3.2.1.3

Annihilation converts mass into photon energy

Particles, Antiparticles & Photons — AQA A-Level Physics

Key Definition
Annihilation — When a particle meets its corresponding antiparticle, both are destroyed and their mass is converted into energy in the form of two gamma-ray photons.
$$E = 2m_e c^2$$
$$\begin{aligned} E_{\min} &= hf_{\min} \\ &= E_0 \end{aligned}$$
  • $E_min$: minimum energy of one photonA quantum (discrete packet) of electromagnetic radiation. Its energy is proportional to its frequency. produced (J)
  • $h$: Planck's constant (J s)
  • $f_min$: minimum frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz). of one photonA quantum (discrete packet) of electromagnetic radiation. Its energy is proportional to its frequency. (Hz)
  • $E_0$: rest mass-energy of one of the particles (J)
Worked Example
Calculate the maximum wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). of one of the photons produced when a proton and antiproton annihilate.
Show Solution
1
List known values

Rest mass-energy of a proton $= 938.3$ MeV

$1$ MeV $= 1.60 \times 10^{-13}$ J

2
Write minimum photon energy

$$E_{\min} = E_0 = 938.3 \times 1.60 \times 10^{-13} \text{ J}$$

3
Rearrange for wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m).

Maximum wavelengthThe minimum distance between two points on a wave that are in phase (e.g. crest to crest). Measured in metres (m). corresponds to minimum energy:

$$\lambda_{\max} = \frac{hc}{E_0}$$
4
Substitute values

$$\lambda_{\max} = \frac{(6.63 \times 10^{-34}) \times (3.0 \times 10^{8})}{938.3 \times (1.60 \times 10^{-13})} = 1.32 \times 10^{-15} \text{ m}$$

Answer
$\lambda_{\max} = 1.32 \times 10^{-15}$ m
Common Mistake MEDIUM
Students often: Using MeV directly in $E = hc/\lambda$ without converting to joules.
Instead: Planck's constant is in J s, so all energies must be in joules. Multiply MeV by 1.60 × 10⁻¹³ to convert.
Particles, Antiparticles & Photons Overview