Worked example 1
Calculate the energy, in MeV, of a single gamma photon produced during a positron-electron annihilation event. Use the following data: mass of an electron/positron = 9.11 × 10⁻³¹ kg; speed of light c = 3.00 × 10⁸ m s⁻¹; elementary charge e = 1.60 × 10⁻¹⁹ C.
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- Identify the principle: The total rest mass of the positron and electron is converted into the energy of two gamma photons, governed by E = mc².
- Calculate total mass: The total mass annihilated is the sum of the electron's mass and the positron's mass. m_total = m_electron + m_positron = 2 × (9.11 × 10⁻³¹ kg) = 1.822 × 10⁻³⁰ kg.
- Calculate total energy in Joules: Substitute the total mass into the mass-energy equivalence formula. E_total = m_total × c² = (1.822 × 10⁻³⁰ kg) × (3.00 × 10⁸ m s⁻¹)² = 1.6398 × 10⁻¹³ J.
- Calculate energy per photon in Joules: This total energy is shared equally between the two gamma photons. E_photon = E_total / 2 = (1.6398 × 10⁻¹³ J) / 2 = 8.199 × 10⁻¹⁴ J.
- Convert energy to eV: To express the energy in electron-volts (eV), divide the energy in Joules by the elementary charge, e. E_photon (eV) = (8.199 × 10⁻¹⁴ J) / (1.60 × 10⁻¹⁹ J/eV) = 512,437.5 eV.
- Convert to MeV: Convert from eV to Mega-electron-volts (MeV) by dividing by 10⁶. E_photon (MeV) = 512,437.5 eV / 10⁶ = 0.512 MeV.
Final Answer: The energy of each gamma photon is 0.512 MeV (to 3 significant figures).