understand that an antiparticle has the same mass but opposite charge to the corresponding particle, and that a positron is the antiparticle of an electron

Antiparticles

In modern particle physics every fundamental particle has a corresponding antiparticle.

The antiparticle has exactly the same rest‑mass, spin and intrinsic angular momentum as its partner, but opposite electric charge and opposite values of all additive quantum numbers (electric charge, lepton number, baryon number, flavour numbers, etc.).

Key properties

  • Identical rest mass: \(m{\text{particle}} = m{\text{antiparticle}}\)
  • Opposite electric charge: \(q{\text{antiparticle}} = -\,q{\text{particle}}\)
  • Opposite additive quantum numbers: lepton number \(L\), baryon number \(B\), flavour numbers (strangeness, charm, …), colour charge (for quarks) all change sign.
  • Same spin: the intrinsic spin value is unchanged.

Particle–antiparticle table (Cambridge 9702)

ParticleSymbolRest mass
(MeV \(c^{-2}\))
Electric chargeLepton / Baryon numberAntiparticleAntiparticle symbol
Electron\(e^{-}\)0.511\(-1e\)\(L=+1\)Positron\(e^{+}\)
Positron\(e^{+}\)0.511\(+1e\)\(L=-1\)Electron\(e^{-}\)
Muon\(\mu^{-}\)105.7\(-1e\)\(L_{\mu}=+1\)Antimuon\(\mu^{+}\)
Antimuon\(\mu^{+}\)105.7\(+1e\)\(L_{\mu}=-1\)Muon\(\mu^{-}\)
Tau\(\tau^{-}\)1776.9\(-1e\)\(L_{\tau}=+1\)Antitau\(\tau^{+}\)
Antitau\(\tau^{+}\)1776.9\(+1e\)\(L_{\tau}=-1\)Tau\(\tau^{-}\)
Electron‑type neutrino\(\nu_{e}\)≈00\(L=+1\)Electron‑type antineutrino\(\bar{\nu}_{e}\)
Muon‑type neutrino\(\nu_{\mu}\)≈00\(L_{\mu}=+1\)Muon‑type antineutrino\(\bar{\nu}_{\mu}\)
Tau‑type neutrino\(\nu_{\tau}\)≈00\(L_{\tau}=+1\)Tau‑type antineutrino\(\bar{\nu}_{\tau}\)
Proton\(p\)938.27\(+1e\)\(B=+1\)Antiproton\(\bar p\)
Antiproton\(\bar p\)938.27\(-1e\)\(B=-1\)Proton\(p\)
Neutron\(n\)939.570\(B=+1\)Antineutron\(\bar n\)
Antineutron\(\bar n\)939.570\(B=-1\)Neutron\(n\)
Up‑quark\(u\)≈2.2\(+\tfrac{2}{3}e\)\(B=+\tfrac{1}{3}\)Anti‑up‑quark\(\bar u\)
Down‑quark\(d\)≈4.7\(-\tfrac{1}{3}e\)\(B=+\tfrac{1}{3}\)Anti‑down‑quark\(\bar d\)
Strange‑quark\(s\)≈95\(-\tfrac{1}{3}e\)\(B=+\tfrac{1}{3}\)Anti‑strange‑quark\(\bar s\)
Charm‑quark\(c\)≈1270\(+\tfrac{2}{3}e\)\(B=+\tfrac{1}{3}\)Anti‑charm‑quark\(\bar c\)
Bottom‑quark\(b\)≈4180\(-\tfrac{1}{3}e\)\(B=+\tfrac{1}{3}\)Anti‑bottom‑quark\(\bar b\)
Top‑quark\(t\)≈173 000\(+\tfrac{2}{3}e\)\(B=+\tfrac{1}{3}\)Anti‑top‑quark\(\bar t\)

Quark flavours (summary)

FlavourSymbolChargeBaryon number
Up\(u\)\(+\tfrac{2}{3}e\)\(+\tfrac{1}{3}\)
Down\(d\)\(-\tfrac{1}{3}e\)\(+\tfrac{1}{3}\)
Strange\(s\)\(-\tfrac{1}{3}e\)\(+\tfrac{1}{3}\)
Charm\(c\)\(+\tfrac{2}{3}e\)\(+\tfrac{1}{3}\)
Bottom\(b\)\(-\tfrac{1}{3}e\)\(+\tfrac{1}{3}\)
Top\(t\)\(+\tfrac{2}{3}e\)\(+\tfrac{1}{3}\)

Hadron classification

  • Baryons: made of three quarks (e.g. proton = \(u u d\), neutron = \(u d d\)). They have baryon number \(B=+1\).
  • Mesons: made of a quark + antiquark pair (e.g. \(\pi^{+}=u\bar d\), \(K^{0}=d\bar s\)). They have baryon number \(B=0\).

Antiparticles in nuclear decay – β⁺ (positron) decay

In β⁺ decay a proton inside the nucleus is transformed into a neutron, emitting a positron and an electron‑type neutrino:

\[

^{A}{Z}\!X \;\longrightarrow\; ^{A}{Z-1}\!Y \;+\; e^{+} \;+\; \nu_{e}

\]

  • Mass number \(A\): unchanged (same total number of nucleons).
  • Atomic number \(Z\): decreases by one because a proton (\(+1e\)) becomes a neutron (0 e).
  • Charge conservation: the emitted positron carries \(+1e\); the rest of the nucleus loses one positive charge, so the total charge is the same before and after.
  • Lepton number: the parent nucleus has \(L=0\). The positron contributes \(L=-1\) and the neutrino contributes \(L=+1\); net lepton number remains 0.
  • Baryon number: unchanged (\(B=+1\) for both the original proton and the resulting neutron).

Conservation laws relevant to particle–antiparticle processes

  • Mass‑energy (E = mc²): total energy, including rest‑mass energy, is conserved.
  • Electric charge: the algebraic sum of charges is unchanged.
  • Baryon number (B): conserved in strong, electromagnetic and weak interactions.
  • Lepton number (L): conserved separately for each lepton family (electron, muon, tau).
  • Momentum and angular momentum: always conserved.

The Positron (\(e^{+}\))

The positron is the antiparticle of the electron. It has the same rest mass (0.511 MeV \(c^{-2}\)) but a positive elementary charge \((+1e)\) and lepton number \(-1\).

Discovery (1932)

C. Anderson observed cloud‑chamber tracks that curved in the opposite direction to electrons when a magnetic field was applied. The curvature indicated a particle of electron mass with positive charge – the positron.

Positron annihilation

When a positron meets an electron they can annihilate, converting their total rest‑mass energy into photons. The most common reaction is

\[

e^{-} + e^{+} \;\rightarrow\; \gamma + \gamma

\]

Each photon carries an energy equal to the electron rest‑mass energy:

\[

E{\gamma}=m{e}c^{2}=511\ \text{keV}

\]

Conservation of momentum forces the two photons to travel in opposite directions (back‑to‑back).

Applications

  • Positron Emission Tomography (PET): Radioisotopes that undergo β⁺ decay emit positrons; the resulting 511 keV photons are detected to produce images of metabolic activity.
  • Positron annihilation spectroscopy – probes vacancies and defects in solids.
  • Precision tests of charge–parity (CP) symmetry and other fundamental symmetries.

Conceptual understanding – what students need to know

  1. State that an antiparticle has the same rest mass and spin as its corresponding particle.
  2. Explain why electric charge and all additive quantum numbers (lepton number, baryon number, flavour numbers) are opposite for particle–antiparticle pairs.
  3. Identify the positron as the antiparticle of the electron and give its mass, charge and lepton number.
  4. Write the β⁺‑decay equation using the \(\,^{A}_{Z}X\) notation and show that charge, nucleon number, baryon number and lepton number are conserved.
  5. Describe electron–positron annihilation and calculate the energy of the emitted photons using \(E=mc^{2}\). Include the effect of any kinetic energy the positron may have.

Sample exam‑style question

Q: A positron with kinetic energy \(200\ \text{keV}\) annihilates with a stationary electron. Assuming the annihilation produces two photons, calculate the energy of each photon.

A:

  1. Rest‑mass energy of each particle: \(511\ \text{keV}\).
  2. Total initial energy

    \[

    E_{\text{total}} = 511\ \text{keV} \;(\text{electron}) + 511\ \text{keV} \;(\text{positron rest}) + 200\ \text{keV} \;(\text{positron KE}) = 1222\ \text{keV}.

    \]

  3. Momentum conservation forces the two photons to have equal energy, so

    \[

    E_{\gamma}= \frac{1222\ \text{keV}}{2}= 611\ \text{keV}.

    \]

Suggested diagrams: (a) Cloud‑chamber tracks showing opposite curvature for an electron and a positron in a magnetic field; (b) Schematic of electron–positron annihilation producing two back‑to‑back 511 keV photons.