mistake in Particle Interactions and Conservation?

Why it happens: Students remember 'a neutral, near-massless particle is emitted' but not that lepton number fixes whether it is matter or antimatter. How to avoid it: In ^- decay an **electron** (L = +1) is emitted, so an **antineutrino** (L = -1) must accompany it to keep total lepton number at 0. Always finish by checking L before and after. Source: Edexcel IAL Physics Unit 4 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Particle Interactions and Conservation

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Short Notes - Particle Interactions and Conservation

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Detailed Study Notes

Detailed notes on Further Mechanics, Fields and Particles for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Particle Interactions and Conservation — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 4)

Mass–energy equivalence $\Delta E = c^2\Delta m$ and the mass units u and MeV/ $c^2$ ; antimatter, annihilation and pair production; the Standard Model — quarks, leptons, and the hadrons (baryons and mesons); and the conservation laws (charge, baryon number, lepton number) that decide whether an interaction is allowed, including beta-minus decay described at the quark level.

At a glance

Mass and energy are equivalent: $\Delta E = c^2\Delta m$ . A small mass change gives a large energy because $c^2 = 9.00\times10^{16}$ . Always put $\Delta m$ in kg.
Mass units for particles: $1\ \text{u} = 1.66\times10^{-27}\ \text{kg} = 931.5\ \text{MeV}/c^2$ ; the electron is $0.511\ \text{MeV}/c^2$ .
Every particle has an antiparticle — same mass, opposite charge. The positron is the anti-electron.
Annihilation: particle + antiparticle $\rightarrow$ two photons; total energy $= 2mc^2$ (two photons conserve momentum). Pair production: photon $\rightarrow$ particle + antiparticle; minimum photon energy $= 2mc^2$ .
Standard Model matter particles = quarks + leptons. Quarks: up $+\tfrac{2}{3}e$ , down and strange $-\tfrac{1}{3}e$ . Leptons: electron, muon, neutrinos (and antiparticles).
Hadrons are made of quarks: baryons = 3 quarks (proton uud, neutron udd); mesons = quark–antiquark pair.
Conservation laws for any interaction: charge $Q$ , baryon number $B$ , lepton number $L$ (and energy/momentum) must each be equal before and after. One violation ⇒ forbidden.
Beta-minus decay at quark level: $\text{d} \rightarrow \text{u}$ (neutron udd → proton uud), emitting a $\text{W}^-$ boson that becomes $\text{e}^- + \bar{\nu}_e$ ; the antineutrino keeps $L = 0$ .

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

Use $\Delta E = c^2\Delta m$ to calculate the energy released in an interaction, and use the atomic mass unit u and MeV/ $c^2$ as units of mass.
Describe antimatter: every particle has an antiparticle of the same mass and opposite charge; identify the positron as the anti-electron.
Explain annihilation (particle + antiparticle → two photons, energy $2mc^2$ ) and pair production (photon → particle + antiparticle, minimum energy $2mc^2$ ), with contexts such as PET scanning.
State the Standard Model classification: quarks (up, down, strange and antiquarks with charges $+\tfrac{2}{3}e$ , $-\tfrac{1}{3}e$ ), leptons (electron, muon, neutrinos), and hadrons as baryons (3 quarks) or mesons (quark–antiquark). Identify force-carrying particles (photon, W/Z bosons) qualitatively.
Apply conservation of charge, baryon number and lepton number (with energy and momentum) to determine whether a proposed interaction is allowed.
Describe beta-minus decay in quark terms ( $\text{d} \rightarrow \text{u}$ , emission of a $\text{W}^-$ boson producing an electron and an electron antineutrino) and interpret simple particle-interaction equations.

Mass–energy equivalence and particle mass units

Mass is a form of energy: $\Delta E = c^2\Delta m$ . Learn the u and MeV/ $c^2$ units.

Einstein's insight is that mass and energy are two forms of the same thing. A change in mass $\Delta m$ is equivalent to an energy

$\Delta E = c^2\,\Delta m$

with $c = 3.00\times10^{8}\ \text{m s}^{-1}$ , so $c^2 = 9.00\times10^{16}\ \text{m}^2\text{s}^{-2}$ . Because $c^2$ is so large, an almost imperceptible mass change corresponds to an enormous energy — this is why nuclear and particle processes release so much energy while the mass change is tiny.

Two special mass units are used for atoms and particles:

The atomic mass unit, $1\ \text{u} = 1.66\times10^{-27}\ \text{kg}$ (one-twelfth of a carbon-12 atom's mass). Using $\Delta E = c^2\Delta m$ , $1\ \text{u}$ is equivalent to $931.5\ \text{MeV}/c^2$ .
The MeV/ $c^2$ : since $E = mc^2$ , dividing a particle's rest energy (in MeV) by $c^2$ gives its mass. The electron has a rest energy of $0.511\ \text{MeV}$ , so its mass is $0.511\ \text{MeV}/c^2$ .

Working method for energy released. (1) Convert the mass change to kilograms. (2) Multiply by $c^2$ to get the energy in joules. (3) If a MeV answer is wanted, divide by $1\ \text{MeV} = 1.60\times10^{-13}\ \text{J}$ — note this is the MeV conversion, not the eV one ( $1.60\times10^{-19}\ \text{J}$ ).

For example, a mass change of $3.00\times10^{-3}\ \text{u}$ is $4.98\times10^{-30}\ \text{kg}$ , giving $\Delta E = 9.00\times10^{16}\times4.98\times10^{-30} = 4.48\times10^{-13}\ \text{J} = 2.80\ \text{MeV}$ — which matches the shortcut $3.00\times10^{-3}\times931.5\ \text{MeV} = 2.79\ \text{MeV}$ .

$\Delta E = c^2\Delta m$ ; put $\Delta m$ in kg and use $c^2 = 9.00\times10^{16}$ .
$1\ \text{u} = 1.66\times10^{-27}\ \text{kg} = 931.5\ \text{MeV}/c^2$ .
Electron mass $= 0.511\ \text{MeV}/c^2$ (rest energy $0.511\ \text{MeV}$ ).
Joules → MeV: divide by $1.60\times10^{-13}$ (not $1.60\times10^{-19}$ ).

See the full worked example for particle interactions and conservation →

Antimatter, annihilation and pair production

Every particle has an antiparticle. Matter + antimatter → 2 photons; photon → matter + antimatter (min $2mc^2$ ).

Every particle has an antiparticle with the same mass but opposite charge (and opposite baryon/lepton number). The antiparticle of the electron is the positron — same mass, charge $+1$ . Antiprotons, antineutrons and antineutrinos exist too.

Annihilation. When a particle meets its antiparticle they are converted entirely into energy, producing two photons (gamma rays). If they meet at rest the total momentum is zero, so the two photons must fly off in opposite directions with equal momenta — that is why two photons are produced, not one (a single photon always carries momentum). The total photon energy equals the rest energy of both particles:

$E_{\text{total}} = 2mc^2,\qquad \text{each photon } = mc^2.$

For the electron–positron case each photon carries $0.511\ \text{MeV}$ .

Pair production. The reverse process: a single photon is converted into a particle–antiparticle pair (e.g. an electron and a positron). To create both particles the photon must supply at least their combined rest energy, so the minimum photon energy is

$E_{\text{min}} = 2mc^2 = 1.022\ \text{MeV}\ \text{for an electron–positron pair.}$

Annihilation (left to right): an electron and positron meet and become two photons flying apart in opposite directions, total energy 2mc². Pair production (right to left): a photon of at least 2mc² becomes a particle–antiparticle pair.

Real context — PET scanning. In positron emission tomography a tracer emits positrons. Each positron quickly meets an electron in the body's tissue and they annihilate, producing two $0.511\ \text{MeV}$ gamma photons that travel in opposite directions. A ring of detectors registers pairs of photons arriving on opposite sides simultaneously, and from many such events the scanner builds up an image of where the tracer is concentrated.

Antiparticle: same mass, opposite charge (positron = anti-electron).
Annihilation: matter + antimatter → 2 photons; total energy $= 2mc^2$ ; two photons conserve momentum.
Pair production: photon → matter + antimatter; minimum photon energy $= 2mc^2$ ( $1.022\ \text{MeV}$ for $e^-e^+$ ).
PET scan: positron annihilates with tissue electron → two opposite $0.511\ \text{MeV}$ gammas.

See the full worked example for particle interactions and conservation →

The Standard Model — quarks, leptons and hadrons

Matter = quarks + leptons. Hadrons are quark-built: baryons (3 quarks), mesons (quark–antiquark).

The Standard Model sorts all matter particles into two fundamental families — quarks and leptons — together with the force-carrying bosons that transmit the interactions between them.

Quarks are fundamental particles with fractional charge:

Quark	Charge
up (u)	$+\tfrac{2}{3}e$
down (d)	$-\tfrac{1}{3}e$
strange (s)	$-\tfrac{1}{3}e$

Each has an antiquark ( $\bar{\text{u}}$ , $\bar{\text{d}}$ , $\bar{\text{s}}$ ) with the opposite charge. Quarks are never found alone — they always combine to form hadrons.

Leptons are fundamental particles that are not made of quarks: the electron, the muon, and their associated neutrinos (plus all their antiparticles). Leptons carry lepton number and do not feel the strong interaction.

Hadrons (particles built from quarks) come in two types:

Baryons — made of three quarks. The proton is uud and the neutron is udd. Baryons have baryon number $+1$ .
Mesons — made of a quark–antiquark pair. The $\pi^+$ meson is $\text{u}\bar{\text{d}}$ . Mesons have baryon number $0$ .

A proton (uud) has charge (+2/3) + (+2/3) + (−1/3) = +1; a neutron (udd) has charge (+2/3) + (−1/3) + (−1/3) = 0. Both are baryons — three quarks each.

Force-carrying (exchange) particles transmit the fundamental interactions. You need to recognise these qualitatively: the photon carries the electromagnetic force, and the W and Z bosons carry the weak interaction. (The gluon carries the strong force between quarks.)

Matter particles = quarks + leptons; forces carried by bosons (photon, W/Z).
Quark charges: up $+\tfrac{2}{3}e$ , down and strange $-\tfrac{1}{3}e$ ; antiquarks opposite.
Baryon = 3 quarks (proton uud, neutron udd); meson = quark–antiquark.
Leptons (electron, muon, neutrinos) are fundamental — NOT made of quarks.

See the full worked example for particle interactions and conservation →

Conservation laws — deciding if an interaction is allowed

Charge, baryon number and lepton number must each balance before and after. One failure ⇒ forbidden.

In every particle interaction, several quantities must be conserved — they are the same before and after. As well as energy and momentum, you must check three quantum numbers:

Charge $Q$ — the total charge (in units of $e$ ) is unchanged.
Baryon number $B$ — $+1$ for each baryon, $-1$ for each antibaryon, $0$ for mesons and leptons.
Lepton number $L$ — $+1$ for each lepton, $-1$ for each antilepton, $0$ for all hadrons.

Method. Tabulate $Q$ , $B$ and $L$ for every particle, sum each column on both sides, and compare. If all three (plus energy and momentum) balance, the interaction is allowed; if any one fails, it is forbidden — a single violation is enough to rule it out.

Worked check — is $\text{p} + \text{n} \rightarrow \text{p} + \text{p} + \pi^-$ allowed? (The $\pi^-$ has $Q=-1$ , $B=0$ , $L=0$ .)

Quantity	Before	After	Conserved?
Charge $Q$	$(+1)+0 = +1$	$(+1)+(+1)+(-1) = +1$	✓
Baryon $B$	$1+1 = 2$	$1+1+0 = 2$	✓
Lepton $L$	$0$	$0$	✓

All three balance, so the interaction is allowed (given enough energy). Contrast this with the forbidden neutron decay $\text{n}\rightarrow\text{p}+\text{e}^-+\nu_e$ : charge and baryon number balance, but the neutrino ( $L=+1$ ) alongside the electron ( $L=+1$ ) gives $L=+2\neq0$ , so lepton number is violated and the decay cannot occur — the real decay emits an antineutrino.

Always check charge, baryon number AND lepton number (plus energy/momentum).
$B$ : baryon $+1$ , antibaryon $-1$ , mesons and leptons $0$ .
$L$ : lepton $+1$ , antilepton $-1$ , hadrons $0$ .
One violated law ⇒ interaction forbidden; all conserved ⇒ allowed.

See the full worked example for particle interactions and conservation →

Beta-minus decay at the quark level

Neutron → proton means one down quark turns into an up quark, via a $\text{W}^-$ boson → electron + antineutrino.

Beta-minus ( $\beta^-$ ) decay is written

$\text{n} \rightarrow \text{p} + \text{e}^- + \bar{\nu}_e.$

Looking inside the nucleons explains it. A neutron is udd and a proton is uud, so exactly one down quark changes into an up quark:

$\text{d} \rightarrow \text{u} + \text{e}^- + \bar{\nu}_e.$

This is a weak interaction: the down quark emits a $\text{W}^-$ boson (becoming an up quark), and the $\text{W}^-$ then decays into the electron and the electron antineutrino.

In beta-minus decay a down quark becomes an up quark (so udd → uud, neutron → proton), emitting a W⁻ boson that decays into an electron (L = +1) and an electron antineutrino (L = −1). Total lepton number stays zero.

Checking the conservation laws confirms the decay is allowed:

Charge: $0 \rightarrow (+1)+(-1)+0 = 0$ ✓
Baryon number: $1 \rightarrow 1+0+0 = 1$ ✓ (the electron and antineutrino are leptons, $B = 0$ )
Lepton number: $0 \rightarrow (+1)+(-1) = 0$ ✓

The lepton-number check is the key: the electron carries $L = +1$ , so an antineutrino ( $L = -1$ ) — not a neutrino — must be emitted to keep the total at zero. This is why $\beta^-$ decay produces an antineutrino.

$\beta^-$ decay: $\text{n}\rightarrow\text{p}+\text{e}^-+\bar{\nu}_e$ ; at quark level $\text{d}\rightarrow\text{u}$ .
Weak interaction: down quark emits a $\text{W}^-$ boson → $\text{e}^- + \bar{\nu}_e$ .
Charge, baryon number and lepton number all conserved.
An antineutrino ( $L=-1$ ) is needed to balance the electron's $L=+1$ .

See the full worked example for particle interactions and conservation →

Quick recap

Mass and energy are equivalent: $\Delta E = c^2\Delta m$ (put $\Delta m$ in kg, square $c$ ); $1\ \text{u} = 1.66\times10^{-27}\ \text{kg} = 931.5\ \text{MeV}/c^2$ ; electron $= 0.511\ \text{MeV}/c^2$ .
Every particle has an antiparticle of equal mass and opposite charge; the positron is the anti-electron.
Annihilation: matter + antimatter → two photons, total energy $2mc^2$ (two photons conserve momentum). Pair production: photon → matter + antimatter, minimum energy $2mc^2$ ( $1.022\ \text{MeV}$ for $e^-e^+$ ).
Standard Model matter = quarks (u $+\tfrac{2}{3}e$ , d and s $-\tfrac{1}{3}e$ ) + leptons (electron, muon, neutrinos); hadrons are baryons (3 quarks: proton uud, neutron udd) or mesons (quark–antiquark).
An interaction is allowed only if charge, baryon number and lepton number (plus energy and momentum) are each conserved; one violation forbids it.
Beta-minus decay: $\text{d}\rightarrow\text{u}$ (neutron → proton) via a $\text{W}^-$ boson → electron + electron antineutrino; the antineutrino keeps lepton number zero.
Convert joules to MeV by dividing by $1.60\times10^{-13}$ ; quote answers to 3 s.f. with units.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$\Delta E = c^2\Delta m$ — mass–energy equivalence; $\Delta m$ in kg, $c^2 = 9.00\times10^{16}$ .
$1\ \text{u} = 1.66\times10^{-27}\ \text{kg} = 931.5\ \text{MeV}/c^2$ ; electron mass $= 0.511\ \text{MeV}/c^2$ .
Annihilation → 2 photons, energy $= 2mc^2$ . Pair production: photon minimum energy $= 2mc^2$ .
Quark charges: up $= +\tfrac{2}{3}e$ ; down, strange $= -\tfrac{1}{3}e$ ; antiquarks opposite.
Proton = uud (charge $+1$ ); neutron = udd (charge $0$ ); baryon = 3 quarks, meson = quark–antiquark.
Conserved in every interaction: charge $Q$ , baryon number $B$ , lepton number $L$ (plus energy and momentum).
$B$ : baryon $+1$ , antibaryon $-1$ , mesons/leptons $0$ . $L$ : lepton $+1$ , antilepton $-1$ , hadrons $0$ .
$\beta^-$ decay: $\text{n}\rightarrow\text{p}+\text{e}^-+\bar{\nu}_e$ ; $\text{d}\rightarrow\text{u}$ via $\text{W}^-$ ; antineutrino keeps $L=0$ .

How it’s examined

Particle physics is examined on the Unit 4 paper (WPH14, Further Mechanics, Fields and Particles), a structured paper with short calculations, definitions and applied reasoning. Typical formats: define antiparticle / annihilation / pair production or classify a particle (1–2 marks, AO1); a mass–energy calculation using $\Delta E = c^2\Delta m$ with u → kg and J → MeV conversions (2–4 marks, M and A marks); an annihilation or pair-production energy calculation ( $2mc^2$ , often combined with $E = hf$ ) (3–4 marks); and a 'determine whether this interaction is allowed / explain why this decay cannot occur' question requiring you to tabulate and check charge, baryon number and lepton number (4–6 marks, AO3). Beta decay at quark level ( $\text{d}\rightarrow\text{u}$ , $\text{W}^-$ boson, antineutrino for lepton conservation) is a recurring 4–6 mark item. ALWAYS show the before/after tally for each conservation law — a bare verdict scores nothing — and keep full precision, converting to kilograms before using $\Delta E = c^2\Delta m$ , and quoting answers to 3 s.f. with units.

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 4 (WPH14) examiner reports; Edexcel IAL Physics Unit 4 sample assessment materials and mark schemes. Last reviewed 2026-07-05.

Take this whole topic with you

Step-by-step worked examples — Particle Interactions and Conservation

Step-by-step solutions to past-paper-style questions on particle interactions and conservation, written exactly the way a tutor would explain them at the board.

Question type:

1Energy equivalent of the electron mass (2 marks)
Getting startedDirect calculation• mass–energy equivalence, AO2, Unit 4
Question
The mass of an electron is $9.11 \times 10^{-31}\ \text{kg}$ . Calculate the energy, in joules, equivalent to this mass. Use $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ . (2 marks)
Step-by-step solution
1. Step 1
  Use Einstein's mass–energy relation. The energy equivalent to a mass $m$ is $E = mc^2$ .
  $E = mc^2$
2. Step 2
  Substitute the electron mass and $c$ , then evaluate to 3 significant figures.
  $E = (9.11 \times 10^{-31}) \times (3.00 \times 10^{8})^2 = 8.20 \times 10^{-14}\ \text{J}$
Answer
$E = 8.20 \times 10^{-14}\ \text{J}$ .
Examiner tip
Mark scheme: (1) use of $E = mc^2$ ; (2) $8.20 \times 10^{-14}\ \text{J}$ to 3 s.f. A very common slip is forgetting to square $c$ (giving $2.7 \times 10^{-22}$ ) — the $c^2$ is what makes even a tiny mass equate to a large energy.
2Converting an electron's energy to MeV (2 marks)
Getting startedDirect calculation• MeV, unit conversion, AO2, Unit 4
Question
The energy equivalent of an electron's mass is $8.20 \times 10^{-14}\ \text{J}$ . Express this energy in MeV, given that $1\ \text{MeV} = 1.60 \times 10^{-13}\ \text{J}$ . (2 marks)
Step-by-step solution
1. Step 1
  Convert from joules to MeV by dividing the energy in joules by the number of joules in 1 MeV.
  $E\ (\text{MeV}) = \frac{E\ (\text{J})}{1.60 \times 10^{-13}}$
2. Step 2
  Substitute and evaluate to 3 significant figures.
  $E = \frac{8.20 \times 10^{-14}}{1.60 \times 10^{-13}} = 0.512\ \text{MeV} \approx 0.511\ \text{MeV}$
Answer
$E \approx 0.511\ \text{MeV}$ — this is the well-known rest energy of the electron (so the electron mass is $0.511\ \text{MeV}/c^2$ ).
Examiner tip
Mark scheme: (1) divide by $1.60 \times 10^{-13}$ ; (2) $0.511\ \text{MeV}$ (accept $0.512$ ). Dividing by $1.60 \times 10^{-19}$ (the electron charge, for eV not MeV) is the classic error — use the MeV conversion, not the eV one.
3Classifying particles: hadron, baryon, meson or lepton (3 marks)
Getting startedIdentify & classify• standard model, classification, AO1, Unit 4
Question
Classify each of the following: (a) a proton (quark content uud); (b) a $\pi^+$ meson (quark content $\text{u}\bar{\text{d}}$ ); (c) an electron. For each, say whether it is a hadron or a lepton, and if a hadron whether it is a baryon or a meson. (3 marks)
Step-by-step solution
1. Step 1
  A hadron is made of quarks. A baryon contains three quarks; a meson contains a quark–antiquark pair. A lepton is a fundamental particle that is NOT made of quarks (e.g. the electron).
2. Step 2
  (a) The proton (uud) has three quarks, so it is a hadron and specifically a baryon.
3. Step 3
  (b) The $\pi^+$ ( $\text{u}\bar{\text{d}}$ ) is a quark–antiquark pair, so it is a hadron and specifically a meson. (c) The electron is not made of quarks, so it is a lepton.
Answer
(a) proton = hadron, baryon (three quarks); (b) $\pi^+$ = hadron, meson (quark–antiquark); (c) electron = lepton (fundamental, no quarks).
Examiner tip
Mark scheme: (1) proton baryon; (2) $\pi^+$ meson; (3) electron lepton. Remember baryon = 3 quarks, meson = quark + antiquark, and leptons (electron, muon, neutrinos) are fundamental — calling the electron a 'baryon' or a 'hadron' loses the mark.
4Photon energy from electron–positron annihilation (4 marks)
Building confidenceMulti-step problem• annihilation, antimatter, AO2, Unit 4
Question
An electron and a positron, each of mass $9.11 \times 10^{-31}\ \text{kg}$ , are effectively at rest when they annihilate, producing two identical photons. Calculate the energy of each photon in joules. Use $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ . (4 marks)
Step-by-step solution
1. Step 1
  In annihilation the total energy released equals the rest energy of both particles: $E_{total} = 2mc^2$ (the electron and the positron have equal mass).
  $E_{total} = 2mc^2$
2. Step 2
  Substitute the mass and $c$ .
  $E_{total} = 2 \times (9.11 \times 10^{-31}) \times (3.00 \times 10^{8})^2 = 1.64 \times 10^{-13}\ \text{J}$
3. Step 3
  Two identical photons are produced (required so that total momentum stays zero — they travel in opposite directions). The energy is shared equally.
  $E_{photon} = \frac{E_{total}}{2}$
4. Step 4
  Evaluate the energy of each photon to 3 significant figures.
  $E_{photon} = \frac{1.64 \times 10^{-13}}{2} = 8.20 \times 10^{-14}\ \text{J}$
Answer
Each photon has energy $8.20 \times 10^{-14}\ \text{J}$ (equivalently $0.511\ \text{MeV}$ ) — so each photon carries the rest energy of one particle.
Examiner tip
Mark scheme: (1) total energy $= 2mc^2$ ; (2) $1.64 \times 10^{-13}\ \text{J}$ ; (3) two photons so halve; (4) $8.20 \times 10^{-14}\ \text{J}$ . Two photons (not one) are required to conserve momentum — a single photon cannot be produced from two particles at rest, since a lone photon must carry momentum.
5Minimum photon energy for pair production (4 marks)
Building confidenceMulti-step problem• pair production, antimatter, AO2, Unit 4
Question
A single photon creates an electron–positron pair (pair production). The electron rest energy is $0.511\ \text{MeV}$ . (a) Calculate the minimum energy of the photon, in MeV. (b) Calculate the corresponding minimum frequency, using $h = 6.63 \times 10^{-34}\ \text{J s}$ and $1\ \text{MeV} = 1.60 \times 10^{-13}\ \text{J}$ . (4 marks)
Step-by-step solution
1. Step 1
  (a) To create both particles the photon must supply at least the rest energy of the electron AND the positron. Their masses are equal, so the minimum energy is $2mc^2$ .
  $E_{min} = 2mc^2 = 2 \times 0.511 = 1.022\ \text{MeV}$
2. Step 2
  (b) Convert this minimum energy into joules ready for the photon-energy equation.
  $E_{min} = 1.022 \times 1.60 \times 10^{-13} = 1.635 \times 10^{-13}\ \text{J}$
3. Step 3
  A photon's energy is $E = hf$ . Rearrange for the frequency $f$ .
  $f = \frac{E_{min}}{h}$
4. Step 4
  Substitute and evaluate to 3 significant figures.
  $f = \frac{1.635 \times 10^{-13}}{6.63 \times 10^{-34}} = 2.47 \times 10^{20}\ \text{Hz}$
Answer
(a) $E_{min} = 1.022\ \text{MeV}$ ( $= 2mc^2$ ). (b) $f = 2.47 \times 10^{20}\ \text{Hz}$ (a gamma-ray photon).
Examiner tip
Mark scheme: (a) 1 mark — $E_{min} = 2 \times 0.511 = 1.022\ \text{MeV}$ . (b) 3 marks — convert to J; use $E = hf$ ; $f = 2.47 \times 10^{20}\ \text{Hz}$ . Using $0.511\ \text{MeV}$ (one particle) instead of $1.022\ \text{MeV}$ is the standard error — pair production makes TWO particles, so you need $2mc^2$ .
6Is a proposed interaction allowed? (5 marks)
Building confidenceShow that / prove• conservation laws, charge, baryon number, AO3, Unit 4
Question
A student proposes the interaction $\text{p} + \text{n} \rightarrow \text{p} + \text{p} + \pi^-$ , where the $\pi^-$ meson has charge $-1$ , baryon number $0$ and lepton number $0$ . By checking the relevant conservation laws, determine whether this interaction is allowed. (5 marks)
Step-by-step solution
1. Step 1
  List the quantities to check: charge $Q$ , baryon number $B$ and lepton number $L$ must each be the same before and after.
2. Step 2
  Charge. Before: $\text{p}(+1) + \text{n}(0) = +1$ . After: $\text{p}(+1) + \text{p}(+1) + \pi^-(-1) = +1$ . Charge is conserved ( $+1 = +1$ ). ✓
  $Q:\quad (+1) + 0 = (+1) + (+1) + (-1) \Rightarrow +1 = +1$
3. Step 3
  Baryon number. Proton and neutron each have $B = +1$ ; the meson has $B = 0$ . Before: $1 + 1 = 2$ . After: $1 + 1 + 0 = 2$ . Baryon number is conserved. ✓
  $B:\quad 1 + 1 = 1 + 1 + 0 \Rightarrow 2 = 2$
4. Step 4
  Lepton number. No leptons appear on either side, so $L = 0$ before and after. Lepton number is conserved. ✓
  $L:\quad 0 = 0$
5. Step 5
  All three conservation laws (charge, baryon number, lepton number) are satisfied, so the interaction is allowed (provided enough energy is available).
Answer
Charge ( $+1 = +1$ ), baryon number ( $2 = 2$ ) and lepton number ( $0 = 0$ ) are all conserved, so the interaction $\text{p} + \text{n} \rightarrow \text{p} + \text{p} + \pi^-$ is allowed.
Examiner tip
Mark scheme: (1) state the laws to check; (2) charge conserved; (3) baryon number conserved (mesons have $B=0$ ); (4) lepton number conserved; (5) conclusion 'allowed'. You MUST show the before-and-after tally for each quantity — a bare 'it's allowed' scores nothing.
7Beta-minus decay at quark level (5 marks)
StretchMulti-step problem• beta decay, quarks, W boson, conservation laws, AO2, Unit 4
Question
In beta-minus ( $\beta^-$ ) decay a neutron changes into a proton: $\text{n} \rightarrow \text{p} + \text{e}^- + \bar{\nu}_e$ . (a) State the change in quark content that occurs. (b) Show that charge, baryon number and lepton number are all conserved. (Neutron = udd, proton = uud; the electron has $L = +1$ and the electron antineutrino $\bar{\nu}_e$ has $L = -1$ .) (5 marks)
Step-by-step solution
1. Step 1
  (a) The neutron (udd) becomes a proton (uud), so exactly one down quark changes into an up quark ( $\text{d} \rightarrow \text{u}$ ). This happens via the weak interaction, mediated by a $\text{W}^-$ boson that then decays to $\text{e}^- + \bar{\nu}_e$ .
  $\text{d} \rightarrow \text{u} + \text{W}^- \rightarrow \text{u} + \text{e}^- + \bar{\nu}_e$
2. Step 2
  (b) Charge. Before: $\text{n}(0)$ . After: $\text{p}(+1) + \text{e}^-(-1) + \bar{\nu}_e(0) = 0$ . Conserved. ✓
  $Q:\quad 0 = (+1) + (-1) + 0 \Rightarrow 0 = 0$
3. Step 3
  Baryon number. Neutron $B = +1$ ; proton $B = +1$ ; the electron and antineutrino are leptons with $B = 0$ . Conserved. ✓
  $B:\quad 1 = 1 + 0 + 0 \Rightarrow 1 = 1$
4. Step 4
  Lepton number. Before: neutron has $L = 0$ . After: $\text{e}^-(L = +1) + \bar{\nu}_e(L = -1) = 0$ . Conserved. ✓ (This is precisely WHY an antineutrino, not a neutrino, must be emitted.)
  $L:\quad 0 = 0 + (+1) + (-1) \Rightarrow 0 = 0$
5. Step 5
  All three quantities balance, confirming the decay is allowed. The quark-level change $\text{d} \rightarrow \text{u}$ underlies the neutron-to-proton conversion.
Answer
(a) One down quark turns into an up quark ( $\text{d} \rightarrow \text{u}$ ), emitting a $\text{W}^-$ boson that becomes $\text{e}^- + \bar{\nu}_e$ . (b) Charge ( $0=0$ ), baryon number ( $1=1$ ) and lepton number ( $0=0$ ) are all conserved.
Examiner tip
Mark scheme: (a) $\text{d} \rightarrow \text{u}$ (1 mark, W $^-$ boson a bonus); (b) charge, baryon and lepton number each shown conserved (3 marks) plus the reason the emitted particle is an antineutrino to keep $L = 0$ . The lepton-number check is the discriminator: $\text{e}^-$ gives $+1$ , so a $\bar{\nu}_e$ ( $-1$ ) is needed.
8Energy released from a mass change in u and MeV (6 marks)
StretchMulti-step problem• mass–energy equivalence, atomic mass unit, MeV, AO2, Unit 4
Question
In an interaction the total mass decreases by $\Delta m = 3.00 \times 10^{-3}\ \text{u}$ , where $1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}$ . (a) Calculate the energy released in joules. (b) Convert this to MeV, using $1\ \text{MeV} = 1.60 \times 10^{-13}\ \text{J}$ . Use $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ . (6 marks)
Step-by-step solution
1. Step 1
  (a) First convert the mass change from atomic mass units into kilograms.
  $\Delta m = (3.00 \times 10^{-3}) \times (1.66 \times 10^{-27}) = 4.98 \times 10^{-30}\ \text{kg}$
2. Step 2
  Apply mass–energy equivalence. The energy released is $\Delta E = c^2 \Delta m$ .
  $\Delta E = c^2 \Delta m$
3. Step 3
  Substitute the mass change in kilograms and $c$ .
  $\Delta E = (3.00 \times 10^{8})^2 \times (4.98 \times 10^{-30}) = 4.48 \times 10^{-13}\ \text{J}$
4. Step 4
  (b) Convert the energy in joules to MeV by dividing by $1.60 \times 10^{-13}$ .
  $\Delta E\ (\text{MeV}) = \frac{\Delta E\ (\text{J})}{1.60 \times 10^{-13}}$
5. Step 5
  Substitute and evaluate to 3 significant figures.
  $\Delta E = \frac{4.48 \times 10^{-13}}{1.60 \times 10^{-13}} = 2.80\ \text{MeV}$
6. Step 6
  Cross-check using the shortcut $1\ \text{u} \equiv 931.5\ \text{MeV}/c^2$ : $3.00 \times 10^{-3} \times 931.5 = 2.79\ \text{MeV}$ , which agrees within rounding.
Answer
(a) $\Delta E = 4.48 \times 10^{-13}\ \text{J}$ . (b) $\Delta E = 2.80\ \text{MeV}$ (consistent with the $1\ \text{u} = 931.5\ \text{MeV}/c^2$ shortcut).
Examiner tip
Mark scheme: (a) convert u → kg; use $\Delta E = c^2\Delta m$ ; $4.48 \times 10^{-13}\ \text{J}$ (3 marks). (b) divide by $1.60 \times 10^{-13}$ ; $2.80\ \text{MeV}$ (3 marks). The frequent unit slip is putting $\Delta m$ in u (not kg) straight into $c^2\Delta m$ — always convert to kilograms first.
9Why a proposed decay is forbidden (6 marks)
StretchShow that / prove• conservation laws, lepton number, forbidden, AO3, Unit 4
Question
A student proposes the neutron decay $\text{n} \rightarrow \text{p} + \text{e}^- + \nu_e$ (emitting a neutrino, $L = +1$ , rather than an antineutrino). By checking the conservation laws, state with reasons which law this proposal violates and hence whether it can occur. (6 marks)
Step-by-step solution
1. Step 1
  Check the three quantities: charge $Q$ , baryon number $B$ and lepton number $L$ .
2. Step 2
  Charge. Before: $\text{n}(0)$ . After: $\text{p}(+1) + \text{e}^-(-1) + \nu_e(0) = 0$ . Charge is conserved. ✓
  $Q:\quad 0 = (+1) + (-1) + 0 = 0$
3. Step 3
  Baryon number. Before: $B = +1$ (neutron). After: $\text{p}(+1) + \text{e}^-(0) + \nu_e(0) = +1$ . Baryon number is conserved. ✓
  $B:\quad 1 = 1 + 0 + 0 = 1$
4. Step 4
  Lepton number. Before: neutron has $L = 0$ . After: $\text{e}^-(+1) + \nu_e(+1) = +2$ . So $L$ goes from $0$ to $+2$ — lepton number is NOT conserved. ✗
  $L:\quad 0 \neq (+1) + (+1) = +2$
5. Step 5
  Because lepton number is violated ( $0 \neq +2$ ), this decay cannot occur, even though charge and baryon number are fine.
6. Step 6
  To conserve lepton number the emitted particle must be an antineutrino $\bar{\nu}_e$ ( $L = -1$ ), giving $L = +1 + (-1) = 0$ after — which is exactly the real $\beta^-$ decay.
Answer
Charge ( $0=0$ ) and baryon number ( $1=1$ ) are conserved, but lepton number is not ( $0 \rightarrow +2$ ). The proposed decay is therefore forbidden; the real $\beta^-$ decay emits an antineutrino ( $L=-1$ ) so that lepton number stays $0$ .
Examiner tip
Mark scheme: charge conserved; baryon number conserved; lepton number NOT conserved ( $0 \neq 2$ ); conclusion 'forbidden'; and the fix — an antineutrino is required. Even one violated law forbids an interaction; you gain marks by naming which law fails and giving the before/after numbers.

Model Answers — Particle Interactions and Conservation

High-scoring sample answers for particle interactions and conservation on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). State what is meant by the antiparticle of a particle, giving the electron as an example.
Model answer
Every particle has a corresponding antiparticle with the same mass but opposite charge (and opposite values of quantities such as baryon or lepton number).

For the electron ( $-1$ charge), the antiparticle is the positron, which has the same mass as the electron but a charge of $+1$ .
Why this scores
Why this scores 2/2. (1) same mass, opposite charge; (2) electron → positron as the named example. Saying only 'opposite charge' misses the equal-mass point; both are needed.
Question 2
2 marks
Q (2 marks). An electron meets a positron and they annihilate. State what is produced, and explain why two photons (rather than one) are emitted.
Model answer
The electron and positron are converted entirely into energy, producing two photons (gamma rays).

Two photons are needed so that momentum is conserved: if the particles are (nearly) at rest the total momentum is zero, so the two photons must travel in opposite directions with equal momenta to keep the total momentum zero. A single photon always carries momentum, so it could not be produced from a system at rest.
Why this scores
Why this scores 2/2. (1) two photons produced; (2) required to conserve momentum (opposite directions from a system at rest). The conservation-of-momentum reason is the marked point — 'just because' does not score.
Question 3
3 marks
Q (3 marks). Distinguish between a baryon and a meson, and state which class a proton and a $\pi^-$ meson each belong to.
Model answer
Both baryons and mesons are hadrons (particles made of quarks), but they differ in quark content:

a baryon is made of three quarks (for example the proton, uud, or the neutron, udd);

a meson is made of a quark and an antiquark (a quark–antiquark pair).

Therefore the proton is a baryon (three quarks) and the $\pi^-$ is a meson (a quark–antiquark pair).
Why this scores
Why this scores 3/3. (1) baryon = 3 quarks; (2) meson = quark–antiquark; (3) correct classification of proton (baryon) and $\pi^-$ (meson). Stating both are hadrons frames the answer well.
Question 4
4 marks
Q (4 marks). The proton has quark content uud and the neutron udd. Using quark charges (up $= +\tfrac{2}{3}e$ , down $= -\tfrac{1}{3}e$ ), show that the proton has charge $+1$ and the neutron is neutral.
Model answer
The charge of a hadron is the sum of its quark charges.

Proton (uud): $Q = \left(+\tfrac{2}{3}\right) + \left(+\tfrac{2}{3}\right) + \left(-\tfrac{1}{3}\right) = +\tfrac{3}{3} = +1$ So the proton has charge $+1$ (i.e. $+e$ ). ✓

Neutron (udd): $Q = \left(+\tfrac{2}{3}\right) + \left(-\tfrac{1}{3}\right) + \left(-\tfrac{1}{3}\right) = 0$ So the neutron is electrically neutral. ✓
Why this scores
Why this scores 4/4. (1) sum quark charges; (2) proton $+\tfrac{2}{3}+\tfrac{2}{3}-\tfrac{1}{3} = +1$ ; (3) neutron $+\tfrac{2}{3}-\tfrac{1}{3}-\tfrac{1}{3} = 0$ ; (4) correct arithmetic to $+1$ and $0$ . Getting the fractions to combine correctly is the discriminator — show the addition explicitly.
Question 5
4 marks
Q (4 marks). In a PET (positron emission tomography) scan, a tracer inside the body emits positrons. Explain what happens to an emitted positron and how this produces the gamma rays detected by the scanner.
Model answer
A positron is the antiparticle of the electron. When an emitted positron meets one of the many electrons in the surrounding body tissue, the two annihilate.

In the annihilation the positron and electron are converted entirely into energy, producing two gamma-ray photons that travel in (almost exactly) opposite directions to conserve momentum. Each photon carries energy $mc^2 = 0.511\ \text{MeV}$ .

The scanner's ring of detectors picks up these pairs of gamma rays arriving on opposite sides at the same time, and from many such detections it reconstructs where in the body the annihilations (and hence the tracer) are concentrated.
Why this scores
Why this scores 4/4. (1) positron = antiparticle of electron; (2) annihilates with a tissue electron; (3) produces two gamma photons in opposite directions (momentum); (4) each photon $= 0.511\ \text{MeV}$ / detected on opposite sides to locate the tracer. Linking the opposite-direction emission to detection is the higher-value point.
Question 6
5 marks
Q (5 marks). In a nuclear process the total rest mass decreases by $2.00 \times 10^{-29}\ \text{kg}$ . (a) Calculate the energy released in joules. (b) Express this energy in MeV. Use $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ and $1\ \text{MeV} = 1.60 \times 10^{-13}\ \text{J}$ .
Model answer
(a) Using mass–energy equivalence, $\Delta E = c^2 \Delta m$ : $\Delta E = (3.00 \times 10^{8})^2 \times (2.00 \times 10^{-29})$ $\Delta E = (9.00 \times 10^{16}) \times (2.00 \times 10^{-29}) = 1.80 \times 10^{-12}\ \text{J}$

(b) Convert to MeV by dividing by $1.60 \times 10^{-13}\ \text{J per MeV}$ : $\Delta E = \frac{1.80 \times 10^{-12}}{1.60 \times 10^{-13}} = 11.3\ \text{MeV}$

So the energy released is $1.80 \times 10^{-12}\ \text{J}$ , or equivalently $11.3\ \text{MeV}$ .
Why this scores
Why this scores 5/5. (a) 3 marks — use of $\Delta E = c^2\Delta m$ ; correct $c^2$ ; $1.80 \times 10^{-12}\ \text{J}$ . (b) 2 marks — divide by $1.60 \times 10^{-13}$ ; $11.3\ \text{MeV}$ . Squaring $c$ correctly and using the MeV (not eV) conversion are the two places marks are lost.
Question 7
5 marks
Q (5 marks). An electron and a positron, each of mass $9.11 \times 10^{-31}\ \text{kg}$ , annihilate from rest and produce two identical photons. (a) Calculate the energy of each photon in joules. (b) Explain why exactly two photons are produced rather than one. Use $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ .
Model answer
(a) The total energy released is the rest energy of both particles, $E_{total} = 2mc^2$ : $E_{total} = 2 \times (9.11 \times 10^{-31}) \times (3.00 \times 10^{8})^2 = 1.64 \times 10^{-13}\ \text{J}$ Two identical photons share this equally, so each has energy: $E_{photon} = \frac{1.64 \times 10^{-13}}{2} = 8.20 \times 10^{-14}\ \text{J}$ (This is $0.511\ \text{MeV}$ — the rest energy of one particle.)

(b) The electron and positron are at rest, so the total momentum before annihilation is zero. Momentum must be conserved, so the products must also have zero total momentum. Two photons travelling in opposite directions with equal momenta achieve this, whereas a single photon would always carry momentum in one direction, which cannot be balanced.
Why this scores
Why this scores 5/5. (a) 3 marks — $E_{total} = 2mc^2 = 1.64 \times 10^{-13}\ \text{J}$ , then halved to $8.20 \times 10^{-14}\ \text{J}$ . (b) 2 marks — total momentum is zero, so two photons go in opposite directions to conserve momentum. The momentum argument (not just 'energy is shared') is the marked point in part (b).
Question 8
5 marks
Q (5 marks). Determine whether the interaction $\text{p} + \text{p} \rightarrow \text{p} + \text{n} + \pi^+$ is allowed. The $\pi^+$ has charge $+1$ , baryon number $0$ and lepton number $0$ . Show your reasoning by checking the conservation laws.
Model answer
Check charge, baryon number and lepton number, comparing the totals before and after.

Charge $Q$ : $\text{before: } (+1) + (+1) = +2 \qquad \text{after: } (+1) + 0 + (+1) = +2$ Charge is conserved ( $+2 = +2$ ). ✓

Baryon number $B$ : each proton and neutron has $B = +1$ ; the meson has $B = 0$ . $\text{before: } 1 + 1 = 2 \qquad \text{after: } 1 + 1 + 0 = 2$ Baryon number is conserved ( $2 = 2$ ). ✓

Lepton number $L$ : no leptons are involved, so $L = 0$ on both sides. ✓

All three quantities are conserved, so the interaction is allowed (given sufficient energy).
Why this scores
Why this scores 5/5. (1) check charge $+2 = +2$ ; (2) baryon number $2 = 2$ ; (3) meson has $B=0$ ; (4) lepton number $0 = 0$ ; (5) conclusion 'allowed'. Showing the before/after totals for each law is essential — the marker awards the checks, not the verdict alone.
Question 9
6 marks
Q (6 marks). Describe how the Standard Model classifies matter particles. In your answer refer to quarks, leptons, and the two types of hadron, and give one example of each of a baryon and a meson.
Model answer
The Standard Model divides matter into two families of fundamental particles: quarks and leptons.

Quarks (e.g. up, down and strange) carry fractional charges — the up quark is $+\tfrac{2}{3}e$ , while the down and strange quarks are each $-\tfrac{1}{3}e$ . Quarks are never found alone; they combine to form composite particles called hadrons.

Leptons are fundamental particles that are not made of quarks. They include the electron, the muon, and their associated neutrinos (and the corresponding antiparticles). Leptons carry lepton number but baryon number zero.

Hadrons (particles made of quarks) come in two types:

Baryons — made of three quarks. Example: the proton (uud) (the neutron udd is another).

Mesons — made of a quark–antiquark pair. Example: the $\pi^+$ meson ( $\text{u}\bar{\text{d}}$ ).

Interactions between these particles are carried by force-carrying (exchange) particles — for example the photon for the electromagnetic force and the W and Z bosons for the weak interaction.
Why this scores
Why this reaches the top band (5–6/6). Names both fundamental families (quarks and leptons); gives fractional quark charges; states quarks combine into hadrons; distinguishes baryons (3 quarks) from mesons (quark–antiquark) with a correct example of each; and adds the exchange particles (photon, W/Z). Confusing the electron for a hadron, or a meson for 'three quarks', is a level-capping error.
Question 10
6 marks
Q (6 marks). Beta-minus decay is written $\text{n} \rightarrow \text{p} + \text{e}^- + \bar{\nu}_e$ . (a) Explain, in terms of quarks, what change occurs inside the neutron. (b) Show that charge and lepton number are conserved, and explain why the emitted particle must be an antineutrino rather than a neutrino.
Model answer
(a) A neutron has quark content udd and a proton uud, so in $\beta^-$ decay one down quark changes into an up quark ( $\text{d} \rightarrow \text{u}$ ). This is a weak interaction: the down quark emits a $\text{W}^-$ boson (turning into an up quark), and the $\text{W}^-$ then decays into the electron and the electron antineutrino.

(b) Charge: before, the neutron is neutral ( $0$ ); after, $\text{p}(+1) + \text{e}^-(-1) + \bar{\nu}_e(0) = 0$ . Charge is conserved ( $0 = 0$ ). ✓

Lepton number: before, $L = 0$ . After, the electron has $L = +1$ and the antineutrino has $L = -1$ , so the total is $+1 + (-1) = 0$ . Lepton number is conserved. ✓

The emitted particle must be an antineutrino ( $L = -1$ ): if a neutrino ( $L = +1$ ) were emitted, the final lepton number would be $+1 + (+1) = +2 \neq 0$ , which violates conservation of lepton number. Only an antineutrino keeps $L$ at zero.
Why this scores
Why this scores 6/6. (a) 2 marks — $\text{d} \rightarrow \text{u}$ with a $\text{W}^-$ boson via the weak interaction. (b) 4 marks — charge conserved; lepton number conserved with the electron $+1$ and antineutrino $-1$ ; and the clinching reason that a neutrino would give $L = +2$ , so an antineutrino is required. The lepton-number argument is what separates a top-band answer.

Key Formulae — Particle Interactions and Conservation

The formulae you need to memorise for particle interactions and conservation on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Mass–energy equivalence
$\Delta E = c^2 \Delta m$
$\Delta E$
energy released or required / $\text{J}$
$\Delta m$
change in (rest) mass / $\text{kg}$
$c$
speed of light $= 3.00 \times 10^{8}\ \text{m s}^{-1}$
When to use
To find the energy equivalent to any mass change — the energy released when mass decreases, or the mass corresponding to a rest energy. ALWAYS put $\Delta m$ in kilograms and remember to square $c$ .
Example
A mass change of $4.98 \times 10^{-30}\ \text{kg}$ releases $\Delta E = (3.00 \times 10^{8})^2 \times 4.98 \times 10^{-30} = 4.48 \times 10^{-13}\ \text{J}$ .
Annihilation / pair-production energy
$E = 2mc^2$
$E$
total energy of the two photons (annihilation) OR minimum photon energy (pair production) / $\text{J}$
$m$
rest mass of ONE particle (particle and antiparticle have equal mass) / $\text{kg}$
$c$
speed of light / $\text{m s}^{-1}$
When to use
For a particle–antiparticle event. Annihilation: the two photons together carry $2mc^2$ , so each carries $mc^2$ . Pair production: the incoming photon needs a MINIMUM of $2mc^2$ to create both particles.
Example
Electron–positron pair production needs at least $E_{min} = 2 \times 0.511 = 1.022\ \text{MeV}$ .
Quark charges (in units of $e$ )
$q_{\text{u}} = +\tfrac{2}{3},\quad q_{\text{d}} = q_{\text{s}} = -\tfrac{1}{3}$
$q_{\text{u}}$
charge of the up quark $= +\tfrac{2}{3}e$
$q_{\text{d}}$
charge of the down quark $= -\tfrac{1}{3}e$
$q_{\text{s}}$
charge of the strange quark $= -\tfrac{1}{3}e$
When to use
To work out the charge of a hadron from its quark content, or to confirm a quark composition. Antiquarks have the opposite sign of charge. Proton (uud) $= +\tfrac{2}{3}+\tfrac{2}{3}-\tfrac{1}{3} = +1$ ; neutron (udd) $= +\tfrac{2}{3}-\tfrac{1}{3}-\tfrac{1}{3} = 0$ .
Example
$\pi^+ = \text{u}\bar{\text{d}} = +\tfrac{2}{3} + (+\tfrac{1}{3}) = +1$ (the $\bar{\text{d}}$ antiquark has charge $+\tfrac{1}{3}e$ ).
Conservation-law checks for interactions
$\sum Q,\ \sum B,\ \sum L \ \text{unchanged before} \rightarrow \text{after}$
$Q$
charge (in units of $e$ )
$B$
baryon number: $+1$ per baryon, $-1$ per antibaryon, $0$ for mesons and leptons
$L$
lepton number: $+1$ per lepton, $-1$ per antilepton, $0$ for hadrons
When to use
To decide whether a proposed interaction is allowed. Total charge, total baryon number AND total lepton number must each be identical on both sides (along with energy and momentum). If any one fails, the interaction cannot occur.
Example
For $\text{n} \rightarrow \text{p} + \text{e}^- + \bar{\nu}_e$ : $Q$ : $0=0$ ; $B$ : $1=1$ ; $L$ : $0 = +1 + (-1) = 0$ — all conserved, so allowed.

Key Definitions and Keywords — Particle Interactions and Conservation

Definitions to memorise and the exact keywords mark schemes credit for particle interactions and conservation answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Mass–energy equivalence
Examiner keyword
The principle that mass and energy are interchangeable: a change in mass $\Delta m$ corresponds to an energy $\Delta E = c^2 \Delta m$ . Even a tiny mass is equivalent to a large energy because $c^2$ is enormous.
Atomic mass unit (u)
Examiner keyword
A unit of mass used for atoms and particles, defined as one-twelfth of the mass of a carbon-12 atom: $1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}$ , equivalent to $931.5\ \text{MeV}/c^2$ .
MeV/ $c^2$ (mass unit)
Examiner keyword
A unit of mass derived from mass–energy equivalence: the mass whose rest energy is 1 MeV. Since $E = mc^2$ , a rest energy expressed in MeV divided by $c^2$ gives the mass. The electron has a mass of $0.511\ \text{MeV}/c^2$ .
Antiparticle
Examiner keyword
A particle with the same mass as a given particle but opposite charge (and opposite baryon/lepton number). Every particle has an antiparticle — for example the positron is the antiparticle of the electron.
Annihilation
Examiner keyword
The process in which a particle and its antiparticle meet and are converted entirely into energy, producing (at least) two photons. The total photon energy equals $2mc^2$ ; two photons are needed to conserve momentum.
Pair production
Examiner keyword
The process in which a photon is converted into a particle–antiparticle pair (e.g. an electron and a positron). The photon must have a minimum energy of $2mc^2$ to create both particles.
Quark
Examiner keyword
A fundamental (indivisible) matter particle with fractional charge: up $= +\tfrac{2}{3}e$ , down and strange $= -\tfrac{1}{3}e$ . Quarks are never found in isolation; they combine to form hadrons.
Lepton
Examiner keyword
A fundamental matter particle that is NOT made of quarks — the electron, the muon and their neutrinos (and antiparticles). Leptons carry lepton number and do not feel the strong interaction.
Hadron
Examiner keyword
A particle made of quarks and therefore subject to the strong interaction. Hadrons divide into baryons (three quarks) and mesons (a quark–antiquark pair).
Baryon
Examiner keyword
A hadron made of three quarks, with baryon number $+1$ (antibaryons have $-1$ ). Examples: the proton (uud) and the neutron (udd).
Meson
Examiner keyword
A hadron made of a quark and an antiquark, with baryon number $0$ . Example: the $\pi^+$ meson ( $\text{u}\bar{\text{d}}$ ).
Baryon number ( $B$ )
Examiner keyword
A quantum number that is $+1$ for every baryon, $-1$ for every antibaryon, and $0$ for mesons and leptons. Total baryon number is conserved in all particle interactions.
Lepton number ( $L$ )
Examiner keyword
A quantum number that is $+1$ for every lepton, $-1$ for every antilepton, and $0$ for all hadrons. Total lepton number is conserved in all particle interactions.
Standard Model
Examiner keyword
The framework that classifies fundamental particles into quarks and leptons (matter particles) and the force-carrying (exchange) bosons — the photon, the W and Z bosons, and the gluon — that mediate the fundamental interactions.
Exchange (force-carrying) particle
A boson exchanged between particles to transmit a fundamental force — the photon carries the electromagnetic force and the W and Z bosons carry the weak interaction (e.g. the $\text{W}^-$ in beta-minus decay).

Common Mistakes and Misconceptions — Particle Interactions and Conservation

The traps other students keep falling into on particle interactions and conservation questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Getting the baryon-number or lepton-number bookkeeping wrong when checking an interaction
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
Students forget that mesons have baryon number $0$ , that antiparticles carry the OPPOSITE sign (e.g. an antineutrino has $L = -1$ ), or they mix up which particles are leptons.
How to avoid it
Tabulate $B$ and $L$ explicitly for every particle: baryons $B = +1$ , antibaryons $-1$ , mesons and leptons $B = 0$ ; leptons $L = +1$ , antileptons $L = -1$ , hadrons $L = 0$ . Sum each column before and after and compare.
✕Saying annihilation produces a single photon (ignoring momentum conservation)
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
Students focus only on energy ( $E = 2mc^2$ ) and forget that momentum must also be conserved.
How to avoid it
Annihilation of a particle–antiparticle pair at rest produces two photons travelling in opposite directions, so the total momentum stays zero. A lone photon always carries momentum, so it cannot be the sole product.
✕Using $mc^2$ instead of $2mc^2$ for the minimum photon energy in pair production
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
Students calculate the rest energy of only one particle, forgetting that a pair (particle AND antiparticle) is created.
How to avoid it
Pair production creates two particles of equal mass, so the minimum photon energy is $E_{min} = 2mc^2$ (e.g. $1.022\ \text{MeV}$ for an electron–positron pair), not $mc^2$ .
✕Misremembering quark charges or the proton/neutron quark composition
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
The fractional charges ( $+\tfrac{2}{3}$ , $-\tfrac{1}{3}$ ) and which of uud/udd is the proton are easy to swap.
How to avoid it
Learn that up $= +\tfrac{2}{3}e$ , down $= -\tfrac{1}{3}e$ ; the proton is uud (charge $+1$ ) and the neutron is udd (charge $0$ ). Check by adding the quark charges — a proton MUST come out to $+1$ .
✕Unit slips in $\Delta E = c^2\Delta m$ — leaving mass in u, or forgetting to square $c$
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
Masses are often quoted in u or MeV/ $c^2$ , and $c$ is large so squaring is easy to overlook.
How to avoid it
Convert $\Delta m$ to kilograms first ( $1\ \text{u} = 1.66 \times 10^{-27}\ \text{kg}$ ), and always use $c^2 = 9.00 \times 10^{16}$ . To convert the resulting joules to MeV, divide by $1.60 \times 10^{-13}$ (NOT $1.60 \times 10^{-19}$ ).
✕Emitting a neutrino instead of an antineutrino in beta-minus decay
Edexcel IAL Physics Unit 4 examiner reports
Why it happens
Students remember 'a neutral, near-massless particle is emitted' but not that lepton number fixes whether it is matter or antimatter.
How to avoid it
In $\beta^-$ decay an electron ( $L = +1$ ) is emitted, so an antineutrino ( $L = -1$ ) must accompany it to keep total lepton number at $0$ . Always finish by checking $L$ before and after.

Particle Interactions and Conservation — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Particle Interactions and Conservation

Short Notes - Particle Interactions and Conservation

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Energy equivalent of the electron mass (2 marks)

2Converting an electron's energy to MeV (2 marks)

3Classifying particles: hadron, baryon, meson or lepton (3 marks)

4Photon energy from electron–positron annihilation (4 marks)

5Minimum photon energy for pair production (4 marks)

6Is a proposed interaction allowed? (5 marks)

7Beta-minus decay at quark level (5 marks)

8Energy released from a mass change in u and MeV (6 marks)

9Why a proposed decay is forbidden (6 marks)

Mass–energy equivalence

Annihilation / pair-production energy

Quark charges (in units of eee)

Conservation-law checks for interactions

Mass–energy equivalence

Atomic mass unit (u)

MeV/c2c^2c2 (mass unit)

Antiparticle

Annihilation

Pair production

Quark

Lepton

Hadron

Baryon

Meson

Baryon number (BBB)

Lepton number (LLL)

Standard Model

Exchange (force-carrying) particle

✕Getting the baryon-number or lepton-number bookkeeping wrong when checking an interaction

✕Saying annihilation produces a single photon (ignoring momentum conservation)

✕Using mc2mc^2mc2 instead of 2mc22mc^22mc2 for the minimum photon energy in pair production

✕Misremembering quark charges or the proton/neutron quark composition

✕Unit slips in ΔE=c2Δm\Delta E = c^2\Delta mΔE=c2Δm — leaving mass in u, or forgetting to square ccc

✕Emitting a neutrino instead of an antineutrino in beta-minus decay

Particle Interactions and Conservation — frequently asked questions

Quark charges (in units of $e$ )

MeV/ $c^2$ (mass unit)

Baryon number ( $B$ )

Lepton number ( $L$ )

✕Using $mc^2$ instead of $2mc^2$ for the minimum photon energy in pair production

✕Unit slips in $\Delta E = c^2\Delta m$ — leaving mass in u, or forgetting to square $c$