The Standard Model classifies all known fundamental particles and their interactions. It describes:
Quarks are fundamental particles that experience the strong force. They carry fractional electric Charge and a colour charge (red, green, or blue).
Quarks are never observed in isolation. They are always bound into colour-neutral combinations:
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).
Hadrons are composite particles made of quarks that experience the strong force.
Baryons consist of three quarks. They have baryon number B = + 1 B = +1 B = + 1 B = − 1 B = -1 B = − 1
Mesons consist of a quark—antiquark pair. They have baryon number B = 0 B = 0 B = 0
(associated production) Via the strong interaction (conserving $S$) but decay individually via the weak interaction.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}$$\pi^- = \bar{u}d .
Charge: − 1 + 1 = 1 + ( − 1 ) = 0 -1 + 1 = 1 + (-1) = 0 − 1 + 1 = 1 + ( − 1 ) = 0 Baryon number: 0 + 1 = 0 + 0 = 1 0 + 1 = 0 + 0 = 1 0 + 1 = 0 + 0 = 1 1 ≠ 0 1 \neq 0 1 = 0
This reaction cannot occur because baryon number is not conserved.
Corrected reaction: K − + p → Λ 0 + π 0 K^- + p \to \Lambda^0 + \pi^0 K − + p → Λ 0 + π 0
\Lambda^0 = uds$$\pi^0 = u\bar{u} or d d ˉ d\bar{d} d d ˉ
Charge: − 1 + 1 = 0 + 0 = 0 -1 + 1 = 0 + 0 = 0 − 1 + 1 = 0 + 0 = 0 Baryon number: 0 + 1 = 1 + 0 = 1 0 + 1 = 1 + 0 = 1 0 + 1 = 1 + 0 = 1 Strangeness: + 1 + 0 = − 1 + 0 = − 1 +1 + 0 = -1 + 0 = -1 + 1 + 0 = − 1 + 0 = − 1
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 − e^- e − Positron (e + e^+ e + Charge reversed Proton (p p p Antiproton (p ˉ \bar{p} p ˉ Charge and baryon number reversed Neutrino (ν e \nu_e ν e Antineutrino (ν ˉ e \bar{\nu}_e ν ˉ e Lepton number reversed
Pair production: A photon with energy at least 2 m e c 2 = 1.022 2m_e c^2 = 1.022 2 m e c 2 = 1.022
γ → e − + e + \gamma \to e^- + e^+ γ → e − + e +
Annihilation: When a particle meets its antiparticle, they annihilate, converting their combined Rest mass energy into photons:
e − + e + → 2 γ e^- + e^+ \to 2\gamma e − + e + → 2 γ
Two photons are required (not one) to conserve both energy and momentum.
Dirac (1928) combined quantum mechanics with special relativity, obtaining an equation that Predicted antiparticles. The positron (e + e^+ e +
Feynman diagrams are pictorial representations of particle interactions. Each diagram corresponds to A mathematical term in the perturbation theory expansion of the interaction amplitude.
Straight lines: fermions (quarks, leptons). Wavy lines: photons (electromagnetic interaction). Wavy/spring lines with arrow: W ± W^\pm W ± Z 0 Z^0 Z 0 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). n → p + e − + ν ˉ e n \to p + e^- + \bar{\nu}_e n → p + e − + ν ˉ e
Diagram: A d d d W − W^- W − u u u W − W^- W −
At the quark level: d → u + W − d \to u + W^- d → u + W − W − → e − + ν ˉ e W^- \to e^- + \bar{\nu}_e W − → e − + ν ˉ e
e − + e + → γ → μ − + μ + e^- + e^+ \to \gamma \to \mu^- + \mu^+ e − + e + → γ → μ − + μ +
Diagram: An e − e^- e − e + e^+ e + μ − \mu^- μ − μ + \mu^+ μ +
Each vertex in a Feynman diagram represents a fundamental interaction and must conserve all Applicable quantum numbers:
QED vertex: A fermion line, an antifermion line, and a photon line meet. Charge is conserved.Weak vertex: A fermion line changes flavour (e.g., d → u d \to u d → u W W W Z Z Z QCD vertex: A quark line emits or absorbs a gluon, changing colour but not flavour.When photons of frequency f f f h f > ϕ hf \gt \phi h f > ϕ ϕ \phi ϕ
h f = E k max + ϕ \boxed{hf = E_k^{\max} + \phi} h f = E k m a x + ϕ
Where E k max E_k^{\max} E k m a x
Threshold frequency: f 0 = ϕ / h f_0 = \phi/h f 0 = ϕ / h Instantaneous emission: Electrons are emitted within 10 − 9 10^{-9} 1 0 − 9 Intensity effect: Increasing intensity increases the number of photoelectrons (more photons) but not their maximum kinetic energy.Frequency effect: Increasing frequency above threshold increases E k max E_k^{\max} E k m a x At the threshold, E k max = 0 E_k^{\max} = 0 E k m a x = 0 h f 0 = ϕ hf_0 = \phi h f 0 = ϕ
f 0 = ϕ h \boxed{f_0 = \frac{\phi}{h}} f 0 = h ϕ
For frequencies below f 0 f_0 f 0
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 p h o t o n = h f = h c / λ = ( 6.63 × 10 − 34 × 3.00 × 10 8 ) / ( 400 × 10 − 9 ) = 4.97 × 10 − 19 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} E photon = h f = h c / λ = ( 6.63 × 1 0 − 34 × 3.00 × 1 0 8 ) / ( 400 × 1 0 − 9 ) = 4.97 × 1 0 − 19 = 3.11 = 3.11 = 3.11
Since 3.11 e V < 4.30 e V = ϕ 3.11\ \mathrm{eV} \lt 4.30\ \mathrm{eV} = \phi 3.11 eV < 4.30 eV = ϕ
For emission, the minimum wavelength is: λ min = h c / ϕ = ( 6.63 × 10 − 34 × 3.00 × 10 8 ) / ( 4.30 × 1.60 × 10 − 19 ) = 2.89 × 10 − 7 \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} λ m i n = h c / ϕ = ( 6.63 × 1 0 − 34 × 3.00 × 1 0 8 ) / ( 4.30 × 1.60 × 1 0 − 19 ) = 2.89 × 1 0 − 7 = 289 = 289 = 289
Hypothesis. Every particle with momentum p p p
λ = h p \boxed{\lambda = \frac{h}{p}} λ = p h
Where h = 6.63 × 10 − 34 h = 6.63 \times 10^{-34} h = 6.63 × 1 0 − 34
This unifies the wave—particle duality: matter particles exhibit wave-like properties (diffraction, Interference) just as electromagnetic waves exhibit particle-like properties (photoelectric effect).
An electron accelerated through potential difference V V V
1 2 m e v 2 = e V ⟹ v = 2 e V m e \frac{1}{2}m_e v^2 = eV \implies v = \sqrt{\frac{2eV}{m_e}} 2 1 m e v 2 = e V ⟹ v = m e 2 e V
p = m e v = 2 m e e V p = m_e v = \sqrt{2m_e eV} p = m e v = 2 m e e V
λ = h 2 m e e V \boxed{\lambda = \frac{h}{\sqrt{2m_e eV}}} λ = 2 m e e V h
For V = 100 V = 100 V = 100 λ = 6.63 × 10 − 34 / 2 × 9.11 × 10 − 31 × 1.60 × 10 − 19 × 100 = 1.23 × 10 − 10 \lambda = 6.63 \times 10^{-34}/\sqrt{2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 100} = 1.23 \times 10^{-10} λ = 6.63 × 1 0 − 34 / 2 × 9.11 × 1 0 − 31 × 1.60 × 1 0 − 19 × 100 = 1.23 × 1 0 − 10 = 0.123 = 0.123 = 0.123
This is comparable to atomic spacing, explaining why electron diffraction can resolve crystal Structures.
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 λ = 2 d sin θ n\lambda = 2d\sin\theta nλ = 2 d sin θ
This provided direct experimental confirmation of wave—particle duality for matter.
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 C_{60} C 60 Interference fringes (Arndt, 1999)
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) λ = h / 2 m e e V = 6.63 × 10 − 34 / 2 × 9.11 × 10 − 31 × 1.60 × 10 − 19 × 150 = 6.63 × 10 − 34 / 4.37 × 10 − 47 = 6.63 × 10 − 34 / 6.61 × 10 − 24 = 1.00 × 10 − 10 \lambda = h/\sqrt{2m_e eV} = 6.63 \times 10^{-34}/\sqrt{2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 150} = 6.63 \times 10^{-34}/\sqrt{4.37 \times 10^{-47}} = 6.63 \times 10^{-34}/6.61 \times 10^{-24} = 1.00 \times 10^{-10} λ = h / 2 m e e V = 6.63 × 1 0 − 34 / 2 × 9.11 × 1 0 − 31 × 1.60 × 1 0 − 19 × 150 = 6.63 × 1 0 − 34 / 4.37 × 1 0 − 47 = 6.63 × 1 0 − 34 /6.61 × 1 0 − 24 = 1.00 × 1 0 − 10 = 0.100 = 0.100 = 0.100
(b) λ = h / ( m p v ) = 6.63 × 10 − 34 / ( 1.67 × 10 − 27 × 2.0 × 10 6 ) = 1.98 × 10 − 13 \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} λ = h / ( m p v ) = 6.63 × 1 0 − 34 / ( 1.67 × 1 0 − 27 × 2.0 × 1 0 6 ) = 1.98 × 1 0 − 13
The proton’s wavelength is about 500 times shorter than the electron’s for comparable energies, Because of its much larger mass.
The de Broglie relation λ = h / p \lambda = h/p λ = h / p E = h f E = hf E = h f
E = h f = h c λ = p c E = hf = \frac{hc}{\lambda} = pc E = h f = λ h c = p c
For massless particles (photons). For massive particles in the non-relativistic limit:
E k = p 2 2 m = h 2 2 m λ 2 E_k = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2} E k = 2 m p 2 = 2 m λ 2 h 2
These relations are the foundation of quantum mechanics. The wave function Ψ \Psi Ψ ∣ Ψ ∣ 2 |\Psi|^2 ∣Ψ ∣ 2
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 = u u d p = uud p = uu d 2 ( + 2 e / 3 ) + ( − e / 3 ) = + e 2(+2e/3) + (-e/3) = +e 2 ( + 2 e /3 ) + ( − e /3 ) = + e ✓ \checkmark ✓ n = u d d n = udd n = u dd ( + 2 e / 3 ) + 2 ( − e / 3 ) = 0 (+2e/3) + 2(-e/3) = 0 ( + 2 e /3 ) + 2 ( − e /3 ) = 0 ✓ \checkmark ✓ π + = u d ˉ \pi^+ = u\bar{d} π + = u d ˉ ( + 2 e / 3 ) + ( + e / 3 ) = + e (+2e/3) + (+e/3) = +e ( + 2 e /3 ) + ( + e /3 ) = + e ✓ \checkmark ✓ K − = u ˉ s K^- = \bar{u}s K − = u ˉ s ( − 2 e / 3 ) + ( − e / 3 ) = − e (-2e/3) + (-e/3) = -e ( − 2 e /3 ) + ( − e /3 ) = − e ✓ \checkmark ✓
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 = 2 m e c 2 + E k = 2 × 0.511 + 2.0 = 3.022 = 2m_e c^2 + E_k = 2 \times 0.511 + 2.0 = 3.022 = 2 m e c 2 + E k = 2 × 0.511 + 2.0 = 3.022
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 -1 + 1 = 0 + 0 = 0 − 1 + 1 = 0 + 0 = 0 Baryon number: 0 + 1 = 0 + 1 = 1 0 + 1 = 0 + 1 = 1 0 + 1 = 0 + 1 = 1 Strangeness: 0 + 0 = + 1 + ( − 1 ) = 0 0 + 0 = +1 + (-1) = 0 0 + 0 = + 1 + ( − 1 ) = 0
The reaction is allowed by all conservation laws (proceeds via the strong interaction).
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 p h o t o n = h c / λ = 2.26 E_{\mathrm{photon}} = hc/\lambda = 2.26 E photon = h c / λ = 2.26 E k = h f − ϕ = 2.26 − 2.0 = 0.26 E_k = hf - \phi = 2.26 - 2.0 = 0.26 E k = h f − ϕ = 2.26 − 2.0 = 0.26 = 4.16 × 10 − 20 = 4.16 \times 10^{-20} = 4.16 × 1 0 − 20 v = 2 E k / m e = 2 × 4.16 × 10 − 20 / 9.11 × 10 − 31 = 3.02 × 10 5 v = \sqrt{2E_k/m_e} = \sqrt{2 \times 4.16 \times 10^{-20}/9.11 \times 10^{-31}} = 3.02 \times 10^5 v = 2 E k / m e = 2 × 4.16 × 1 0 − 20 /9.11 × 1 0 − 31 = 3.02 × 1 0 5 − 1 ^{-1} − 1
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 u u d d d W + W^+ W + W + W^+ W + e + + ν e e^+ + \nu_e e + + ν e
Diagram structure:
u u u At the first vertex: u u u W + W^+ W + d d d At the second vertex: W + W^+ W + e + e^+ e + ν e \nu_e ν e 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 × 1.60 × 10 − 19 = 4.0 × 10 − 21 E_k = 0.025 \times 1.60 \times 10^{-19} = 4.0 \times 10^{-21} E k = 0.025 × 1.60 × 1 0 − 19 = 4.0 × 1 0 − 21 v = 2 E k / m n = 2 × 4.0 × 10 − 21 / 1.67 × 10 − 27 = 2189 v = \sqrt{2E_k/m_n} = \sqrt{2 \times 4.0 \times 10^{-21}/1.67 \times 10^{-27}} = 2189 v = 2 E k / m n = 2 × 4.0 × 1 0 − 21 /1.67 × 1 0 − 27 = 2189 − 1 ^{-1} − 1 λ = h / ( m n v ) = 6.63 × 10 − 34 / ( 1.67 × 10 − 27 × 2189 ) = 1.82 × 10 − 10 \lambda = h/(m_n v) = 6.63 \times 10^{-34}/(1.67 \times 10^{-27} \times 2189) = 1.82 \times 10^{-10} λ = h / ( m n v ) = 6.63 × 1 0 − 34 / ( 1.67 × 1 0 − 27 × 2189 ) = 1.82 × 1 0 − 10 = 0.182 = 0.182 = 0.182
This wavelength is comparable to interatomic spacing, which is why thermal neutrons are used for Neutron diffraction studies of crystal structures.
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 + K^+ K + s s s s s s ∼ 10 − 8 \sim 10^{-8} ∼ 1 0 − 8 ∼ 10 − 23 \sim 10^{-23} ∼ 1 0 − 23
The ρ 0 \rho^0 ρ 0 u u ˉ u\bar{u} u u ˉ π + + π − \pi^+ + \pi^- π + + π −
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 n = 1 n = 1 λ = 2 d sin θ = 2 × 2.0 × 10 − 10 × sin 50 ∘ = 3.06 × 10 − 10 \lambda = 2d\sin\theta = 2 \times 2.0 \times 10^{-10} \times \sin 50^\circ = 3.06 \times 10^{-10} λ = 2 d sin θ = 2 × 2.0 × 1 0 − 10 × sin 5 0 ∘ = 3.06 × 1 0 − 10
From λ = h / 2 m e V \lambda = h/\sqrt{2meV} λ = h / 2 m e V m = h 2 / ( 2 e V λ 2 ) = ( 6.63 × 10 − 34 ) 2 / ( 2 × 1.60 × 10 − 19 × 500 × ( 3.06 × 10 − 10 ) 2 ) = 4.40 × 10 − 67 / ( 1.50 × 10 − 37 ) = 2.93 × 10 − 30 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} m = h 2 / ( 2 e V λ 2 ) = ( 6.63 × 1 0 − 34 ) 2 / ( 2 × 1.60 × 1 0 − 19 × 500 × ( 3.06 × 1 0 − 10 ) 2 ) = 4.40 × 1 0 − 67 / ( 1.50 × 1 0 − 37 ) = 2.93 × 1 0 − 30
Comparing with known masses: m e = 9.11 × 10 − 31 m_e = 9.11 \times 10^{-31} m e = 9.11 × 1 0 − 31 m p = 1.67 × 10 − 27 m_p = 1.67 \times 10^{-27} m p = 1.67 × 1 0 − 27 3.2 m e 3.2\,m_e 3.2 m e m μ = 1.88 × 10 − 28 m_\mu = 1.88 \times 10^{-28} m μ = 1.88 × 1 0 − 28 λ = 6.63 × 10 − 34 / 2 × 1.88 × 10 − 28 × 1.60 × 10 − 19 × 500 = 6.63 × 10 − 34 / 3.01 × 10 − 44 = 6.63 × 10 − 34 / 1.74 × 10 − 22 = 3.82 × 10 − 12 \lambda = 6.63 \times 10^{-34}/\sqrt{2 \times 1.88 \times 10^{-28} \times 1.60 \times 10^{-19} \times 500} = 6.63 \times 10^{-34}/\sqrt{3.01 \times 10^{-44}} = 6.63 \times 10^{-34}/1.74 \times 10^{-22} = 3.82 \times 10^{-12} λ = 6.63 × 1 0 − 34 / 2 × 1.88 × 1 0 − 28 × 1.60 × 1 0 − 19 × 500 = 6.63 × 1 0 − 34 / 3.01 × 1 0 − 44 = 6.63 × 1 0 − 34 /1.74 × 1 0 − 22 = 3.82 × 1 0 − 12
The calculated λ = 3.06 × 10 − 10 \lambda = 3.06 \times 10^{-10} λ = 3.06 × 1 0 − 10 λ e = h / 2 m e e V = 6.63 × 10 − 34 / 2 × 9.11 × 10 − 31 × 1.60 × 10 − 19 × 500 = 5.49 × 10 − 11 \lambda_e = h/\sqrt{2m_e eV} = 6.63 \times 10^{-34}/\sqrt{2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 500} = 5.49 \times 10^{-11} λ e = h / 2 m e e V = 6.63 × 1 0 − 34 / 2 × 9.11 × 1 0 − 31 × 1.60 × 1 0 − 19 × 500 = 5.49 × 1 0 − 11
Rounding intermediate answers too early, which compounds errors in multi-step calculations.
Forgetting to include units in final answers, especially when working with derived units like N kg − 1 m 2 \text{N}\,\text{kg}^{-1}\,\text{m}^2 N kg − 1 m 2
Neglecting air resistance or assuming ideal conditions when the question specifies a real-world scenario.
Confusing scalar and vector quantities. Always check whether direction matters for the quantity in question.
The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.
Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.
Physics explores the fundamental rules governing matter, energy, space, and time. At its heart lies the principle that complex phenomena emerge from simple interactions - gravity shapes orbits, electromagnetism binds atoms, and quantum mechanics governs the subatomic realm. Understanding these laws allows us to build technologies from smartphones to spacecraft and to comprehend our place in the cosmos.
