Hadrons: Baryons and Mesons

Now that we know about quarks and their properties, we can understand the composite particles they form. Any particle made of quarks is called a hadron. Hadrons experience the strong nuclear force (unlike leptons, which do not). There are two types of hadrons: baryons (made of three quarks) and mesons (made of a quark and an antiquark).

Baryons

A baryon is a hadron composed of three quarks (qqq). The most familiar baryons are the proton and the neutron.

Baryon Number

Every quark has a baryon number of +1/3, so a baryon (three quarks) has a total baryon number of +1. Antibaryons (three antiquarks) have baryon number −1. Baryon number is conserved in all particle interactions.

Common Baryons and Their Quark Compositions

Baryon	Symbol	Quark Content	Charge (e)	B	S	Mass (MeV/c²)
Proton	p	uud	+1	+1	0	938.3
Neutron	n	udd	0	+1	0	939.6
Lambda	Λ⁰	uds	0	+1	−1	1115.7
Sigma-plus	Σ⁺	uus	+1	+1	−1	1189.4
Sigma-zero	Σ⁰	uds	0	+1	−1	1192.6
Sigma-minus	Σ⁻	dds	−1	+1	−1	1197.4
Xi-zero	Ξ⁰	uss	0	+1	−2	1314.9
Xi-minus	Ξ⁻	dss	−1	+1	−2	1321.7
Omega-minus	Ω⁻	sss	−1	+1	−3	1672.5

Checking Charges

You can verify the charge of any baryon by adding up the quark charges:

Proton (uud): +2/3 + 2/3 − 1/3 = +1e ✔
Neutron (udd): +2/3 − 1/3 − 1/3 = 0 ✔
Sigma-minus (dds): −1/3 − 1/3 − 1/3 = −1e ✔
Omega-minus (sss): −1/3 − 1/3 − 1/3 = −1e, S = 3 × (−1) = −3 ✔

Worked Example: Determining Unknown Quark Content

Problem: A baryon X has charge 0, baryon number +1, and strangeness −2. What is its quark content?

B = +1 → three quarks
S = −2 → two strange quarks (each s gives S = −1)
Charge = 0 → two s quarks contribute 2 × (−1/3) = −2/3, so the third quark must have charge +2/3 → up quark

Answer: uss (this is the Ξ⁰ baryon)

Proton Stability

The proton is the lightest baryon and is believed to be stable (or at least have a half-life exceeding 10³⁴ years). It cannot decay into anything lighter while conserving baryon number, because there is no lighter baryon for it to become. The neutron, by contrast, is unstable as a free particle — it undergoes beta-minus decay with a half-life of about 10.3 minutes:

n (udd) → p (uud) + e⁻ + ν̄ₑ

However, neutrons bound inside stable nuclei do not decay, because the nuclear binding energy makes the decay energetically forbidden.

Mesons

A meson is a hadron composed of a quark and an antiquark (qq̄). Mesons have baryon number zero (+1/3 from the quark and −1/3 from the antiquark = 0).

Common Mesons and Their Quark Compositions

Meson	Symbol	Quark Content	Charge (e)	S	Mass (MeV/c²)
Pion-plus	π⁺	ud̄	+1	0	139.6
Pion-minus	π⁻	ūd	−1	0	139.6
Pion-zero	π⁰	uū or dd̄	0	0	135.0
Kaon-plus	K⁺	us̄	+1	+1	493.7
Kaon-minus	K⁻	ūs	−1	−1	493.7
Kaon-zero	K⁰	ds̄	0	+1	497.6
Anti-kaon-zero	K̄⁰	d̄s	0	−1	497.6

Checking Charges (Mesons)

π⁺ (ud̄): +2/3 + 1/3 = +1e ✔
K⁺ (us̄): +2/3 + 1/3 = +1e ✔ (note: s̄ has charge +1/3)
K⁻ (ūs): −2/3 − 1/3 = −1e ✔

Strangeness of Kaons

Kaons contain strange or antistrange quarks and therefore carry non-zero strangeness:

Kaon	Quark content	Strangeness	Why
K⁺	us̄	+1	s̄ has S = +1
K⁻	ūs	−1	s has S = −1
K⁰	ds̄	+1	s̄ has S = +1
K̄⁰	d̄s	−1	s has S = −1

Mesons Are Unstable

All mesons are unstable. They decay via the strong force (if strangeness is conserved) or the weak force (if strangeness changes). The pion-plus, for example, decays via the weak force:

π⁺ → μ⁺ + ν_μ

The kaon-plus also decays via the weak force (strangeness changes from +1 to 0):

K⁺ → μ⁺ + ν_μ or K⁺ → π⁺ + π⁰

Antiparticles

Every particle has a corresponding antiparticle with the same mass and spin but opposite values of charge, baryon number, lepton number, and strangeness.

Particle	Antiparticle	Quark content	Q	B
Proton (p)	Antiproton (p̄)	uud → ūūd̄	+1 → −1	+1 → −1
Neutron (n)	Antineutron (n̄)	udd → ūd̄d̄	0 → 0	+1 → −1
π⁺ (ud̄)	π⁻ (ūd)	—	+1 → −1	0 → 0
K⁺ (us̄)	K⁻ (ūs)	—	+1 → −1	0 → 0

When a particle meets its antiparticle, they can annihilate, converting all their mass into energy (usually gamma-ray photons). The reverse process — pair production — creates a particle-antiparticle pair from a sufficiently energetic photon.

Pair Production: Minimum Photon Energy

For pair production to occur, the photon must have at least enough energy to create the rest mass of both particles:

E_min = 2mc²

Pair	Minimum photon energy
e⁻ e⁺	2 × 0.511 = 1.022 MeV
p p̄	2 × 938.3 = 1876.6 MeV
μ⁻ μ⁺	2 × 105.7 = 211.4 MeV

Pair production must also occur near a nucleus to conserve momentum — a free photon cannot produce a pair in empty space.

Classifying Particles: A Summary

graph TD
    P["All Particles"] --> H["Hadrons (feel strong force)"]
    P --> L["Leptons (do NOT feel strong force)"]
    P --> B["Gauge Bosons (force carriers)"]
    H --> Bar["Baryons (qqq, B = +1)"]
    H --> Mes["Mesons (qq̄, B = 0)"]
    Bar --> Pro["Proton (uud)"]
    Bar --> Neu["Neutron (udd)"]
    Bar --> Lam["Λ⁰ (uds)"]
    Mes --> Pi["π⁺ (ud̄)"]
    Mes --> Ka["K⁺ (us̄)"]
    L --> CL["Charged: e⁻, μ⁻, τ⁻"]
    L --> NL["Neutral: ν_e, ν_μ, ν_τ"]

	Hadrons	Leptons
Made of	Quarks	Not made of quarks (fundamental)
Feel strong force?	Yes	No
Types	Baryons (qqq) and Mesons (qq̄)	Charged leptons and neutrinos
Baryon number	Baryons: +1, Mesons: 0	0
Examples	p, n, π, K	e⁻, μ⁻, νₑ

How to Determine Quark Composition

To figure out the quark content of a hadron:

Determine if it is a baryon or meson — does it have baryon number +1 (or −1 for antibaryons) or 0?
Check the charge — the quark charges must add up to the total charge of the hadron.
Check strangeness — if strangeness is non-zero, include the appropriate number of strange quarks (s has S = −1) or antistrange quarks (s̄ has S = +1).
Fill in with up and down quarks to get the right charge.

Worked Example: Unknown Meson

Problem: A meson has charge −1e and strangeness −1. Find its quark content.

Meson → one quark + one antiquark
S = −1 → contains one strange quark (s, with S = −1). The antiquark must have S = 0.
The s quark has charge −1/3. We need total charge = −1. So the antiquark must have charge −1 − (−1/3) = −2/3. An antiquark with charge −2/3 is ū (anti-up).
Composition: ūs → this is the K⁻ meson ✔

Common Exam Mistakes

Confusing baryons and mesons. Baryons have three quarks (B = +1); mesons have a quark-antiquark pair (B = 0). If the question says B = 0 and it is a hadron, it must be a meson.
Forgetting that the pion-zero is its own antiparticle. π⁰ = uū or dd̄ — the particle and antiparticle are the same.
Getting strangeness signs wrong for antiquarks. s has S = −1 and s̄ has S = +1. Many students reverse these.
Thinking mesons have baryon number. Mesons always have B = 0 because the quark and antiquark contributions cancel.

Summary

Hadrons are composite particles made of quarks. Baryons contain three quarks and have baryon number +1 (e.g., proton = uud, neutron = udd). Mesons contain a quark-antiquark pair and have baryon number 0 (e.g., π⁺ = ud̄, K⁺ = us̄). Every particle has an antiparticle with opposite quantum numbers. The proton is the only stable baryon; all mesons and free neutrons are unstable. Quark compositions can be deduced from charge, baryon number, and strangeness.

A-Level Deep Dive: Hadrons, Baryons and Mesons

Spec mapping

Edexcel 9PH0 specification Topic 8 — Nuclear and Particle Physics covers the classification of particles into hadrons (baryons and mesons) and leptons; the quark composition of common hadrons (proton uud, neutron udd, pions, kaons); the conservation laws governing particle interactions, including charge, baryon number, lepton number and strangeness; and the contrast between strong and weak interactions in respect of strangeness conservation (refer to the official specification document for exact wording). Although this lesson sits within Topic 8, the underlying logic of conservation arguments and quark bookkeeping is examined synoptically across Paper 2 and Paper 3, particularly where decay processes intersect with energy-mass conservation from Topic 7. The Edexcel data and formulae booklet does not list quark compositions — these must be memorised — but it does provide rest-mass values that are commonly used in conservation-of-energy follow-up questions.

Worked example with full mark scheme

Question (8 marks):

(a) A particle has charge +1e, baryon number 0, and strangeness +1. State its quark composition and name the particle. (2)

(b) Consider the proposed reaction:

$K^- + p \to \Lambda^0 + \pi^0$

where $K^-$ has quark composition $s\bar{u}$ , the proton is $uud$ , $\Lambda^0$ has composition $uds$ , and $\pi^0$ has composition $u\bar{u}$ or $d\bar{d}$ . Determine, with full working, whether baryon number, charge and strangeness are each conserved, and hence state whether the reaction is allowed by the strong interaction. (6)

Solution with mark scheme:

(a) Step 1 — deduce composition from quantum numbers.

Baryon number $B = 0$ implies a meson (quark–antiquark pair), since each quark contributes $+\tfrac{1}{3}$ and each antiquark $-\tfrac{1}{3}$ , and only $q\bar{q}$ gives net zero. Strangeness $S = +1$ requires an anti-strange quark $\bar{s}$ (because $s$ has $S = -1$ and $\bar{s}$ has $S = +1$ ). Charge $+1e$ then requires a partner of charge $+\tfrac{2}{3}e$ , namely an up quark $u$ . Composition: $u\bar{s}$ .

M1 — recognising that $B = 0$ forces a meson and that $S = +1$ requires $\bar{s}$ .

A1 — correct composition $u\bar{s}$ ; particle is the $K^+$ kaon.

(b) Step 1 — baryon number balance.

LHS: $K^-$ is a meson ( $B = 0$ ); proton has $B = +1$ . Total $B_\text{LHS} = 0 + 1 = +1$ .

RHS: $\Lambda^0$ contains three quarks ( $uds$ ) so $B = +1$ ; $\pi^0$ is a meson ( $B = 0$ ). Total $B_\text{RHS} = 1 + 0 = +1$ .

M1 — correctly assigning $B = 0$ to mesons and $B = +1$ to baryons on both sides; baryon number conserved.

Step 2 — charge balance.

LHS: $Q_{K^-} = -1e$ , $Q_p = +1e$ ; total $Q_\text{LHS} = 0$ .

RHS: $\Lambda^0$ neutral, $\pi^0$ neutral; total $Q_\text{RHS} = 0$ .

M1 — charge conserved.

Step 3 — strangeness balance.

LHS: $K^-$ contains $\bar{u}s$ , but $s$ is the strange quark with $S = -1$ , so $K^-$ has $S = -1$ . Proton has $S = 0$ . Total $S_\text{LHS} = -1$ .

RHS: $\Lambda^0$ contains one $s$ , so $S = -1$ . $\pi^0$ has $S = 0$ . Total $S_\text{RHS} = -1$ .

M1 — strangeness correctly assigned to each hadron.

A1 — strangeness conserved ( $\Delta S = 0$ ).

Step 4 — conclusion.

Charge, baryon number and strangeness are all conserved. The reaction is therefore allowed by the strong interaction (which conserves all these quantities).

A1 — clear conclusion linked to the conservation laws, citing the strong interaction explicitly.

Total: 8 marks (M4 A4, split as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A neutral particle X is observed to decay via the strong interaction into a $\Lambda^0$ baryon and a $K^0$ meson. The $K^0$ has quark composition $d\bar{s}$ .

(a) Determine the strangeness of X, justifying your reasoning. (3)

(b) Suggest one possible quark composition for X, and state the resulting baryon number. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1) — identifying that the strong interaction conserves strangeness, so $S_X = S_{\Lambda^0} + S_{K^0}$ .
M1 (AO2) — assigning $S_{\Lambda^0} = -1$ (one $s$ quark) and $S_{K^0} = +1$ ( $\bar{s}$ contributes $+1$ ).
A1 (AO1) — concluding $S_X = -1 + 1 = 0$ .

(b)

M1 (AO2) — recognising that X must be neutral, with $B = +1$ (since $\Lambda^0$ is a baryon and $K^0$ is a meson, total RHS baryon number is $+1$ ).
M1 (AO1) — proposing a three-quark composition with net charge zero and $S = 0$ , e.g. $udd$ (the neutron) or another light baryon.
A1 (AO3) — correctly stating the resulting baryon number $B = +1$ and noting that the candidate baryon must have sufficient rest energy to produce $\Lambda^0 + K^0$ .

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. Edexcel Paper 2 questions on hadron classification typically blend AO1 recall (quark contents, conservation laws) with AO2 reasoning (applying conservation to a specific reaction), and reserve AO3 for the final synthesis step.

Synoptic links

Connects to:

Standard Model (Topic 8): baryons and mesons together constitute the hadrons, distinguishing them from the leptons (electron, muon, tau and their neutrinos). The hadron/lepton division is the first major branching of the Standard Model and is reinforced every time a conservation law is applied.