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Radiation and Medical Imaging

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Senior 6 · Physics

Radiation and Medical Imaging

5 lessons

Lesson 1 of 5

Ionising and Non-Ionising Radiation in Matter

25 min

Fifty kilojoules that do no harm

A microwave oven running at 800 W pours about 50 kJ of energy into a cup of tea in one minute, and the tea simply gets hot. A dental X-ray delivers to the patient's jaw only a tiny fraction of a joule, yet the dentist steps out of the room and the patient may wear a lead collar over the thyroid gland. If total energy decided the danger, the oven would be the threat. It is not, because what matters for living cells is how the energy arrives: in how large a packet, and how densely along its path. In Units 5 and 7 you met alpha, beta and gamma radiation and X-rays. This lesson asks what any radiation actually does to the atoms of the body. The answer underlies every medical image in this unit, and every safety rule.

Two families of radiation

Medicine uses waves and particles across an enormous range of energies. They split into two families according to one question: can a single photon or particle knock an electron out of an atom or molecule?

Figure 1.1: The electromagnetic spectrum on a photon-energy scale. Only photons to the right of the dashed line carry enough energy to ionise the molecules of the body.

The energy needed to remove the outermost electron from the atoms and molecules of living tissue is about 10 eV to 15 eV; for a water molecule it is 12.6 eV. A photon carries $E = hf = \frac{hc}{\lambda}$ , so photons can ionise only if $\lambda \lesssim 100\,\mathrm{nm}$ : far ultraviolet, X-rays and gamma rays. Making the beam more intense adds more photons but does not make any one photon more energetic, which is why a powerful radio transmitter cannot ionise a single atom while a single 60 keV X-ray photon, through the electron it sets moving, can ionise nearly two thousand molecules.

Definition

Ionising radiation is radiation whose individual photons or particles carry enough energy to remove electrons from atoms or molecules, producing ions. Examples: X-rays, gamma rays, alpha and beta particles, neutrons, protons.

Non-ionising radiation is radiation whose photons or particles cannot ionise; it can still deposit energy, mainly as heat or by exciting molecules. Examples: radio waves, microwaves, infrared, visible light, most ultraviolet, and ultrasound (a mechanical wave, not an electromagnetic one).

Directly ionising radiation (charged particles: α, β, protons) ionises atoms along its track through the electric force. Indirectly ionising radiation (photons, neutrons) first gives its energy to charged particles, which then do the ionising.

Radiation	Family	Medical use (this unit)	Main effect in tissue
radio-frequency waves	non-ionising	MRI (Lesson 4)	slight heating
visible and infrared light	non-ionising	endoscopy (Lesson 4)	heating, excitation
ultrasound	non-ionising (mechanical)	scanning (Lesson 3)	slight heating
X-rays	ionising, indirectly	radiography, CT, therapy	ionisation by secondary electrons
gamma rays	ionising, indirectly	nuclear medicine, therapy (Lesson 5)	ionisation by secondary electrons
β⁻ and β⁺ particles	ionising, directly	therapy; β⁺ in PET (Lesson 5)	ionisation along a track
α particles, protons	ionising, directly	targeted therapy	dense ionisation along a short track

Ultraviolet sits on the border. UV-A and UV-B (wavelengths 280 nm to 400 nm) cannot ionise, but a UV-B photon (about 4 eV) can still excite DNA strongly enough to join neighbouring bases together. That is how sunlight causes sunburn and skin cancer without ionising anything: "non-ionising" does not mean "harmless".

Worked example 1: three photons compared with the threshold

Find the photon energy, in eV, of (a) a mobile-phone signal at 900 MHz, (b) green light of wavelength 550 nm, and (c) find the longest wavelength that can ionise a water molecule (ionisation energy 12.6 eV). Use $h = 6.63 \times 10^{-34}\,\mathrm{J\,s}$ , $c = 3.00 \times 10^{8}\,\mathrm{m\,s^{-1}}$ , $e = 1.60 \times 10^{-19}\,\mathrm{C}$ .

Solution

(a) $E = hf = (6.63 \times 10^{-34})(9.00 \times 10^{8}) = 5.97 \times 10^{-25}\,\mathrm{J}$ . In eV: $\frac{5.97 \times 10^{-25}}{1.60 \times 10^{-19}} = 3.7 \times 10^{-6}\,\mathrm{eV}$ , over three million times too small to ionise.

(b) $E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{550 \times 10^{-9}} = 3.62 \times 10^{-19}\,\mathrm{J} = 2.26\,\mathrm{eV}$ . Enough to trigger the retina, not enough to ionise.

(c) $\lambda_{\max} = \frac{hc}{E} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{12.6 \times 1.60 \times 10^{-19}} = 9.87 \times 10^{-8}\,\mathrm{m} \approx 99\,\mathrm{nm}$ , in the far ultraviolet.

Conclusion: the danger of a photon is set by its frequency, not by the power of the source.

Charged particles: ionisation along a track

A charged particle moving through matter pulls on the electrons it passes, ionising or exciting atom after atom until it has lost all its kinetic energy and stops. It therefore has a definite range. In air, about 34 eV is used up, on average, for every ion pair formed (this average is called the W-value; it is larger than the ionisation energy because some energy goes into excitation).

Alpha particles are emitted with one or a few discrete energies characteristic of the nuclide, typically 4 MeV to 9 MeV. They are slow and doubly charged, so they ionise very densely and stop within a few centimetres of air, or about 40 µm of tissue: less than the thickness of the dead outer layer of skin.
Beta particles are emitted with a continuous spread of energies up to a maximum. They are fast and light, so they ionise sparsely, are scattered easily, and the most energetic travel a few metres in air and up to about 1 cm in tissue.

The energy a particle deposits per unit length of its track is its linear energy transfer (LET). Alpha particles in tissue have an LET of the order of 100 keV per micrometre; fast electrons, of the order of 0.2 keV per micrometre. High-LET radiation packs its damage so closely that both strands of a DNA molecule are often broken at once, which a cell finds much harder to repair. You will see this again in Lesson 2 as the radiation weighting factor.

Worked example 2: dense and sparse ionisation

Americium-241 emits alpha particles mainly at 5.49 MeV, which travel about 4.1 cm in air. A beta particle of 1.0 MeV travels a path of about 3.5 m in air. Taking W = 35 eV per ion pair for the alpha particle and 34 eV for the beta particle, find for each (a) the total number of ion pairs and (b) the average number per millimetre of path.

Solution

(a) Alpha: $N = \frac{5.49 \times 10^{6}\,\mathrm{eV}}{35\,\mathrm{eV}} = 1.6 \times 10^{5}$ . Beta: $N = \frac{1.0 \times 10^{6}}{34} = 2.9 \times 10^{4}$ .

(b) Alpha: $\frac{1.57 \times 10^{5}}{41\,\mathrm{mm}} \approx 3.8 \times 10^{3}$ per mm. Beta: $\frac{2.94 \times 10^{4}}{3500\,\mathrm{mm}} \approx 8$ per mm.

Interpretation: the two particles carry similar energies, but the alpha particle spends it about 450 times more densely. (Intermediate values were kept unrounded; the path lengths are typical values, so the answers are estimates to 2 significant figures.)

Photons: three ways to interact

An X-ray or gamma photon has no charge, so it travels in a straight line until a single event removes it from the beam. Three events matter in medicine (Figure 1.2).

Figure 1.2: The three main interactions of X-ray and gamma photons with matter.

Photoelectric absorption. The photon is absorbed completely by a tightly bound inner electron, which is ejected with kinetic energy $E - E_{\text{binding}}$ . It is most likely when the photon energy is only a little above the binding energy, and its probability per atom rises steeply with atomic number, roughly as $\frac{Z^3}{E^3}$ . This is why bone (calcium, $Z = 20$ ) and iodine or barium contrast agents absorb diagnostic X-rays so much more strongly than soft tissue (effective $Z \approx 7.4$ ).
Compton scattering. The photon collides with a loosely bound outer electron, gives it part of its energy and continues in a new direction with a longer wavelength: $\lambda' - \lambda = \frac{h}{m_e c}\,(1 - \cos\theta), \qquad \frac{h}{m_e c} = 2.43\,\mathrm{pm}$ Its probability depends on the number of electrons per unit volume, almost independent of $Z$ . In soft tissue it dominates from about 25 keV to about 25 MeV, which covers most of nuclear medicine and radiotherapy.
Pair production. Near a nucleus, a photon of energy at least $2m_ec^2 = 1.022\,\mathrm{MeV}$ can turn into an electron and a positron, as you saw in Unit 6. In tissue it becomes important only at the highest energies used in radiotherapy.

Worked example 3: why scattered photons blur an image

Technetium-99m, the most used tracer in nuclear medicine, emits gamma photons of 140 keV. One such photon is Compton-scattered through 90° inside the patient. Find its energy after scattering and the energy given to the electron. Take $m_ec^2 = 511\,\mathrm{keV}$ .

Solution

Dividing the Compton equation into energies (using $E = \frac{hc}{\lambda}$ ) gives $E' = \frac{E}{1 + \frac{E}{m_ec^2}(1 - \cos\theta)}$ .

With $\theta = 90^\circ$ , $\cos\theta = 0$ : $E' = \frac{140}{1 + \frac{140}{511}} = \frac{140}{1.274} = 110\,\mathrm{keV}$ .

Electron: $140 - 109.9 = 30\,\mathrm{keV}$ (subtracting the unrounded 109.9 keV).

Why it matters: the scattered photon leaves the body in a new direction, so if it were detected it would appear to come from the wrong place. Because it has lost about a fifth of its energy, the gamma camera of Lesson 5 can reject it by accepting only photons close to 140 keV.

Attenuation of a photon beam

Consider a narrow, parallel beam of photons, all of one energy, of intensity $I$ entering a thin layer of thickness $\mathrm{d}x$ . Each of the three processes removes photons at random, so the fraction removed is proportional to the thickness:

$\frac{\mathrm{d}I}{I} = -\mu\,\mathrm{d}x \quad\Longrightarrow\quad \int_{I_0}^{I}\frac{\mathrm{d}I}{I} = -\mu\int_{0}^{x}\mathrm{d}x \quad\Longrightarrow\quad I = I_0\,e^{-\mu x}$

The constant $μ \mu$ is the linear attenuation coefficient (unit $\mathrm{m^{-1}}$ or $\mathrm{cm^{-1}}$ ); it is the sum of the contributions of the three processes. Setting $I = \frac{I_0}{2}$ gives the half-value layer (HVL), the half-value thickness you used in Unit 5:

$x_{\frac{1}{2}} = \frac{\ln 2}{\mu}$

Because $μ \mu$ is proportional to the density, tables list the mass attenuation coefficient $\mu_m = \frac{\mu}{\rho}$ , in $\mathrm{cm^{2}\,g^{-1}}$ ; then $\mu = \mu_m\rho$ . This has the same form as the decay law $N = N_0e^{-\lambda t}$ , with thickness in place of time.

Material (density)	$\mu_m$ at 60 keV	$\mu_m$ at 1.25 MeV
skeletal muscle ( $1.05\,\mathrm{g\,cm^{-3}}$ )	$0.205\,\mathrm{cm^{2}\,g^{-1}}$	$0.0627\,\mathrm{cm^{2}\,g^{-1}}$
cortical bone ( $1.92\,\mathrm{g\,cm^{-3}}$ )	$0.315\,\mathrm{cm^{2}\,g^{-1}}$	$0.0587\,\mathrm{cm^{2}\,g^{-1}}$
water ( $1.00\,\mathrm{g\,cm^{-3}}$ )	$0.206\,\mathrm{cm^{2}\,g^{-1}}$	$0.0632\,\mathrm{cm^{2}\,g^{-1}}$

Values from the NIST X-ray mass attenuation coefficient tables (Hubbell and Seltzer), rounded to 3 significant figures.

Conditions. The exponential law holds for a narrow beam of photons of a single energy. In a broad beam, some scattered photons still reach the detector, so the measured intensity is higher than $I_0e^{-\mu x}$ . Charged particles do not obey it at all: they have a definite range.

Worked example 4: where contrast comes from

A narrow 60 keV X-ray beam crosses 6.0 cm of a limb along two paths: (A) 6.0 cm of muscle; (B) 4.0 cm of muscle and 2.0 cm of cortical bone. Use the table above. (a) Find the fraction transmitted along each path and their ratio. (b) Repeat for 1.25 MeV gamma rays and comment.

Solution

(a) $\mu_{\text{muscle}} = 0.2048 \times 1.05 = 0.2150\,\mathrm{cm^{-1}}$ ; $\mu_{\text{bone}} = 0.3148 \times 1.92 = 0.6044\,\mathrm{cm^{-1}}$ (NIST values before rounding).

A: $e^{-0.2150 \times 6.0} = e^{-1.290} = 0.275$ . B: $e^{-(0.2150 \times 4.0 + 0.6044 \times 2.0)} = e^{-2.069} = 0.126$ . Ratio: $\frac{0.275}{0.126} = 2.2$ .

(b) $\mu_{\text{muscle}} = 0.06265 \times 1.05 = 0.0658\,\mathrm{cm^{-1}}$ ; $\mu_{\text{bone}} = 0.05871 \times 1.92 = 0.1127\,\mathrm{cm^{-1}}$ . A: $e^{-0.395} = 0.674$ . B: $e^{-0.489} = 0.613$ . Ratio: 1.10.

Comment: at 60 keV, photoelectric absorption by calcium makes bone stop about twice as many photons, so it stands out clearly. At 1.25 MeV, Compton scattering dominates and depends only on electron density, so bone and muscle look almost alike. Diagnostic X-ray tubes therefore run at about 25 kV to 150 kV, while radiotherapy uses MeV beams that pass through bone without being strongly absorbed by it.

Experiment 1.1: measuring μ for lead

This is a teacher demonstration with a sealed caesium-137 source (gamma rays of 662 keV), handled with tongs and returned to its lead container straight after use. A Geiger–Müller tube faces the source at a fixed distance, with a collimator so that the beam is narrow. The background count rate is measured first, with the source in its container, and subtracted from every reading. Lead sheets are then added one at a time.

thickness x / mm	0	2	4	6	8	10	12
corrected count rate C / s⁻¹	846	661	503	395	299	236	178
ln (C ÷ s⁻¹)	6.74	6.49	6.22	5.98	5.70	5.46	5.18

Illustrative readings of the kind this demonstration gives.

Taking logarithms of $C = C_0e^{-\mu x}$ gives $\ln C = \ln C_0 - \mu x$ : a straight line of gradient $- μ -\mu$ (Figure 1.3). A straight line confirms the exponential law; a curve that flattens at large $x$ would show scattered photons reaching the detector.

Figure 1.3: ln C against thickness of lead, with the line of best fit.

Worked example 5: analysing Experiment 1.1

Use the line of best fit in Figure 1.3 to find (a) the linear attenuation coefficient of lead for 662 keV gamma rays, (b) the half-value layer, and (c) the thickness of lead that would cut the count rate to 1% of its unshielded value.

Solution

(a) The line passes through (0, 6.75) and (12, 5.19). Gradient $= \frac{5.19 - 6.75}{12\,\mathrm{mm}} = -0.130\,\mathrm{mm^{-1}}$ , so $\mu = 0.130\,\mathrm{mm^{-1}} = 1.30\,\mathrm{cm^{-1}}$ . (Published tables give about $1.3\,\mathrm{cm^{-1}}$ at this energy.)

(b) $x_{\frac{1}{2}} = \frac{\ln 2}{\mu} = \frac{0.693}{0.130\,\mathrm{mm^{-1}}} = 5.3\,\mathrm{mm}$ .

(c) $\frac{C}{C_0} = 0.01 \Rightarrow x = \frac{\ln 100}{\mu} = \frac{4.605}{0.130\,\mathrm{mm^{-1}}} = 35\,\mathrm{mm}$ , about 6.6 half-value layers.

Warning

Common mistakes. (1) Thinking that a stronger (more intense) source of light or radio waves becomes ionising: intensity adds photons; only frequency raises the energy per photon. (2) Writing $I = I_0e^{-\mu x}$ for alpha or beta particles: charged particles have a range, not an exponential tail. (3) Forgetting to subtract the background count before taking logarithms; the error is largest for the thickest absorbers. (4) Mixing units: if $μ \mu$ is in $\mathrm{cm^{-1}}$ , $x$ must be in cm.

Tip

Match each photon process to its job in medicine. Photoelectric: contrast (it depends strongly on $Z$ , so it separates bone from flesh in a radiograph). Compton: dose and blur (it happens everywhere, deposits energy and scatters photons the wrong way). Pair production: high-energy therapy, and the positrons behind PET in Lesson 5.

Is radiation dangerous?, by Matt Anticole (TED-Ed). Watch for how the video separates ionising from non-ionising radiation by the energy of each packet, and compare its examples with Figure 1.1.

✦

Practice

State the difference between ionising and non-ionising radiation, and classify: infrared, gamma rays, ultrasound, beta particles, microwaves.
Find the energy in eV of an ultraviolet photon of wavelength 300 nm, and say whether it can ionise a water molecule (12.6 eV).
Explain why alpha particles are harmless outside the body but very dangerous when an alpha emitter is swallowed.
The half-value layer of soft tissue for a certain X-ray beam is 3.2 cm. Find μ, and the fraction of the beam that crosses 15 cm of tissue.
A 662 keV gamma photon is Compton-scattered through 180°. Find its energy after scattering ( $m_ec^2 = 511\,\mathrm{keV}$ ).
The intensity of a narrow gamma beam falls to 20% of its initial value in 4.0 cm of concrete. Find μ and the half-value layer of concrete for these rays.
Explain why a radiograph of a hand shows the bones clearly at 50 kV, but an image made with 6 MV radiotherapy X-rays shows almost no contrast between bone and soft tissue.
Uwase says: "Doubling the thickness of a lead shield doubles the protection." Evaluate her claim using the attenuation law.

Show answersHide answers

Ionising radiation has photons or particles energetic enough to remove electrons from atoms; non-ionising does not. Ionising: gamma rays, beta particles. Non-ionising: infrared, ultrasound, microwaves.
$E = \frac{1240\,\mathrm{eV\,nm}}{300\,\mathrm{nm}} = 4.13\,\mathrm{eV}$ : no, it cannot ionise water (though UV of this energy can still damage DNA).
Outside, the alpha particles are stopped by the dead outer layer of skin (range about 40 µm). Inside, they deposit all their energy, at very high LET, in living cells next to the source.
$\mu = \frac{0.693}{3.2} = 0.217\,\mathrm{cm^{-1}}$ ; $\frac{I}{I_0} = e^{-0.2166 \times 15} = e^{-3.25} = 0.039$ , about 3.9%.
$E' = \frac{662}{1 + \frac{662}{511} \times 2} = \frac{662}{3.591} = 184\,\mathrm{keV}$ .
$\mu = \frac{\ln 5}{4.0} = \frac{1.609}{4.0} = 0.40\,\mathrm{cm^{-1}}$ ; $x_{\frac{1}{2}} = \frac{0.693}{0.4024} = 1.7\,\mathrm{cm}$ .
At 50 kV photoelectric absorption dominates and rises steeply with Z, so calcium-rich bone absorbs much more. At MeV energies Compton scattering dominates; it depends on electron density, which is similar in bone and tissue, so contrast almost vanishes.
Not as stated. If one thickness transmits the fraction $f = e^{-\mu x}$ , double the thickness transmits $e^{-2\mu x} = f^2$ . A shield that lets through 10% becomes one that lets through 1%: the dose behind it falls by a further factor of 10, not 2. How much doubling helps depends on $\mu x$ , and no thickness stops every photon.

Key points

Ionising radiation removes electrons from atoms (photon energy above about 10 eV to 15 eV, $\lambda \lesssim 100\,\mathrm{nm}$ ); the energy per photon, not the beam power, decides it.
Charged particles ionise along a track and have a range; α: high LET, short range; β: low LET, up to a few metres in air.
Photons: photoelectric (low $E$ , high $Z$ : contrast), Compton (most of the medical range: scatter and dose), pair production ( $E \ge 1.022\,\mathrm{MeV}$ ).
Narrow monoenergetic beam: $I = I_0e^{-\mu x}$ , $x_{\frac{1}{2}} = \frac{\ln 2}{\mu}$ , $\mu = \mu_m\rho$ .

Sources and credits

32.2: Biological Effects of Ionizing Radiation, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Attenuation coefficients: X-Ray Mass Attenuation Coefficients (NIST Standard Reference Database 126) by J. H. Hubbell and S. M. Seltzer, National Institute of Standards and Technology. Data quoted, rounded by Ganzaa.
Photo: Gamma Ray Lead Measurement.jpg by Michael Krahe, Wikimedia Commons, licensed under CC BY-SA 3.0.
Video: Is radiation dangerous? - Matt Anticole, TED-Ed, embedded from YouTube.

The three diagrams, the experiment readings (illustrative), the worked examples and the practice exercises are written by Ganzaa. This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Fifty kilojoules that do no harm

2Two families of radiation

3Charged particles: ionisation along a track

4Photons: three ways to interact

5Attenuation of a photon beam

6Experiment 1.1: measuring μ for lead

Fifty kilojoules that do no harm

Two families of radiation

Charged particles: ionisation along a track

Photons: three ways to interact

Attenuation of a photon beam

Experiment 1.1: measuring μ for lead