Particles, Antiparticles and Photons

Antimatter, pair production, annihilation, and the photon model of electromagnetic radiation

SPEC 3.2.1.3 • PARTICLES AND RADIATION

Introduction to Antimatter

Key Principle

For every type of particle, there exists a corresponding antiparticle with the same mass but opposite charge.

In 1928, Paul Dirac predicted the existence of antimatter from his theoretical work combining quantum mechanics and special relativity.

In 1932, Carl Anderson discovered the positron (antielectron) in cosmic ray experiments, confirming Dirac's prediction.

Key Fact

When a particle meets its antiparticle, they annihilate each other, converting their mass entirely into energy (photons).

Properties of Antiparticles

Same Properties

Mass - identical to particle

Rest energy - E = mc²

Magnitude of charge

Spin magnitude

Opposite Properties

Charge - opposite sign

Lepton number

Baryon number

Strangeness

Notation

Antiparticles are shown with:

A bar over the symbol: p (antiproton), n (antineutron), ν (antineutrino)

Opposite charge sign: e⁺ (positron), e⁻ (electron)

Key Particle-Antiparticle Pairs

ParticleSymbolAntiparticleSymbolRest Mass (kg)Charge

Electrone⁻Positrone⁺9.11 × 10⁻³¹∓e

ProtonpAntiprotonp1.67 × 10⁻²⁷∓e

NeutronnAntineutronn1.67 × 10⁻²⁷0

NeutrinoνₑAntineutrinoν ₑ≈ 00

Important

The neutron and antineutron both have zero charge, but they are still different particles! They have opposite baryon numbers (+1 and −1).

Photons

Definition

A photon is a quantum (discrete packet) of electromagnetic radiation. It is the particle that carries electromagnetic force.

Photon Energy

E = h f

where h = 6.63 × 10⁻³⁴ J s (Planck's constant) and f = frequency (Hz)

Photon Properties

Zero rest mass - photons only exist when moving

Always travels at c (3.00 × 10⁸ m/s in vacuum)

No charge

No antiparticle - the photon is its own antiparticle

Rest Mass Energy

Einstein's Equation

E = m c^{2}

Rest energy (J) = mass (kg) × (speed of light)²

Every particle has energy simply by having mass. This is called rest energy or rest mass energy.

Key Values

ParticleRest Energy (J)Rest Energy (MeV)Electron / Positron8.19 × 10⁻¹⁴ J0.511 MeVProton / Antiproton1.50 × 10⁻¹⁰ J938.3 MeV

Unit Conversion

1 MeV = 1.60 × 10⁻¹³ J | 1 eV = 1.60 × 10⁻¹⁹ J

Annihilation

Definition

Annihilation occurs when a particle meets its antiparticle. Both are destroyed and their mass is converted into energy in the form of photons.

e⁻ + e⁺ → 2γ (two gamma photons)

Key Points

Minimum 2 photons produced (to conserve momentum)

Total energy of photons = rest mass energy of both particles

For electron-positron: E_total = 2 × 0.511 MeV = 1.022 MeV

Each photon carries E = m₀c² = 0.511 MeV (minimum)

Pair Production

Definition

Pair production occurs when a high-energy photon is converted into a particle-antiparticle pair. This is the reverse of annihilation.

γ → e⁻ + e⁺ (near a nucleus)

Threshold Energy

E_{p h o t o n} \geq 2 m_{0} c^{2}

For e⁻/e⁺ pair: E ≥ 2 × 0.511 = 1.022 MeV

Why Pair Production Needs a Nucleus

Pair production cannot occur in empty space - it must happen near a nucleus. Why?

Momentum Conservation

A single photon has momentum

p = \frac{E}{c}

. If it converted directly into an electron-positron pair, both particles would need to share this momentum.

However, mathematically it's impossible to conserve both energy AND momentum when one photon becomes two particles of equal mass.

The Solution

A nearby nucleus can absorb some momentum, allowing both conservation laws to be satisfied. The nucleus recoils slightly, carrying away the excess momentum.

Exam Tip

If asked why pair production needs a nucleus, the answer is: "To conserve momentum" - the nucleus absorbs recoil momentum.

Key Equations

Photon Energy

E = h f

Energy = Planck's constant × frequency

Rest Mass Energy

E = m c^{2}

Energy = mass × (speed of light)²

Annihilation

Minimum photon energy:

E = m₀c²

(per photon, 2 photons produced)

Pair Production

Threshold photon energy:

E ≥ 2m₀c²

(creates particle + antiparticle)

01Worked Example

Calculate the rest mass energy of an electron in MeV.

Mass of electron = 9.11 × 10⁻³¹ kg, c = 3.00 × 10⁸ m/s

1

Apply E = mc²

2

E = (9.11 × 10⁻³¹) × (3.00 × 10⁸)²

3

E = 9.11 × 10⁻³¹ × 9.00 × 10¹⁶ = 8.20 × 10⁻¹⁴ J

4

Convert to MeV: E = 8.20 × 10⁻¹⁴ ÷ (1.60 × 10⁻¹³)

Answer: E = 0.511 MeV

02Worked Example

Calculate the minimum frequency of a photon required to produce an electron-positron pair.

Rest energy of electron = 0.511 MeV

1

Minimum energy = 2m₀c² = 2 × 0.511 = 1.022 MeV

2

Convert to Joules: E = 1.022 × 1.60 × 10⁻¹³ = 1.64 × 10⁻¹³ J

3

Use E = hf, so f = E/h

4

f = (1.64 × 10⁻¹³) ÷ (6.63 × 10⁻³⁴)

Answer: f = 2.47 × 10²⁰ Hz

Worked Example03

An electron and positron annihilate at rest. Calculate the energy of each photon produced.

1

Total rest mass energy = 2 × m₀c² = 2 × 0.511 MeV

2

Total energy = 1.022 MeV

3

Two photons are produced (momentum conservation)

4

Energy per photon = 1.022 ÷ 2 = 0.511 MeV

Answer: Each photon has energy 0.511 MeV (8.19 × 10⁻¹⁴ J)

Worked Example04

A photon has energy 2.5 MeV. Can it produce a proton-antiproton pair? (Rest mass energy of proton = 938.3 MeV)

1

Minimum energy for proton-antiproton pair = 2 × 938.3 MeV

2

Minimum energy = 1876.6 MeV

3

Compare: Photon energy (2.5 MeV) << Required (1876.6 MeV)

Answer: No, the photon energy is far too low. It would need at least 1876.6 MeV.

Worked Example05

A photon with energy 1.8 MeV undergoes pair production near a nucleus, creating an electron-positron pair. Calculate the total kinetic energy of the electron and positron.

1

Photon energy = 1.8 MeV

2

Energy to create pair = 2 × 0.511 = 1.022 MeV

3

Excess energy becomes kinetic energy

4

KE_total = 1.8 − 1.022 = 0.778 MeV

Answer: Total KE = 0.778 MeV (shared between electron and positron)

Worked Example06

Calculate the wavelength of the photons produced when an electron and positron annihilate at rest.

1

Each photon energy = 0.511 MeV = 8.19 × 10⁻¹⁴ J

2

Use E = hc/λ, so λ = hc/E

3

λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (8.19 × 10⁻¹⁴)

4

λ = 1.99 × 10⁻²⁵ ÷ 8.19 × 10⁻¹⁴ = 2.43 × 10⁻¹² m

Answer: λ = 2.43 × 10⁻¹² m (gamma ray wavelength)

Worked Example07

A gamma ray photon with wavelength 8.0 × 10⁻¹³ m passes near a nucleus. (a) Calculate the photon energy in MeV. (b) Can it produce an electron-positron pair? (c) If yes, calculate the KE given to the particles.

1

(a) E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (8.0 × 10⁻¹³)

2

E = 2.49 × 10⁻¹³ J = 2.49 × 10⁻¹³ ÷ 1.60 × 10⁻¹³ = 1.56 MeV

3

(b) Threshold = 1.022 MeV. Since 1.56 > 1.022, Yes, pair production can occur

4

(c) KE = 1.56 − 1.022 = 0.54 MeV (shared between e⁻ and e⁺)

Practice

Question 1: A positron and electron, both at rest, annihilate. Calculate the frequency of each photon produced.

1

Energy per photon = m₀c² = 0.511 MeV = 8.19 × 10⁻¹⁴ J

2

f = E/h = (8.19 × 10⁻¹⁴) ÷ (6.63 × 10⁻³⁴)

Answer: f = 1.24 × 10²⁰ Hz

Practice

Question 2: What is the minimum photon energy required to create a muon-antimuon pair? (Muon rest energy = 105.7 MeV)

1

Minimum energy = 2 × rest energy of muon

2

E_min = 2 × 105.7 MeV

Answer: E_min = 211.4 MeV

Practice

Question 3: A photon has energy 3.2 × 10⁻¹³ J. (a) Express this energy in MeV. (b) Calculate the frequency of this photon. (c) Can this photon undergo pair production to create an electron-positron pair?

1

(a) E = 3.2 × 10⁻¹³ ÷ 1.60 × 10⁻¹³ = 2.0 MeV

2

(b) f = E/h = 3.2 × 10⁻¹³ ÷ 6.63 × 10⁻³⁴ = 4.83 × 10²⁰ Hz

3

(c) Threshold = 1.022 MeV. Since 2.0 > 1.022: Yes

Summary

Antiparticles

ParticleAntiparticleKey Property

Electron (e⁻)Positron (e⁺)Same mass, opposite charge

Proton (p)Antiproton ( p )Same mass, opposite charge

Neutron (n)Antineutron ( n )Both neutral, opposite baryon number

Annihilation

Particle + Antiparticle → 2γ

Min photon energy = m₀c²

Pair Production

γ → Particle + Antiparticle

Threshold: E ≥ 2m₀c²

Needs a nucleus nearby

Key Equations

E = hf | E = mc² | 1 MeV = 1.60 × 10⁻¹³ J