calculate the energy of the gamma-ray photons emitted during the annihilation of an electron-positron pair

Production and Use of X‑rays

What are X‑rays? 🤓

X‑rays are a type of high‑energy electromagnetic radiation. They are produced when fast electrons hit a metal target, causing the electrons to decelerate abruptly. The sudden change in speed releases energy in the form of X‑ray photons.

How are X‑rays produced? 🔧

  • Electrons are accelerated by a high voltage (kV).

  • They strike a metal anode (usually tungsten).

  • Two processes occur: bremsstrahlung (braking radiation) and characteristic radiation.

  • The X‑ray tube emits a spectrum of photon energies.

Electron‑positron annihilation and gamma rays 🌟

When an electron (\$e^-\$) meets its antimatter counterpart, a positron (\$e^+\$), they annihilate each other. The rest mass energy of the pair is converted into two gamma‑ray photons, each travelling in opposite directions.

Energy conservation gives:

\$E{\text{total}} = 2\,me c^2\$

Since the two photons share the energy equally:

\$E{\gamma} = me c^2\$

Calculating the energy of a gamma‑ray photon 📐

  1. Write down the known constants:

    • \$m_e = 9.10938356 \times 10^{-31}\,\text{kg}\$

    • \$c = 2.99792458 \times 10^8\,\text{m/s}\$

  2. Insert them into \$E = m_e c^2\$:

    \$E = (9.10938356 \times 10^{-31}\,\text{kg}) \times (2.99792458 \times 10^8\,\text{m/s})^2\$

  3. Compute the value:

    \$E \approx 8.187 \times 10^{-14}\,\text{J}\$

  4. Convert joules to electron‑volts (1 eV = 1.602 × 10⁻¹⁹ J):

    \$E \approx \frac{8.187 \times 10^{-14}\,\text{J}}{1.602 \times 10^{-19}\,\text{J/eV}} \approx 511\,\text{keV}\$

  5. Result: each photon has an energy of about 511 keV.

StepFormulaValue
1\$E = m_e c^2\$\$9.10938356 \times 10^{-31}\,\text{kg} \times (2.99792458 \times 10^8\,\text{m/s})^2\$
2\$E \approx 8.187 \times 10^{-14}\,\text{J}\$\$8.187 \times 10^{-14}\,\text{J}\$
3\$E \text{ (keV)} = \frac{E \text{ (J)}}{1.602 \times 10^{-19}\,\text{J/eV}}\$\$511\,\text{keV}\$

Exam Tips 📚

  • Remember: each photon carries half the total energy of the annihilation.

  • Use the rest‑mass energy formula: \$E = m c^2\$.

  • Convert joules to electron‑volts carefully; 1 eV = 1.602 × 10⁻¹⁹ J.

  • Show all steps and units to avoid marks lost for missing work.

  • When asked for “energy of a gamma‑ray photon”, answer 511 keV (or 8.187 × 10⁻¹⁴ J).