Interaction of Radiation with Matter

Introduction

Every use of radiation — detection, imaging, therapy and shielding — depends on how radiation gives up its energy to matter. Alpha particles are stopped by a sheet of paper, while gamma rays pass through centimetres of lead; these familiar facts have precise explanations. This page contrasts charged particles with photons, introduces linear energy transfer and attenuation, and follows the chemistry that begins when radiation passes through water, the main component of living cells.

Core explanation

Charged particles: continuous slowing. Alpha and beta particles lose energy in many small steps, mostly by electrical interactions with atomic electrons that cause ionisation and excitation . Each ionisation in air costs on average about 34 eV, so a 5 MeV alpha particle produces roughly 150 000 ion pairs. Because charged particles slow continuously, they have a definite range .

- Alpha particles are heavy and doubly charged. They move slowly for their energy and interact intensely, travelling in almost straight lines. Range: a few centimetres in air and about 40 µm in tissue. The rate of energy loss rises as they slow, producing a sharp peak just before stopping (the Bragg peak ). - Beta particles are light, so they are easily deflected and follow tortuous paths. Ranges are millimetres in tissue and metres in air. When fast electrons are decelerated near nuclei of high atomic number they emit X-rays called bremsstrahlung , which is why beta emitters are shielded first with low-Z materials such as acrylic.

Linear energy transfer. LET is the energy deposited per unit path length. Alpha particles have high LET (about 100 keV µm⁻¹ or more), creating dense tracks of ionisation; electrons and gamma rays have low LET (below about 1 keV µm⁻¹ for fast electrons), with sparse, widely spaced ionisations. At the same absorbed energy, high-LET radiation damages biological molecules more severely because events cluster within a few nanometres.

Photons: all-or-nothing interactions. Gamma and X-ray photons have no charge and interact only occasionally. Each interaction removes the photon from the beam or scatters it. Three processes dominate:

1. Photoelectric absorption — the photon is absorbed and ejects an inner-shell electron. Dominant at low energy and high Z (probability roughly ∝ Z⁴–Z⁵/E³), which is why lead is an excellent shield for low-energy photons. 2. Compton scattering — the photon transfers part of its energy to an outer electron and continues with lower energy. Dominant at intermediate energies (about 0.1–10 MeV in light materials) and depends mainly on electron density. 3. Pair production — above 1.022 MeV, a photon near a nucleus converts into an electron–positron pair. Important at high energy and high Z.

Because each photon interacts or does not, a beam is attenuated exponentially: I = I₀e^(−μx). Photons have no fixed range; a thicker shield reduces intensity by a factor but never to exactly zero.

Radiolysis of water. Radiation ionises and excites water molecules: H₂O → H₂O⁺ + e⁻. Within picoseconds, H₂O⁺ reacts with another water molecule to give H₃O⁺ and the hydroxyl radical •OH; the electron becomes surrounded by water molecules as a hydrated electron , e⁻(aq). Excited molecules break into H• and •OH. These species react in the dense regions of the track ( spurs ) to form H₂ and H₂O₂. The main primary products are e⁻(aq), •OH, H•, H₂, H₂O₂ and H₃O⁺. The hydroxyl radical is highly oxidising and is responsible for much of the indirect damage to DNA. High-LET tracks favour recombination, so they give more H₂ and H₂O₂ and fewer free radicals escaping the track.

Formulae

Attenuation: I = I₀e^(−μx); half-value layer HVL = ln 2 / μ.

Mass attenuation coefficient: μ/ρ (cm² g⁻¹).

G-value: molecules per 100 eV absorbed (SI: mol J⁻¹; 1 molecule per 100 eV ≈ 1.04 × 10⁻⁷ mol J⁻¹).

Step-by-step reasoning

To estimate the shielding needed for a gamma source:

1. Find μ for the absorber at the photon energy. 2. Calculate the half-value layer, HVL = 0.693/μ. 3. Decide the reduction factor required. 4. Find the number of HVLs: n = log₂(reduction factor). 5. Multiply n by the HVL to obtain the thickness.

Visual explanation

Picture an alpha track as a short, thick, straight line crowded with ionisations, ending in a dense knot. A beta track is a long, thin, wandering line. A gamma beam is a stream of dots, most passing straight through the absorber, with only occasional dots stopped or deflected at random points.

Real-world analogy

A charged particle is like a bowling ball rolling through a field of skittles: it knocks many down and slows steadily until it stops. A photon is like a bullet fired through a sparse forest: it travels untouched until, at an unpredictable point, it strikes one tree.

Real-world example

Proton therapy exploits the Bragg peak. By choosing the proton energy, oncologists place the peak of energy deposition inside a tumour, so tissue beyond it receives almost no dose, which is particularly valuable near the spinal cord or in children.

Why?

Why do gamma rays follow an exponential law while alpha particles have a definite range? Each photon has a fixed probability of interacting per unit thickness, independent of how far it has travelled, which gives exponential decrease. Alpha particles all lose energy through many small collisions, so particles of the same energy stop at about the same depth.

Common misconception

"Shielding can stop all gamma rays." Attenuation is exponential, so each half-value layer halves the intensity but never reduces it to zero. Shielding reduces exposure to an acceptable level; it does not eliminate it.

Worked example

Question: For 662 keV gamma rays, lead has μ ≈ 1.2 cm⁻¹. What thickness of lead reduces the intensity to 1/16 of its initial value?

Reasoning: HVL = 0.693 ÷ 1.2 ≈ 0.58 cm. A factor of 16 is 2⁴, so four HVLs are needed: 4 × 0.58 ≈ 2.3 cm.

Answer: About 2.3 cm of lead.

Quick check

1. Why should a strong beta source be shielded first with acrylic rather than lead? Answer: Low-Z acrylic stops electrons while producing little bremsstrahlung, whereas lead would convert more of their energy into penetrating X-rays.

Exam focus

Be ready to compare alpha, beta and gamma in terms of charge, mass, LET, range and penetration, and to explain each photon interaction and where it dominates. Use I = I₀e^(−μx) and HVLs confidently. Know the primary products of water radiolysis and the role of the hydroxyl radical.

Advanced insight

The hydrated electron, discovered by pulse radiolysis in 1962, is one of the strongest reducing agents in aqueous chemistry, with a standard potential of about −2.9 V. Pulse radiolysis — an intense, nanosecond burst of electrons followed by fast spectroscopy — lets chemists generate radicals on demand and measure their rate constants directly, making it a key tool of free-radical chemistry.

Summary

Charged particles lose energy continuously by ionisation, giving a finite range: alpha particles have high LET and short range, beta particles lower LET and longer, twisting paths with bremsstrahlung. Photons interact by photoelectric absorption, Compton scattering and pair production, and are attenuated exponentially. Radiolysis of water produces hydrated electrons, hydroxyl radicals, hydrogen atoms, H₂ and H₂O₂, which cause most indirect biological damage.

Practice questions

1. Define linear energy transfer and state which of alpha, beta and gamma radiation has the highest LET. Answer: LET is energy deposited per unit path length; alpha radiation has the highest LET. 2. A shield of 3 HVLs is used. What fraction of the incident gamma intensity passes through? Answer: (1/2)³ = 1/8, or 12.5%. 3. Which photon interaction dominates for low-energy photons in lead, and why? Answer: Photoelectric absorption, because its probability rises steeply with atomic number and falls steeply with photon energy. 4. Name the primary radiolysis product most responsible for indirect DNA damage and state why. Answer: The hydroxyl radical, •OH, because it is a very reactive oxidising species that abstracts hydrogen from sugars and adds to bases.