Published by Patrick Mutisya · 14 days ago
Calculate the energy of the gamma‑ray photons emitted during the annihilation of an electron–positron pair.
When an electron (\$e^{-}\$) meets its antiparticle, the positron (\$e^{+}\$), they can annihilate, converting their rest mass completely into electromagnetic radiation. Because the rest mass energy of an electron (or positron) is large compared with typical atomic transition energies, the photons produced are in the gamma‑ray region.
The total energy released is the sum of the rest‑mass energies of the two particles:
\$E{\text{total}} = 2\,m{e}c^{2}\$
where \$m_{e}=9.109\times10^{-31}\,\text{kg}\$ is the electron mass and \$c=3.00\times10^{8}\,\text{m s}^{-1}\$ is the speed of light.
Conservation of momentum requires that the two photons are emitted in opposite directions with equal energy. Therefore:
\$E{\gamma}=m{e}c^{2}\$
In electron‑volts (eV) the rest‑mass energy is a well‑known constant:
\$m_{e}c^{2}=511\ \text{keV}\$
| Constant | Symbol | Value | Units |
|---|---|---|---|
| Electron rest mass | \$m_{e}\$ | 9.109 × 10⁻³¹ | kg |
| Speed of light | \$c\$ | 3.00 × 10⁸ | m s⁻¹ |
| Electron charge (for conversion) | \$e\$ | 1.602 × 10⁻¹⁹ | C |
| Rest‑mass energy (electron) | \$m_{e}c^{2}\$ | 511 000 | eV |
\$E{\gamma}=m{e}c^{2}\$
\$\$E_{\gamma}= (9.109\times10^{-31}\,\text{kg})(3.00\times10^{8}\,\text{m s}^{-1})^{2}
=8.187\times10^{-14}\,\text{J}\$\$
\$\$E_{\gamma}= \frac{8.187\times10^{-14}\,\text{J}}{1.602\times10^{-19}\,\text{J eV}^{-1}}
\approx 5.11\times10^{5}\,\text{eV}=511\ \text{keV}\$\$
The annihilation of an electron–positron pair converts the total rest‑mass energy \$2m_{e}c^{2}\$ into two photons. Each photon therefore has an energy equal to the rest‑mass energy of a single electron, \$511\ \text{keV}\$, which is in the gamma‑ray region. This simple calculation underpins many applications, from PET scanners to fundamental particle‑physics experiments.