Annihilation of Matter & Antimatter (Edexcel International A Level (IAL) Physics): Revision Note

Exam code: YPH11

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Annihilation of Matter & Antimatter

Annihilation

  • When a particle meets its antiparticle partner, the two will annihilate

  • Annihilation is:

When a particle meets its equivalent antiparticle, they both are destroyed, and their mass is converted into energy in the form of two gamma ray photons

2.1.5Annihilation

When an electron and positron collide, their mass is converted into energy in the form of two photons emitted in opposite directions

Pair Production

  • Pair production is the opposite of annihilation

  • Pair production is:

When a photon interacts with a nucleus or atom and the energy of the photon is used to create a particle–antiparticle pair

  • The presence of a nearby nucleus is essential in pair production so that the process conserves both energy and momentum

  • A single photon alone cannot produce a particle–antiparticle pair, or the conservation laws would be broken

  • Pair creation is a case of energy being converted into matter

2.2.5 Pair Production

When a photon with enough energy interacts with a nucleus it can produce an electron-positron pair

  • This means the energy of the photon must be above a certain value to provide the total rest mass energy of the particle–antiparticle pair

  • Einstein's famous mass-energy relation showed that energy can be converted into mass, and vice versa

  • It is given by:

ΔE = c2 Δm

  • Where:

    • Δm = rest mass of the particle (kg)

    • c = speed of light (m s–1)

    • ΔE = rest mass energy of the particle (J)

  • Therefore, in order to create a particle & anti-particle pair, the energy carried by a single photon must be at least twice the rest-mass energy required, i.e.

2ΔE = 2(c2 Δm)

  • This also means if a particle meets its anti-particle and annihilates, the energy carried away by each of the two photons Ephoton is given by: 

Ephoton = hf = hcλ= c2 Δm

Worked Example

Calculate the maximum wavelength of one of the photons produced when a proton and antiproton annihilate.

Rest mass of a proton = 938.257 MeV/c2

Answer:

Step 1: Write down the known quantities

  • Rest mass-energy of a proton (and antiproton) = 938.257 MeV

  • 1 MeV = 1.60×10−13 J

Step 2: Determine the total rest mass energy of the proton-antiproton pair

Etotal = 2 × 938.257 = 1876.514 MeV

Step 3: Determine the minimum energy of one of the photons in J

Emin = 1876.514 × (1.60 × 10−13)2 = 1.5012 × 10−10 J

  • Note: since the question asks for the energy of one of the photons, you could skip step 2 and simply use 938.257 MeV

Step 4: Write down the equation for minimum photon energy

Emin = hfmin = hcλmax

Step 5: Rearrange for the maximum photon wavelength

λmax = hcEmin

Step 6: Calculate the maximum photon wavelength

λmax = (6.63 × 10−34)(3.0 × 108)1.5012 × 10−10

λmax = 1.32 × 10−15 m

Examiner Tips and Tricks

Since the Planck constant is in Joules (J s) remember to always convert the rest mass-energy from MeV to J.

Remember that the equation E = hf is only relevant for photons, not for all particles!

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Katie M

Author: Katie M

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Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

Caroline Carroll

Reviewer: Caroline Carroll

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Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.