Particle Physics

Particle Physics

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Board Coverage AQA Paper 2 | Edexcel CP6 | OCR (A) Paper 2 | CIE P4

1. The Standard Model

The Standard Model classifies all known fundamental particles and their interactions. It describes:

• 12 fermions (matter particles): 6 quarks and 6 leptons, each with an antiparticle.
• 5 gauge bosons (force carriers): photon, $W^+$, $W^-$, $Z^0$, gluon (8 types).
• 1 scalar boson: Higgs ($H^0$), responsible for giving mass to $W$, $Z$ bosons and fermions.

The Four Fundamental Interactions

Interaction	Mediator	Acts on	Range	Relative strength
Electromagnetic	Photon ($\gamma$)	Charged particles	Infinite	$\sim 10^{-2}$
Strong (colour)	Gluon ($g$)	Quarks, gluons	$\sim 10^{-15}$ m	$\sim 1$
Weak	$W^\pm$, $Z^0$	All fermions	$\sim 10^{-18}$ m	$\sim 10^{-6}$
Gravitational	Graviton (hypothetical)	All mass/energy	Infinite	$\sim 10^{-39}$

2. Quarks

Quarks are fundamental particles that experience the strong force. They carry fractional electric charge and a colour charge (red, green, or blue).

The Six Flavours

Generation	Up-type	Charge	Down-type	Charge
1	Up ($u$)	$+2e/3$	Down ($d$)	$-e/3$
2	Charm ($c$)	$+2e/3$	Strange ($s$)	$-e/3$
3	Top ($t$)	$+2e/3$	Bottom ($b$)	$-e/3$

Quark Confinement

Quarks are never observed in isolation. They are always bound into colour-neutral combinations:

• Baryons: Three quarks (one of each colour, or colour-anticolour combinations that cancel). Examples: proton ($uud$), neutron ($udd$).
• Mesons: A quark--antiquark pair. Examples: pion ($\pi^+ = u\bar{d}$), kaon ($K^+ = u\bar{s}$).

The strong force increases with distance (unlike gravity and electromagnetism, which decrease). Pulling quarks apart stores energy in the colour field until it is energetically favourable to create a new quark--antiquark pair (quark--antiquark pair production).

Properties of Quarks

Each quark possesses: electric charge, colour charge, baryon number ($+1/3$ each), and flavour quantum numbers (strangeness, charm, etc.). Antiquarks have opposite signs for all these quantities.

3. Leptons

Leptons are fundamental particles that do not experience the strong force.

The Six Leptons

Generation	Charged lepton	Neutrino
1	Electron ($e^-$)	Electron neutrino ($\nu_e$)
2	Muon ($\mu^-$)	Muon neutrino ($\nu_\mu$)
3	Tau ($\tau^-$)	Tau neutrino ($\nu_\tau$)

Each lepton has a corresponding antiparticle ($e^+$, $\bar{\nu}_e$, etc.).

Conservation of Lepton Number

Lepton number $L_e$, $L_\mu$, $L_\tau$ are conserved separately in all interactions. For example, in beta-minus decay:

$n \to p + e^- + \bar{\nu}_e$

$L_e$: $0 \to 0 + 1 + (-1) = 0$. Conserved.

4. Hadrons: Baryons and Mesons

Hadrons are composite particles made of quarks that experience the strong force.

Baryons

Baryons consist of three quarks. They have baryon number $B = +1$ (antibaryons: $B = -1$).

Particle	Quark content	Charge	Strangeness
Proton ($p$)	$uud$	$+e$	$0$
Neutron ($n$)	$udd$	$0$	$0$
$\Sigma^+$	$uus$	$+e$	$-1$
$\Xi^-$	$dss$	$-e$	$-2$
$\Omega^-$	$sss$	$-e$	$-3$

Mesons

Mesons consist of a quark--antiquark pair. They have baryon number $B = 0$.

Particle	Quark content	Charge	Strangeness
$\pi^+$	$u\bar{d}$	$+e$	$0$
$\pi^-$	$\bar{u}d$	$-e$	$0$
$\pi^0$	$u\bar{u}$ or $d\bar{d}$	$0$	$0$
$K^+$	$u\bar{s}$	$+e$	$+1$
$K^-$	$\bar{u}s$	$-e$	$-1$

Verifying Quark Content

Proton charge: $q_p = 2\!\left(\frac{+2e}{3}\right) + \frac{-e}{3} = \frac{4e - e}{3} = +e$. $\checkmark$

Neutron charge: $q_n = \frac{+2e}{3} + 2\!\left(\frac{-e}{3}\right) = \frac{2e - 2e}{3} = 0$. $\checkmark$

Beta-minus decay of a neutron:

$udd \to uud + e^- + \bar{\nu}_e$

A $d$ quark converts to a $u$ quark (via the weak interaction, mediated by a $W^-$ boson):

$d \to u + W^-, \qquad W^- \to e^- + \bar{\nu}_e$

Strangeness changes by $\Delta S = +1$ (a strange quark is destroyed), consistent with the weak interaction (which does not conserve strangeness).

5. Conservation Laws

In all particle interactions, the following quantities are always conserved:

Quantity	Conserved in all interactions?
Energy	Yes
Momentum	Yes
Electric charge	Yes
Lepton number ($L_e$, $L_\mu$, $L_\tau$)	Yes
Baryon number ($B$)	Yes
Strangeness ($S$)	Strong and EM only (not weak)

warning

warning not by the weak interaction. Strange particles are always produced in pairs (associated production) via the strong interaction (conserving $S$) but decay individually via the weak interaction.

Details

Worked Example: Conservation Check

Verify conservation laws for: $K^- + p \to \pi^+ + \pi^-$.

Quark content: $K^- = \bar{u}s$, $p = uud$, $\pi^+ = u\bar{d}$, $\pi^- = \bar{u}d$.

Charge: $-1 + 1 = 1 + (-1) = 0$. Conserved. Baryon number: $0 + 1 = 0 + 0 = 1$. NOT conserved ($1 \neq 0$).

This reaction cannot occur because baryon number is not conserved.

Corrected reaction: $K^- + p \to \Lambda^0 + \pi^0$ (or other baryon + meson combinations).

$\Lambda^0 = uds$, $\pi^0 = u\bar{u}$ or $d\bar{d}$.

Charge: $-1 + 1 = 0 + 0 = 0$. Conserved. Baryon number: $0 + 1 = 1 + 0 = 1$. Conserved. Strangeness: $+1 + 0 = -1 + 0 = -1$. Conserved.

6. Antiparticles

Every particle has a corresponding antiparticle with the same mass but opposite values of all quantum numbers (charge, baryon number, lepton number, strangeness).

Particle	Antiparticle	Key difference
Electron ($e^-$)	Positron ($e^+$)	Charge reversed
Proton ($p$)	Antiproton ($\bar{p}$)	Charge and baryon number reversed
Neutrino ($\nu_e$)	Antineutrino ($\bar{\nu}_e$)	Lepton number reversed

Pair Production and Annihilation

Pair production: A photon with energy at least $2m_e c^2 = 1.022$ MeV can create an electron--positron pair (usually near a nucleus to conserve momentum):

$\gamma \to e^- + e^+$

Annihilation: When a particle meets its antiparticle, they annihilate, converting their combined rest mass energy into photons:

$e^- + e^+ \to 2\gamma$

Two photons are required (not one) to conserve both energy and momentum.

Dirac's Prediction

Dirac (1928) combined quantum mechanics with special relativity, obtaining an equation that naturally predicted antiparticles. The positron ($e^+$) was discovered by Anderson (1932) in cosmic ray photographs, confirming Dirac's prediction.

7. Feynman Diagrams

Feynman diagrams are pictorial representations of particle interactions. Each diagram corresponds to a mathematical term in the perturbation theory expansion of the interaction amplitude.

Conventions

• Straight lines: fermions (quarks, leptons).
• Wavy lines: photons (electromagnetic interaction).
• Wavy/spring lines with arrow: $W^\pm$, $Z^0$ bosons (weak interaction).
• Curly lines: gluons (strong interaction).
• Time flows from left to right.
• Particles are labelled with their symbols.
• Arrows on fermion lines indicate the direction of fermion number flow (forward for particles, backward for antiparticles).

Beta-Minus Decay

$n \to p + e^- + \bar{\nu}_e$

Diagram: A $d$ quark line enters, emits a $W^-$ boson (wavy line), and continues as a $u$ quark line. The $W^-$ decays into an electron line and an antineutrino line.

At the quark level: $d \to u + W^-$, then $W^- \to e^- + \bar{\nu}_e$.

Electron--Positron Annihilation

$e^- + e^+ \to \gamma \to \mu^- + \mu^+$

Diagram: An $e^-$ line and an $e^+$ line (arrow reversed) meet at a vertex, connected by a photon line. The photon line connects to a second vertex where a $\mu^-$ line and $\mu^+$ line emerge.

Key Vertices

Each vertex in a Feynman diagram represents a fundamental interaction and must conserve all applicable quantum numbers:

1. QED vertex: A fermion line, an antifermion line, and a photon line meet. Charge is conserved.
2. Weak vertex: A fermion line changes flavour (e.g., $d \to u$), connected by a $W$ or $Z$ boson.
3. QCD vertex: A quark line emits or absorbs a gluon, changing colour but not flavour.

8. Photoelectric Effect

Einstein's Equation

When photons of frequency $f$ strike a metal surface, electrons are emitted only if $hf \gt \phi$, where $\phi$ is the work function of the metal.

$\boxed{hf = E_k^{\max} + \phi}$

where $E_k^{\max}$ is the maximum kinetic energy of the emitted photoelectrons.

Key Observations

1. Threshold frequency: $f_0 = \phi/h$. No emission below this frequency, regardless of intensity.
2. Instantaneous emission: Electrons are emitted within $10^{-9}$ s of illumination. This rules out a classical energy-accumulation model.
3. Intensity effect: Increasing intensity increases the number of photoelectrons (more photons) but not their maximum kinetic energy.
4. Frequency effect: Increasing frequency above threshold increases $E_k^{\max}$ linearly.

Proof of the Threshold Frequency

At the threshold, $E_k^{\max} = 0$, so $hf_0 = \phi$:

$\boxed{f_0 = \frac◆LB◆\phi◆RB◆◆LB◆h◆RB◆}$

For frequencies below $f_0$, no electron can be emitted regardless of intensity, because each photon carries insufficient energy. Increasing intensity means more photons, not more energy per photon.

Details

Worked Example: Photoelectric Effect

Light of wavelength 400 nm strikes a zinc plate ($\phi = 4.30$ eV). Calculate the maximum kinetic energy of the emitted photoelectrons and determine whether emission occurs.

Answer. $E_{\mathrm{photon}} = hf = hc/\lambda = (6.63 \times 10^{-34} \times 3.00 \times 10^8)/(400 \times 10^{-9}) = 4.97 \times 10^{-19}$ J $= 3.11$ eV.

Since $3.11\ \mathrm{eV} \lt 4.30\ \mathrm{eV} = \phi$, no photoelectrons are emitted.

For emission, the minimum wavelength is: $\lambda_{\min} = hc/\phi = (6.63 \times 10^{-34} \times 3.00 \times 10^8)/(4.30 \times 1.60 \times 10^{-19}) = 2.89 \times 10^{-7}$ m $= 289$ nm (UV).

9. Electron Diffraction and de Broglie Wavelength

de Broglie's Hypothesis

Hypothesis. Every particle with momentum $p$ has an associated wavelength:

$\boxed{\lambda = \frac{h}{p}}$

where $h = 6.63 \times 10^{-34}$ J s is Planck's constant.

This unifies the wave--particle duality: matter particles exhibit wave-like properties (diffraction, interference) just as electromagnetic waves exhibit particle-like properties (photoelectric effect).

Derivation for Non-Relativistic Electrons

An electron accelerated through potential difference $V$ gains kinetic energy:

$\frac{1}{2}m_e v^2 = eV \implies v = \sqrt◆LB◆\frac{2eV}{m_e}◆RB◆$

$p = m_e v = \sqrt{2m_e eV}$

$\boxed{\lambda = \frac◆LB◆h◆RB◆◆LB◆\sqrt{2m_e eV}◆RB◆}$

For $V = 100$ V: $\lambda = 6.63 \times 10^{-34}/\sqrt◆LB◆2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 100◆RB◆ = 1.23 \times 10^{-10}$ m $= 0.123$ nm.

This is comparable to atomic spacing, explaining why electron diffraction can resolve crystal structures.

Davisson--Germer Experiment (1927)

Davisson and Germer directed a beam of electrons at a nickel crystal and observed a diffraction pattern — sharp intensity maxima at specific angles. The angles matched the prediction of the de Broglie wavelength using the Bragg condition:

$n\lambda = 2d\sin\theta$

This provided direct experimental confirmation of wave--particle duality for matter.

Evidence for Wave Nature of Particles

Phenomenon	Particle	Evidence of wave nature
Electron diffraction	Electron	Davisson--Germer (1927)
Neutron diffraction	Neutron	Crystal diffraction patterns
Electron interference	Electron	Double-slit experiment
Molecular diffraction	$C_{60}$ (fullerene)	Interference fringes (Arndt, 1999)

Details

Worked Example: de Broglie Wavelength

Calculate the de Broglie wavelength of (a) an electron with kinetic energy 150 eV, (b) a proton moving at $2.0 \times 10^6$ m s$^{-1}$.

Answer. (a) $\lambda = h/\sqrt{2m_e eV} = 6.63 \times 10^{-34}/\sqrt◆LB◆2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 150◆RB◆ = 6.63 \times 10^{-34}/\sqrt◆LB◆4.37 \times 10^{-47}◆RB◆ = 6.63 \times 10^{-34}/6.61 \times 10^{-24} = 1.00 \times 10^{-10}$ m $= 0.100$ nm.

(b) $\lambda = h/(m_p v) = 6.63 \times 10^{-34}/(1.67 \times 10^{-27} \times 2.0 \times 10^6) = 1.98 \times 10^{-13}$ m.

The proton's wavelength is about 500 times shorter than the electron's for comparable energies, because of its much larger mass.

10. Wave--Particle Duality: Unification

The de Broglie relation $\lambda = h/p$ and the Einstein relation $E = hf$ together imply:

$E = hf = \frac◆LB◆hc◆RB◆◆LB◆\lambda◆RB◆ = pc$

for massless particles (photons). For massive particles in the non-relativistic limit:

$E_k = \frac{p^2}{2m} = \frac◆LB◆h^2◆RB◆◆LB◆2m\lambda^2◆RB◆$

These relations are the foundation of quantum mechanics. The wave function $\Psi$ of a particle satisfies the Schrodinger equation, and the probability of finding the particle in a region is $|\Psi|^2$.

Problem Set

Details

Problem 1

State the quark content of (a) a proton, (b) a neutron, (c) a $\pi^+$ meson, (d) a $K^-$ meson. Verify the electric charge in each case.

Answer. (a) $p = uud$: $2(+2e/3) + (-e/3) = +e$. $\checkmark$ (b) $n = udd$: $(+2e/3) + 2(-e/3) = 0$. $\checkmark$ (c) $\pi^+ = u\bar{d}$: $(+2e/3) + (+e/3) = +e$. $\checkmark$ (d) $K^- = \bar{u}s$: $(-2e/3) + (-e/3) = -e$. $\checkmark$

Details

Problem 2

A positron with kinetic energy 2.0 MeV collides with an electron at rest. Calculate the total energy available for photon production.

Answer. Total energy $= 2m_e c^2 + E_k = 2 \times 0.511 + 2.0 = 3.022$ MeV.

Details

Problem 3

Check whether the following reaction conserves charge, baryon number, and strangeness: $\pi^- + p \to K^0 + \Lambda^0$.

Answer. Quark content: $\pi^- = \bar{u}d$, $p = uud$, $K^0 = d\bar{s}$, $\Lambda^0 = uds$.

Charge: $-1 + 1 = 0 + 0 = 0$. Conserved. Baryon number: $0 + 1 = 0 + 1 = 1$. Conserved. Strangeness: $0 + 0 = +1 + (-1) = 0$. Conserved.

The reaction is allowed by all conservation laws (proceeds via the strong interaction).

Details

Problem 4

Light of wavelength 550 nm falls on a metal with work function 2.0 eV. Calculate the maximum kinetic energy of the photoelectrons and their maximum speed.

Answer. $E_{\mathrm{photon}} = hc/\lambda = 2.26$ eV. $E_k = hf - \phi = 2.26 - 2.0 = 0.26$ eV $= 4.16 \times 10^{-20}$ J. $v = \sqrt{2E_k/m_e} = \sqrt◆LB◆2 \times 4.16 \times 10^{-20}/9.11 \times 10^{-31}◆RB◆ = 3.02 \times 10^5$ m s$^{-1}$.

Details

Problem 5

Draw the Feynman diagram for beta-plus decay: $p \to n + e^+ + \nu_e$. Describe the quark-level process.

Answer. At the quark level: a $u$ quark converts to a $d$ quark by emitting a $W^+$ boson. The $W^+$ then decays to $e^+ + \nu_e$.

Diagram structure:

• $u$ quark line enters from the left.
• At the first vertex: $u$ emits $W^+$ and becomes $d$ (continues right).
• At the second vertex: $W^+$ splits into $e^+$ (forward arrow for antiparticle shown as backward arrow) and $\nu_e$.

Details

Problem 6

Calculate the de Broglie wavelength of a neutron with kinetic energy 0.025 eV (thermal neutron at room temperature).

Answer. $E_k = 0.025 \times 1.60 \times 10^{-19} = 4.0 \times 10^{-21}$ J. $v = \sqrt{2E_k/m_n} = \sqrt◆LB◆2 \times 4.0 \times 10^{-21}/1.67 \times 10^{-27}◆RB◆ = 2189$ m s$^{-1}$. $\lambda = h/(m_n v) = 6.63 \times 10^{-34}/(1.67 \times 10^{-27} \times 2189) = 1.82 \times 10^{-10}$ m $= 0.182$ nm.

This wavelength is comparable to interatomic spacing, which is why thermal neutrons are used for neutron diffraction studies of crystal structures.

Details

Problem 7

Explain why the $K^+$ meson ($u\bar{s}$) decays via the weak interaction with a lifetime of $\sim 10^{-8}$ s, while the $\rho^0$ meson ($u\bar{u}$) decays via the strong interaction with a lifetime of $\sim 10^{-23}$ s.

Answer. The $K^+$ contains a strange quark. Decaying the $s$ quark requires changing its flavour (since there is no lighter meson containing an $s$ quark that conserves mass-energy). Flavour change requires the weak interaction, which is much weaker than the strong force, hence the much longer lifetime ($\sim 10^{-8}$ s vs $\sim 10^{-23}$ s).

The $\rho^0$ ($u\bar{u}$) can decay to $\pi^+ + \pi^-$ via the strong interaction (no flavour change needed), so it decays almost instantaneously on the nuclear timescale.

Details

Problem 8

A particle of unknown mass is accelerated through 500 V and produces a first-order diffraction maximum at $50^\circ$ when scattered by a crystal with lattice spacing $2.0 \times 10^{-10}$ m. Identify the particle.

Answer. From Bragg's law (first order, $n = 1$): $\lambda = 2d\sin\theta = 2 \times 2.0 \times 10^{-10} \times \sin 50^\circ = 3.06 \times 10^{-10}$ m.

From $\lambda = h/\sqrt{2meV}$: $m = h^2/(2eV\lambda^2) = (6.63 \times 10^{-34})^2/(2 \times 1.60 \times 10^{-19} \times 500 \times (3.06 \times 10^{-10})^2) = 4.40 \times 10^{-67}/(1.50 \times 10^{-37}) = 2.93 \times 10^{-30}$ kg.

Comparing with known masses: $m_e = 9.11 \times 10^{-31}$ kg, $m_p = 1.67 \times 10^{-27}$ kg. The mass is approximately $3.2\,m_e$, which does not match a known fundamental particle. This suggests a systematic error or that the particle is a muon ($m_\mu = 1.88 \times 10^{-28}$ kg). For a muon: $\lambda = 6.63 \times 10^{-34}/\sqrt◆LB◆2 \times 1.88 \times 10^{-28} \times 1.60 \times 10^{-19} \times 500◆RB◆ = 6.63 \times 10^{-34}/\sqrt◆LB◆3.01 \times 10^{-44}◆RB◆ = 6.63 \times 10^{-34}/1.74 \times 10^{-22} = 3.82 \times 10^{-12}$ m.

The calculated $\lambda = 3.06 \times 10^{-10}$ m is most consistent with an electron ($\lambda_e = h/\sqrt{2m_e eV} = 6.63 \times 10^{-34}/\sqrt◆LB◆2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 500◆RB◆ = 5.49 \times 10^{-11}$ m). The discrepancy suggests an experimental issue or different scattering geometry.