Gamma radiation

Gamma radiation consists of high-energy photons emitted when a nucleus moves from a higher-energy state to a lower-energy state. In a pure gamma transition the numbers of protons and neutrons do not change: what changes is the nucleus's internal energy and often its angular-momentum state.

A nuclear energy-level transition

An excited nucleus can release energy as a gamma photon. The photon energy equals the difference between the two nuclear levels: Eγ = hν = hc/λ. Because nuclear energy levels are quantized, gamma-ray spectra often contain characteristic lines.

Gamma emission lowers nuclear excitation without changing A or Z. The emitted photon carries the energy difference.

Gamma emission often follows another decay

Alpha or beta decay can leave the daughter nucleus in an excited state. The daughter may then emit one or more gamma photons as it settles to lower levels. That means “gamma radiation” often describes the de-excitation step rather than a change of element. Gamma emission should therefore be separated conceptually from the transformation that produced the excited nucleus.

Attenuation is probabilistic. A thicker absorber reduces the transmitted intensity; it does not make each surviving photon gradually lose a fixed fraction of its energy.

Attenuation is exponential

For a narrow beam in a uniform absorber, a common model is I = I₀e^-μx, where μ is the linear attenuation coefficient and x is thickness. The half-value layer is the thickness that reduces the beam intensity by half. The coefficient depends strongly on photon energy and material composition.

Attenuation explorer

μ (cm⁻¹) thickness x (cm)

Photons interact differently from charged particles

Gamma photons carry no electric charge, so they do not continuously ionize along a track in the same way as alpha particles. Instead, individual photons can undergo interactions such as photoelectric absorption, Compton scattering or pair production, depending on energy and material. This is why “penetrating” does not mean “never interacts”: it means the probability of traversing a given thickness can be appreciable.

Worked examples

1. Photon energy from frequency

Solution

For ν = 3.0 × 10¹⁹ Hz, E = hν ≈ 6.63 × 10⁻³⁴ × 3.0 × 10¹⁹ = 1.99 × 10⁻¹⁴ J.

2. What changes in pure gamma emission?

Solution

The nucleus moves to a lower energy state, but A and Z remain unchanged. The emitted photon carries energy and momentum.

3. Use attenuation

Solution

If μx = ln 2, then I/I₀ = e^{-ln 2} = 1/2. That thickness is one half-value layer.