
Bounded modern-physics survey · Interactive lesson
Quarks, leptons, and the Standard Model
You can already balance a decay and price it in MeV. Now open the particles up: a hadron is a set of quarks whose charges add in thirds, the weak force rides on a carrier whose mass fixes its reach, and β decay is one d quark turning into a u. The ledger says what may happen; the carrier's mass says why weak decays are slow.
UPIII-11.09 · Lesson 9 of 9 in this unit. Position is not a completion score.Bounded modern-physics survey. This modern-physics material is an introductory survey, not a complete course in relativity, quantum, atomic, nuclear or particle physics. Your university determines the required depth and assessment.
Build the model
Six quarks, six leptons, four carriers, and a ledger that decides what may happen.
The key idea
The Standard Model sorts matter into six quarks and six leptons, describes the strong, electromagnetic and weak interactions as the exchange of carrier bosons, and supplies the conserved numbers that decide which reactions can occur.
Example. The neutron is udd: −1/3 − 1/3 + 2/3 = 0 in units of e, with baryon number 3 × 1/3 = 1. β⁻ decay turns one d into a u, and the same sums give uud: charge +1, baryon number still 1, a proton.
The complete physical model and assumptions
At the level used here the Standard Model is three things: an inventory, a ledger and a set of carriers. The inventory is six quarks, charged +2/3 or −1/3 in units of e, and six leptons in three families. Every hadron ever observed is a colour-neutral combination of quarks: three for a baryon, a quark–antiquark pair for a meson, and four or five in a few exotic states, so its charge, baryon number and strangeness are sums over its content. The ledger is the set of additive numbers that decays and reactions conserve: charge and baryon number always, the three lepton-family numbers in every decay treated here, and strangeness in everything except weak processes. The carriers are the photon, eight gluons, and the W and Z. For the photon, W and Z the rest energy fixes the reach, R = ħc/Mc²: unbounded for the photon, 2.5 × 10⁻³ fm for the W; the massless gluons are cut off by confinement instead. Together they turn β decay into a single d quark becoming a u by emitting a virtual W⁻, which materialises as an electron and an antineutrino. The model stops short of explaining itself. Masses are inputs, confinement is observed rather than derived, neutrino oscillation shows the family ledger is not exact, and a ledger that balances permits a process without saying how fast it goes: rates need the full field theory.
1A hadron's quantum numbers are sums over its quarks
Six quark flavours fall into two charge families: up-type u, c and t with Q = +2/3, and down-type d, s and b with Q = −1/3, in units of e. Every quark carries baryon number B = +1/3, the strange quark carries strangeness S = −1, and an antiquark reverses every sign. A baryon is three quarks and a meson a quark with an antiquark, and the hadron's additive numbers are simply sums. The proton uud has Q = 2/3 + 2/3 − 1/3 = +1 and B = 1; the Λ⁰ (uds) has Q = 0 and S = −1; the π⁺ (ud̄) has Q = 2/3 + 1/3 = +1 and B = 0; the K⁺ (us̄) has S = +1. Writing each up-type charge as 1 − 1/3 gives the shortcut Q = NU − B. Mass is the exception: it does not add. The u and d quarks weigh about 2.2 and 4.7 MeV/c², so uud sums to 9 MeV/c², under 1% of the proton's 938.3 MeV/c². The rest is the energy of the gluon field and of the quarks' confined motion.
2Colour fixes the Pauli puzzle, and confinement is a rule
The Δ⁺⁺ is uuu with spin 3/2: three identical fermions in the spatial ground state with parallel spins, a state symmetric in every label we have named. Pauli exclusion forbids it unless the quarks differ in one more label. That label is colour, and it takes three values. Every hadron ever observed is colour neutral, one quark of each colour in a baryon and a colour with its anticolour in a meson, and no isolated quark has ever been seen. Colour neutrality needs N(q) − N(q̄) to be a multiple of 3, so B is an integer and, through Q = NU − B, so is Q: that is why no confirmed free charge of e/3 has ever turned up. Here confinement is an observed rule, not a derived one; lattice calculations of the strong interaction reproduce it numerically, but no analytic proof exists. Its signature is a field between separating quarks that stores about 1 GeV per femtometre, a tension of 1.6 × 10⁵ N. At 1 fm that is already over seven pion rest energies, so well before 2 fm the field pays for new quark–antiquark pairs and snaps: a hard collision yields jets of hadrons, never a free quark.
3Leptons keep three separate ledgers
The six leptons feel no strong force and come in three families, (e⁻, νₑ), (μ⁻, νμ) and (τ⁻, ντ), with charged-lepton rest energies of 0.511, 105.7 and 1777 MeV. Give Lₑ = +1 to e⁻ and νₑ, −1 to e⁺ and ν̄ₑ, and define Lμ and Lτ the same way; hadrons carry zero. In every decay in this unit each family total is conserved on its own. Muon decay, μ⁻ → e⁻ ν̄ₑ νμ, closes all three: Lμ goes 1 → 0 + 0 + 1 and Lₑ goes 0 → 1 − 1 + 0. The simpler-looking μ⁻ → e⁻ γ conserves charge, energy and total lepton number, yet changes both Lₑ and Lμ, and no search has ever seen it: fewer than one muon decay in 10¹² can go that way. The rule is empirical rather than fundamental. Neutrino oscillation shows that a νₑ can arrive as a νμ, so family number is not exactly conserved; but the violation this feeds into charged-lepton decays lies far below any measurement, and as a bookkeeping rule for decays it has never failed. Total lepton number, Lₑ + Lμ + Lτ, has shown no violation at all.
4β decay is one quark changing flavour
In hadrons, n → p + e⁻ + ν̄ₑ. In quarks it is udd → uud: one d quark becomes a u by emitting a W⁻, the W⁻ becomes the electron and the antineutrino, and the other two quarks are spectators. Check each vertex on its own. At the quark vertex, charge −1/3 → +2/3 + (−1) and B = 1/3 → 1/3; at the lepton vertex, −1 → −1 + 0 and Lₑ = 0 → +1 − 1. β⁺ decay runs u → d + W⁺ with W⁺ → e⁺ + νₑ. A free proton cannot do it, because the neutron and positron together outweigh it by 1.804 MeV/c², so it happens only inside nuclei whose binding supplies the energy. Electron capture, p + e⁻ → n + νₑ, is the same exchange with an electron absorbed instead of a positron emitted. With s in place of d the vertex drives strange decays, and only the W changes flavour: photon, gluon and Z leave every quark's flavour unchanged. So a change of strangeness marks a weak process, and the clock agrees. A strange hadron that can only decay weakly lives 10⁻¹⁰ to 10⁻⁸ s, against about 10⁻²³ s for a strong decay and 10⁻²⁰ to 10⁻¹⁶ s for an electromagnetic one.
5A carrier's mass sets its range, and makes the weak force weak
The photon, eight gluons, the W± at 80.4 GeV and the Z⁰ at 91.2 GeV carry the electromagnetic, strong and weak interactions. Exchanging a carrier of mass M borrows energy Mc² for a time of order ħ/Mc², so it reaches R = ħc/Mc², and its potential takes the Yukawa form U(r) = −αħc e(−r/R)/r. Let M → 0 and R → ∞: the Coulomb potential returns, which is the correspondence check. Yukawa's own case, the pion at 139.6 MeV, gives R = 1.41 fm, the reach of the force between nucleons; the W gives 2.45 × 10⁻³ fm. The weak coupling itself is not small: αW ≈ 1/29.5, about 4.6 times α. What makes β decay slow is the range. A slow particle, whose wavelength far exceeds R, cannot resolve the potential's shape and feels only its volume integral, ∫U d³r = −4παħcR², so the effective strength goes as αW/(MW c²)². Multiply by π/√2 and you have Fermi's constant: GF/(ħc)³ = 1.166 × 10⁻⁵ GeV⁻². Gluons are massless, yet the strong force does not reach far: there confinement, not mass, cuts it off near 1 fm.
6Where the survey stops
The inventory is short: six quarks, six leptons, four kinds of carrier, and the Higgs boson at 125 GeV, whose field gives the W, the Z and the fermions their masses. Every ledger in this lesson says only whether a process may occur, not how fast. The Λ⁰ and the free neutron both pass theirs, yet the Λ⁰ lives 2.6 × 10⁻¹⁰ s and the neutron about 880 s, a factor of 3 × 10¹² that only the full field theory of rates can account for. Nor are the masses predicted. They are inputs, running from 0.511 MeV for the electron to 172.6 GeV for the top quark, a ratio of 3.4 × 10⁵, with the neutrino masses far smaller still. The number of families is measured, not derived; confinement is observed, not proved; and gravity is absent altogether. Use the model as a classification: a ledger that fails rules a process out, and one that balances only lets it in.
Key equations
Read the units, assumptions and limits before substituting.
Baryon number of a hadron
B = [N(q) − N(q̄)] / 3
No process has ever been seen to change the total, so the proton, the lightest baryon, has nothing lighter with B = 1 to decay into.
B dimensionless; N(q) and N(q̄) count quarks and antiquarks of every flavour. B = 1 for qqq, 0 for qq̄, −1 for q̄q̄q̄Charge from quark content
Q = NU − B
Each up-type charge is 1 − 1/3, so Q is an integer whenever B is. Proton uud: 2 − 1 = +1. Ω⁻ (sss): 0 − 1 = −1. K⁻ (sū): −1 − 0 = −1.
Q in units of e = 1.602 × 10⁻¹⁹ C; NU is the number of up-type quarks (u, c, t) minus up-type antiquarksLepton-family number
Lₑ = N(e⁻) + N(νₑ) − N(e⁺) − N(ν̄ₑ)
Each family total is conserved separately in every decay treated here. It is an empirical accounting rule, and neutrino oscillation shows it is not exact.
dimensionless; Lμ and Lτ are defined the same way for (μ⁻, νμ) and (τ⁻, ντ); every hadron carries zeroFlavour change at a W vertex
d → u + W⁻W⁻ → e⁻ + ν̄ₑ
Only the W changes flavour. With s in place of d the same vertex drives strange decays, so a strangeness change of one unit is the weak interaction's fingerprint.
charge: −1/3 = +2/3 + (−1) at the quark vertex and −1 = −1 + 0 at the lepton vertex; Lₑ: 0 = +1 − 1Range of an exchanged carrier
R = ħ/(Mc) = ħc/(Mc²)
The pion's 139.6 MeV gives 1.41 fm, the reach of the nuclear force. The W's 80.4 GeV gives 2.45 × 10⁻³ fm, some 340 times smaller than the proton's radius.
R in m; M the carrier's mass in kg. With ħc = 197.3 MeV fm and Mc² in MeV, R comes out in fm = 10⁻¹⁵ mYukawa potential
U(r) = −α ħc e(−r/R) / r
As M → 0, R → ∞ and U → −αħc/r, the Coulomb form. For finite R, ∫U d³r = −4παħcR², so a heavy carrier acts as a contact of strength ∝ α/(Mc²)².
U in J; α a dimensionless coupling; r and R in m; attractive sign shownInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at the W, slider 1.905: its plateau stands 4.7 times above the photon line, yet it is gone by 10⁻² fm, deep inside a proton. At −0.855, the pion, R grows to 1.41 fm, the nuclear force. At −4 with α = 0.007 the curve lies along the photon line: the Coulomb limit.
Live results
- REST ENERGY Mc²
- 80352.6 MeV
- RANGE R = ħc/Mc²
- 0.0025 fm
- PROTON RADIUS ÷ R
- 342.0
- COUPLING ÷ 1/137
- 4.66
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. log₁₀ of rest energy Mc² in GeV: 1.905; Coupling α at short range: 0.034
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
A picture carried over from first courses is that a neutron is a proton with an electron stuck to it, and β decay lets the electron out. Three facts rule it out. Spin: a proton and an electron are two spin-½ particles, which combine to total spin 0 or 1, never the neutron's ½. Energy: an electron confined within about 1 fm has momentum of order ħ/Δx ≈ 197 MeV/c, so a kinetic energy near 197 MeV, some 250 times the 0.782 MeV that neutron decay can give it. And the quark picture: the decay changes a d quark into a u, and the electron and antineutrino are created at the W vertex, just as a photon is created when an atom emits. Neither existed before the decay. The neutron is udd, not p + e⁻.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Three hadrons have quark content uss, sū and cd̄. For each, find the charge Q (in units of e), the baryon number B and the strangeness S, say whether it is a baryon or a meson, and check Q with Q = NU − B. Then find what a two-quark state ud would carry, and say why none is seen.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Values per quark: u and c carry Q = +2/3; d and s carry Q = −1/3; every quark has B = +1/3; s has S = −1. An antiquark reverses each sign, so s̄ has S = +1.
- uss: Q = 2/3 − 1/3 − 1/3 = 0, B = 1, S = −2. Three quarks make a baryon, the Ξ⁰. Check: NU = 1, so Q = 1 − 1 = 0.
- sū: Q = −1/3 − 2/3 = −1, B = 1/3 − 1/3 = 0, S = −1. A quark and an antiquark make a meson, the K⁻. Check: NU = −1 (one anti-up), so Q = −1 − 0 = −1.
- cd̄: Q = 2/3 + 1/3 = +1, B = 0, S = 0, since it carries charm rather than strangeness. A meson, the D⁺. Check: NU = 1, so Q = 1 − 0 = +1.
- ud: Q = 2/3 − 1/3 = +1/3 and B = 2/3. Two quarks cannot be colour neutral, and only colour-neutral combinations, with whole-number B and therefore whole-number Q, are ever observed.
Answer. Ξ⁰ (uss): Q = 0, B = 1, S = −2. K⁻ (sū): Q = −1, B = 0, S = −1. D⁺ (cd̄): Q = +1, B = 0, S = 0. A ud pair would carry Q = +1/3 and B = 2/3; it is not colour neutral and has never been seen.
MediumExample 2
In free-neutron decay the only energy available is (mₙ − mₚ)c² = 1.293 MeV, yet the W⁻ that carries the decay has rest energy 80.4 GeV. By what factor is the W short of energy? Estimate how long the virtual W can exist, Δt ≈ ħ/(MW c²), and the farthest it can travel. Compare that distance with the proton's charge radius, 0.84 fm, and say what it implies about where the decay happens.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Shortfall: 80.4 × 10³ MeV ÷ 1.293 MeV = 6.22 × 10⁴. No real W can be produced, so the W must be virtual, borrowed for a time limited by ħ.
- Lifetime of the loan: Δt ≈ ħ/(MW c²) = (6.582 × 10⁻²² MeV s) ÷ (8.04 × 10⁴ MeV) = 8.187 × 10⁻²⁷ s.
- Farthest reach: cΔt = (2.998 × 10⁸ m s⁻¹)(8.187 × 10⁻²⁷ s) = 2.454 × 10⁻¹⁸ m = 2.454 × 10⁻³ fm. This is R = ħc/(MW c²) = 197.3 MeV fm ÷ 80 400 MeV by another route.
- Compare: 0.84 fm ÷ 2.454 × 10⁻³ fm = 342. The W cannot carry anything from one quark to another. The d → u change and the creation of e⁻ and ν̄ₑ happen at a single point, on a single quark, while the other two quarks look on.
- Limiting case: were the W massless, Δt and the reach would be unbounded and the interaction would stretch as far as electromagnetism does. The large mass is what makes it a contact interaction.
Answer. The W is short by a factor of 6.2 × 10⁴. It can exist for about 8.2 × 10⁻²⁷ s and travel at most 2.5 × 10⁻³ fm, some 340 times less than the proton radius, so the decay happens at a single quark.
HardExample 3
A search for μ⁺ → e⁺ γ stops positive muons at rest. Show which ledger entries forbid that decay while μ⁺ → e⁺ νₑ ν̄μ is allowed. If μ⁺ → e⁺ γ did occur, what positron energy would it give? Then show that the allowed decay produces positrons across a range of energies whose top edge is the same value. Take mμc² = 105.658 MeV and mₑc² = 0.511 MeV.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Ledger: μ⁺ is an antilepton, with Lμ = −1, Lₑ = 0 and Q = +1. For e⁺ γ the final state has Lμ = 0 and Lₑ = −1, so total lepton number survives but each family number changes by one: forbidden. For e⁺ νₑ ν̄μ, Lₑ = −1 + 1 + 0 = 0 and Lμ = 0 + 0 − 1 = −1, and the charge stays +1: allowed, as μ⁺ → ν̄μ + W⁺ followed by W⁺ → e⁺ + νₑ.
- Two-body decay at rest: mμc² = Eₑ + Eγ, with equal and opposite momenta, so Eγ = pₑc. Using Eₑ² = (pₑc)² + (mₑc²)², write mμc² − Eₑ = √(Eₑ² − mₑ²c⁴) and square: mμ²c⁴ − 2mμc²Eₑ = −mₑ²c⁴, so Eₑ = (mμ² + mₑ²)c²/(2mμ).
- Numbers: Eₑ = (105.658² + 0.511²) MeV² ÷ (2 × 105.658 MeV) = 11 163.874 MeV² ÷ 211.316 MeV = 52.830 MeV, and Eγ = 105.658 − 52.830 = 52.828 MeV. Limiting case: mₑ → 0 gives exactly mμc²/2 = 52.829 MeV, and the two energies differ by only mₑ²c²/mμ = 2.5 keV.
- Three-body decay: the two neutrinos share whatever the positron leaves. The positron's energy is greatest when the neutrino pair has the smallest possible invariant mass, zero, which happens when both neutrinos move together, opposite the positron. The pair then recoils like one massless particle, and the two-body result applies unchanged: the top edge is 52.830 MeV.
- Spread the neutrinos apart and the positron gets less, down to Eₑ = mₑc² = 0.511 MeV when the neutrinos fly back to back and leave it at rest. So the allowed decay fills 0.511 to 52.830 MeV continuously, and μ⁺ → e⁺ γ would appear as a sharp line at the very top edge, in time with a back-to-back 52.828 MeV photon.
Answer. μ⁺ → e⁺ γ changes Lₑ and Lμ by one each and is forbidden. It would give Eₑ = 52.830 MeV with a 52.828 MeV photon. The allowed decay spreads positron total energies from 0.511 MeV up to the same 52.830 MeV edge.
Quick check
Select an answer and explain it before checking.
Review & continue
The Standard Model sorts matter into six quarks and six leptons, describes the strong, electromagnetic and weak interactions as the exchange of carrier bosons, and supplies the conserved numbers that decide which reactions can occur.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics III completion tasks.