Module 05 · ABR domain II.A · Radiation measurements

Dose Quantities: Kerma, CPE and Cavity Theory

Every dose you prescribe was, at some point, a charge collected in a few tenths of a cubic centimetre of air. This module builds the chain that turns that charge into dose to water: how photons hand energy to electrons (kerma), when the energy handed over equals the energy that stays (charged particle equilibrium), and how a gas-filled detector reports dose in the medium around it (cavity theory). It is the vocabulary physicists use when they explain why a detector reads what it reads.

In this module
  1. 01The one idea
  2. 02Counting the beam: fluence
  3. 03Kerma: energy handed over
  4. 04Absorbed dose: energy that stays
  5. 05Charged particle equilibrium
  6. 06Exposure, W/e and the f-factor
  7. 07Bragg-Gray cavity theory
  8. 08Spencer-Attix and large cavities
  9. 09Cavity bench
  10. 10Worked examples
  11. 11In the planning room
  12. 12Board pearls and traps
  13. 13Question bank
  14. 14Formula card and sources
01 · Start here

The one idea

Photons do not deposit dose. They hand energy to electrons, and the electrons deposit it somewhere else. Every dose quantity in this module is a way of keeping track of where that energy was handed over and where it finally stayed.

photon energy fluence Ψ  →  kerma K (handed to electrons)  →  absorbed dose D (stays in the mass)
Kerma is counted where the photon interacts. Dose is counted where the electron stops giving up energy. The two are equal only when as much electron energy flows into a small volume as flows out of it.

Module 03 followed photons through attenuation and showed that μtr and μen track how much of each interaction's energy goes to electrons. This module takes the next step. Multiply the energy fluence by μtr/ρ and you get kerma. Ask when kerma equals dose and you get charged particle equilibrium. Put a small air cavity in the medium and ask what dose the air receives and you get cavity theory, which is how every ion chamber calibration works.

Three answers carry the whole module:

  1. Under charged particle equilibrium, absorbed dose equals collision kerma. That lets you calculate dose from the photon beam alone.
  2. One coulomb of ionization in air corresponds to 33.97 joules deposited. That number, W/e, converts charge to dose in air.
  3. A small cavity converts with stopping power ratios; a large one with μen ratios. The size is judged against the range of the electrons, not against the patient.
Pearl

When a dose quantity confuses you, ask two questions: whose energy is being counted (photon or electron), and where is it being counted (at the interaction or where it is absorbed). Every definition in this module answers those two questions differently.

02 · Describing the field

Counting the beam: fluence

Before you can say how much energy a beam deposits, you need a way to say how much beam there is at a point. Picture a small sphere centred on the point. Count the photons that cross it, from any direction, and divide by the sphere's cross-sectional area. That is the particle fluence, Φ, in photons per square metre (usually quoted per cm²). Using a sphere rather than a flat plate means a photon counts once whatever direction it travels, which matters inside a patient where scattered photons come from everywhere.

Weight each photon by its energy and you have the energy fluence, Ψ, in joules per square metre. For a single photon energy the two are related simply. For a real beam you sum over the spectrum. The rates, fluence per unit time, are the fluence rate and energy fluence rate.

Φ = dNda     Ψ = dEda = Φ · hν  (single energy)
da is the cross-sectional area of the small sphere. Both fall with the inverse square of distance from a point source in air.

Worked number: 1.0 × 1010 photons/cm² of 1.25 MeV carry Ψ = 1.0 × 1010 × 1.25 MeV × 1.602 × 10−13 J/MeV = 2.0 × 10−3 J/cm².

Board point

Fluence is a property of the radiation field at a point. It is not a dose and has no medium attached. Kerma and dose need a medium, because they depend on what the photons interact with.

03 · Step one of two

Kerma: energy handed over

When a photon interacts, some of its energy becomes kinetic energy of electrons (the Compton electron, the photoelectron, the electron and positron from pair production). The rest leaves as a scattered photon, a characteristic x-ray, or annihilation radiation. Kerma (kinetic energy released per unit mass) counts only the kinetic energy given to charged particles in a small mass, at the place where the photon interacted. Its unit is the gray, J/kg, the same as dose.

Kerma has the shape it has because it is the product of how much energy passes the point (Ψ) and what fraction of that energy is transferred to electrons per unit mass of the medium (μtr/ρ). More photons, or a medium that takes a bigger share of each photon's energy, means more kerma.

K = Ψ · μtrρ
K = Kcol + Krad     Kcol = K (1 − g) = Ψ · μenρ
The electrons set in motion spend most of their energy on ionization and excitation near their track (collision kerma, Kcol) and a fraction g on bremsstrahlung photons that can escape (radiative kerma, Krad). g is the radiative fraction from Module 03.

Worked number: continuing from section 02, Ψ = 2.0 × 10−3 J/cm² of 1.25 MeV photons in water, μen/ρ = 0.0296 cm²/g. Kcol = 2.0 × 10−3 × 0.0296 = 5.9 × 10−5 J/g = 0.059 J/kg = 5.9 cGy.

Increase thisEffect on kermaWhy
Energy fluenceRises in proportionMore energy crossing the point
Distance from a point source (in air)Falls as 1/r²Fluence spreads over a larger area
Depth in a mediumFalls, roughly exponentiallyPrimary photons are attenuated; kerma is greatest at the surface
Z of medium at kV energiesRises steeplyPhotoelectric μen/ρ scales roughly with Z³
Electron energy (radiative fraction g)Kcol/K falls slightlyFaster electrons in higher-Z media lose more to bremsstrahlung
Board point

Kerma is defined only for indirectly ionizing radiation (photons and neutrons). It is maximal at the surface of a phantom and falls with depth from the first millimetre. Radiative fraction g is small in tissue: about 0.3 percent for Co-60 photons in air, rising with energy and with Z.

04 · Step two of two

Absorbed dose: energy that stays

Kerma tells you where electrons were launched. Absorbed dose tells you where their energy ended up. It is the mean energy imparted by all ionizing radiation to a small mass, divided by that mass. Energy imparted is everything that comes in (photons and electrons) minus everything that goes out, so electrons launched upstream that stop inside the mass count, and electrons launched inside that leave do not.

Dose applies to any ionizing radiation, directly or indirectly ionizing, in any medium. It is what biology responds to and what you prescribe.

D = dε̄dm     1 Gy = 1 J/kg = 100 rad
ε̄ is mean energy imparted. Energy carried out by bremsstrahlung photons is not imparted locally, which is why dose is linked to collision kerma rather than total kerma.
QuantityWhat it countsRadiationUnit
Fluence Φ, ΨPhotons (or energy) crossing a small sphereAnym⁻², J/m²
Kerma KKinetic energy given to charged particles, at the interaction sitePhotons, neutronsGy
Collision kerma KcolThe part of K not later radiated as bremsstrahlungPhotons, neutronsGy
Exposure XCharge of one sign produced in air by the electrons a photon beam sets movingPhotons, in air onlyC/kg, R
Absorbed dose DEnergy imparted that stays in the massAnyGy
Equivalent dose HTOrgan dose × radiation weighting factor wR (protection only)AnySv

For protection, ICRP 103 assigns wR = 1 for photons and electrons, 2 for protons, 20 for alpha particles, and an energy-dependent value for neutrons. Effective dose then weights organs by tissue factors. These are risk quantities for populations and belong to Module 21; never use them for a patient's treatment dose.

Trap

Treating kerma and dose as interchangeable because they share a unit. At the skin of a megavoltage patient, kerma is at its maximum and dose is a fraction of its peak. They are equal only under the conditions in the next section.

05 · The bridge

Charged particle equilibrium

Take a small volume in a uniform photon beam. Electrons are launched inside it and some carry energy out. Electrons launched upstream carry energy in. If, for every electron that leaves with some energy, another of the same energy enters, the volume keeps exactly the energy that was transferred inside it, minus the bremsstrahlung that escapes. That balance is charged particle equilibrium (CPE), and under it:

D = Kcol = Ψ · μenρ   (CPE)
This is the most useful equation in dosimetry. It lets you calculate dose from the photon fluence alone, without tracking a single electron.

CPE needs the photon fluence to be uniform over a region at least as large as the maximum electron range around the point, in a uniform medium. In a real megavoltage beam that is never quite true, because the photons are attenuated over the distance the electrons travel. The electrons arriving at a point were launched upstream, where the photon fluence was slightly higher. So beyond the buildup region, dose is a little larger than the local collision kerma, by a nearly constant factor β. Physicists call this transient charged particle equilibrium (TCPE).

D = β · Kcol   (TCPE, beyond dmax, β slightly > 1)
β is about 1.005 for Co-60 and rises to a few percent at high megavoltage, where electrons travel farther and the fluence falls more over their range. D and Kcol then fall in parallel.
Region (MV beam in water)RelationWhy
Surface and buildup, 0 to dmaxD < KcolNot enough upstream electrons yet; more energy leaves than enters
At dmaxD ≈ KcolThe curves cross; this is the closest a real beam gets to CPE
Beyond dmaxD = β Kcol, β > 1Electrons carry energy from upstream, where fluence was higher
Ideal, no attenuationD = KcolTrue CPE; approached for Co-60 more closely than for 18 MV

Kerma and dose through the buildup

Teaching model, not commissioning data
Beam

The amber curve is primary collision kerma, highest at the surface. The blue curve is dose, built from electrons launched upstream. Watch where they cross, then watch the gap beyond the crossing grow from Co-60 to 18 MV. Tick the box and the gap disappears.

Collision kerma KcolAbsorbed dose DReadout depth
Crossing (dmax)0
β beyond dmax0
Kcol at point0
D at point0
D / Kcol0
RegionTCPE

Model: Kcol = 100·e−μd for primary photons with μ fixed by a representative energy (1.25, 2 and 6 MeV); dose is that kerma carried forward by electrons with an exponential range kernel, which gives D = β·Kcol(1 − e−(λ−μ)d) with β = λ/(λ − μ). λ is chosen so the crossing sits at a typical dmax. Scatter and contaminant electrons are left out, so the model starts at zero surface dose; real 6 MV beams start at roughly 15 to 25 percent.

Board point

CPE fails, and D cannot be computed as Kcol, in the buildup region, near interfaces between different media, at field edges (penumbra), in small fields and in low-density lung where electron ranges grow, close to brachytherapy sources, and at high energy. Exposure and kerma-based calibrations need CPE, which is why exposure is not used above about 3 MeV.

In clinic

Lung SBRT with a small field: electrons travel three to five times farther in lung, more leave the beam than return, and dose on axis falls below what a kerma-based algorithm predicts. This is lateral electronic disequilibrium, the reason type B algorithms or Monte Carlo are required in lung (Module 08, section 07).

06 · The historical bridge

Exposure, W/e and the f-factor

Before anyone could measure energy deposited directly, they could measure charge. Exposure, X, is the total charge of one sign produced in a small mass of air when all the electrons liberated by photons in that mass are stopped completely in air. It is the ionization equivalent of collision kerma in air. Its SI unit is C/kg; the old unit, the roentgen, is 1 R = 2.58 × 10−4 C/kg.

To turn charge into energy you need the average energy spent per ion pair formed in dry air, W. Divided by the electron charge, W/e = 33.97 J/C (equivalently 33.97 eV per ion pair). ICRU Report 90 kept this value and widened its uncertainty to about 0.35 percent. Multiply exposure by W/e and you have collision kerma in air.

Kcol,air = X · We   →   1 R × 2.58 × 10−4 C/kg/R × 33.97 J/C = 0.876 cGy
One roentgen corresponds to 0.876 cGy of air collision kerma. Under CPE this is also the dose to air.

To get dose in another medium placed at the same point, under CPE, scale by how much more energy per gram that medium absorbs from the same photon fluence: the ratio of mass energy absorption coefficients. The combination is the f-factor, in cGy per R.

Dmed = X · fmed     fmed = 0.876 × (μen/ρ)med(μen/ρ)air cGy/R
For water at megavoltage the μen ratio is about 1.11, so fwater ≈ 0.97 cGy/R. Older texts quote 0.869 instead of 0.876 because they used W = 33.7 eV.
Mediumf at kV (≈ 30 keV)f at Co-60 and MVWhy
Water, muscle≈ 0.9≈ 0.97Hydrogen gives water more electrons per gram than air
Fatlower than muscle≈ 0.97Low Z; more hydrogen
Cortical boneseveral times muscle (≈ 4×)≈ 0.92Photoelectric Z³ dependence at kV; fewer electrons per gram at MV

Approximate values after Khan and Gibbons and Attix. Bone figures depend on the composition assumed. Learn the direction and the rough size: bone much higher at kV, slightly lower than soft tissue at megavoltage.

Board point

Exposure is defined only for photons, only in air, and is practical only up to about 3 MeV, because above that the secondary electrons travel too far for an ionization chamber to achieve CPE. It cannot be used for electron beams. Modern calibration (TG-51) bypasses exposure and calibrates chambers directly in dose to water.

07 · Small cavities

Bragg-Gray cavity theory

An ion chamber measures dose to the air in its cavity. You want dose to the water around it. Cavity theory connects the two.

Suppose the cavity is so small that photons essentially never interact in it. Every bit of dose in the air then comes from electrons launched in the surrounding water that cross the cavity. If the cavity is also too small to change the number or energy of those electrons, the same electron fluence crosses the air as would have crossed the water it replaced. Dose from an electron fluence is fluence times mass collision stopping power, energy lost per unit path per unit density. The ratio of doses is therefore the ratio of stopping powers.

DmedDgas = (S̄/ρ)med(S̄/ρ)gas = s̄med,gas
Dgas = Qmgas · We
S̄/ρ is the mass collision stopping power averaged over the electron spectrum crossing the cavity. Q is the collected charge and m the mass of air in the cavity, which depends on its temperature and pressure.

The two Bragg-Gray conditions are worth stating exactly, because exams ask for them:

  1. The cavity does not perturb the charged particle fluence in the medium (it is small compared with electron ranges).
  2. The dose in the cavity is deposited entirely by charged particles crossing it; photon interactions in the cavity itself are negligible.

CPE is not one of the conditions. Bragg-Gray works in regions without CPE, which is why it holds for electron beams and why chambers work at depth in megavoltage beams.

Worked number: a cavity collects charge that gives Dair = 0.959 Gy in a Co-60 beam. With s̄water,air = 1.133, Dwater = 0.959 × 1.133 ≈ 1.09 Gy, before perturbation corrections.

Beams̄water,air (approx.)Comment
Co-60≈ 1.13The reference beam for chamber calibration coefficients
6 MV≈ 1.12Falls slowly as photon energy rises
18 MV≈ 1.10Higher-energy electrons; density effect in water
Electron beams≈ 0.97 to 1.10Varies strongly with depth and energy, so TG-51 handles it at a reference depth

Spencer-Attix values, rounded. The exact numbers sit inside the kQ factors your physicist uses; they are given here for direction and size, not for calculation.

Trap

Converting an ion chamber's air dose with the μen ratio (≈ 1.11) instead of the stopping power ratio (≈ 1.13). An air cavity a few millimetres across is small compared with megavoltage electron ranges, so it is a Bragg-Gray cavity. The wrong ratio understates water dose by about 2 percent.

08 · Refinements

Spencer-Attix and large cavities

Spencer-Attix: the delta-ray fix

Bragg-Gray treats every electron as losing energy continuously along its track. Real electrons occasionally knock out a fast secondary electron, a delta ray, that may itself cross the cavity or escape it. Spencer and Attix split electrons at a cutoff energy Δ, the energy an electron needs to just cross the cavity (about 10 keV for a typical chamber). Energy losses that create electrons below Δ are counted as local; electrons above Δ are counted as part of the fluence. The ratio uses restricted stopping powers, LΔ/ρ, plus a track-end term for electrons that fall below Δ.

DmedDgas = (L̄Δ/ρ)med(L̄Δ/ρ)gas
The stopping power ratios inside TG-51 and IAEA TRS-398 are Spencer-Attix ratios with Δ = 10 keV. The difference from Bragg-Gray is usually under 1 percent, but that is the size of the errors calibration protocols exist to remove.

Large cavities: back to μen

Now make the detector large compared with electron ranges, for example a thick LiF chip in a kilovoltage beam, or any detector thick enough to set up its own CPE. Electrons from the surrounding medium deposit their energy in a thin rim; almost all the dose in the detector comes from photons interacting in the detector material itself. Under CPE in both media, each dose is Kcol = Ψ·μen/ρ, and the photon fluence is common, so the ratio is the μen ratio.

DmedDdet = (μ̄en/ρ)med(μ̄en/ρ)det   (large cavity, photons)
DdetDmed = d · s̄det,med + (1 − d) · (μ̄en/ρ)det,med   (Burlin, intermediate)
Burlin's general cavity theory interpolates: d near 1 for a small cavity, near 0 for a large one. TLDs, diodes and film in some beams sit in this intermediate zone, which is part of why they need beam-specific calibration.
Detector size vs electron rangeDose comes fromConversionExample
SmallElectrons from the medium crossing itStopping power ratioAir-filled ion chamber in MV photons or electrons
IntermediateBothBurlin weightingTLD chip in Co-60; diode in MV beams
LargePhotons interacting in the detectorμen ratioThick TLD or dosimeter in a kV beam
Board point

"Small" and "large" are relative to the range of the secondary electrons, so the same detector can be a small cavity at 18 MV and a large one at 50 kV. Energy dependence of a detector is largest when its effective Z differs from water and the beam is kilovoltage, where μen ratios swing with photoelectric absorption.

Pearl

Small cavity, stopping powers. Large cavity, μen. CPE needed for large cavities, not for Bragg-Gray.

09 · Run it

Cavity bench

Follow one chamber reading all the way to dose to water. Change the room temperature and pressure and watch the air mass, and therefore the dose, move. Then compare the Bragg-Gray answer with what you would get from the wrong (μen) ratio.

Charge to dose to water

Teaching model, not a calibration

0.6 cm³ is a Farmer chamber; 0.01 to 0.1 cm³ is a small-volume or micro chamber. Dry air density at 20 °C and 101.325 kPa is taken as 1.2045 kg/m³ and scaled by the ideal gas law.

0Gy to water (Bragg-Gray, uncorrected)
Exposure equivalent0
With μen ratio 1.1120
Error from wrong ratio0
Real calibrations also correct for ion recombination, polarity, the chamber wall and stem, and the displacement of water by the cavity. TG-51 folds the stopping power ratio and those perturbations into one factor, kQ, applied to a dose-to-water calibration coefficient (Module 06).
In clinic

The temperature-pressure correction you see on every output check, PTP, is this same air-mass effect. A vented chamber on a hot, low-pressure day contains less air, collects less charge, and would under-read by a percent or more if the correction were skipped.

10 · Practice the method

Worked examples

Work each one on paper first. The constants are given in the problem, as they are on the exam.

EX 1

Exposure to air kerma to water dose

A Co-60 beam delivers an exposure of 100 R at a point in air. W/e = 33.97 J/C, 1 R = 2.58 × 10−4 C/kg, and (μen/ρ)water/(μen/ρ)air = 1.112. Find the air collision kerma and the dose to a small mass of water at that point, just large enough for CPE.

Show solution
Kcol,air = 100 × 2.58 × 10−4 × 33.97 = 0.876 Gy = 87.6 cGy
Dwater = 87.64 × 1.112 = 97.5 cGy  (fwater = 0.974 cGy/R)

87.6 cGy in air; ≈ 97 cGy to water

Sanity check: water absorbs about 11 percent more per gram than air at megavoltage, so the water dose must be a little higher than the air kerma and a little below 100 cGy. The common wrong answer is "100 R = 100 cGy", which ignores both W/e and the medium.

EX 2

Fluence to kerma to dose

A point in water receives a fluence of 1.0 × 1010 photons/cm² of 1.25 MeV. (μen/ρ)water = 0.0296 cm²/g and the radiative fraction g = 0.003. 1 MeV = 1.602 × 10−13 J. Find the collision kerma, the total kerma, and the dose under CPE.

Show solution

Energy fluence: Ψ = 1.0 × 1010 × 1.25 × 1.602 × 10−13 = 2.0025 × 10−3 J/cm².

Kcol = 2.0025 × 10−3 × 0.0296 = 5.93 × 10−5 J/g = 5.93 cGy
K = Kcol / (1 − g) = 5.93 / 0.997 = 5.95 cGy
D (CPE) = Kcol = 5.93 cGy

Kcol ≈ 5.93 cGy, K ≈ 5.95 cGy, D ≈ 5.93 cGy

Sanity check: the units chain J/cm² × cm²/g gives J/g; multiply by 1000 for J/kg (Gy). Forgetting that factor gives 0.0059 cGy. Dose equals collision kerma, not total kerma; in water at Co-60 the difference is tiny, which is why it is a recall question more than a calculation one.

EX 3

Chamber charge to air dose

A 0.600 cm³ vented chamber collects 20.0 nC at 22 °C and 100.0 kPa. Dry air density is 1.2045 kg/m³ at 20 °C and 101.325 kPa. W/e = 33.97 J/C. Find the mean dose to the air in the cavity.

Show solution

Air density at room conditions: ρ = 1.2045 × (293.15/295.15) × (100.0/101.325) = 1.181 kg/m³. Mass: m = 1.181 × 0.600 × 10−6 m³ = 7.08 × 10−7 kg (0.708 mg).

Dair = (20.0 × 10−9 / 7.08 × 10−7) × 33.97 = 0.959 Gy

≈ 0.96 Gy to air

Sanity check: a Farmer chamber collects roughly 20 nC per Gy, so tens of nanocoulombs means about a gray. Using the reference density without correcting for 22 °C and 100 kPa gives 0.940 Gy, 2 percent low; that 2 percent is the PTP correction.

EX 4

Bragg-Gray to dose to water

The chamber in EX 3 sat at depth in water in a Co-60 beam. Take s̄water,air = 1.133 and (μen/ρ)water,air = 1.112, and ignore perturbation corrections. What is the dose to water? What would you get with the wrong ratio?

Show solution

An air cavity a few millimetres across is small compared with Co-60 electron ranges in water, so it is a Bragg-Gray cavity and the stopping power ratio applies.

Dwater = 0.959 × 1.133 = 1.087 Gy
with μen ratio: 0.959 × 1.112 = 1.066 Gy, 1.9% low

≈ 1.09 Gy to water

Sanity check: water dose is above air dose by about 13 percent, because water has more electrons per gram than air and stops electrons faster. Dividing instead of multiplying gives 0.846 Gy, an answer below the air dose, which cannot be right.

EX 5

Bone at kilovoltage and megavoltage

A soft-tissue point receives 2.00 Gy. A small piece of cortical bone at the same location sees the same photon fluence, and CPE holds in both. The ratio (μen/ρ)bone/(μen/ρ)tissue is about 4.0 for a beam with effective energy 30 keV and about 0.96 at 1.25 MeV. Find the bone dose in each beam.

Show solution
30 keV: Dbone = 2.00 × 4.0 = 8.0 Gy
1.25 MeV: Dbone = 2.00 × 0.96 = 1.92 Gy

8.0 Gy at kV; ≈ 1.9 Gy at Co-60

Sanity check: at kV the photoelectric effect (Z³) makes calcium-rich bone a strong absorber; at megavoltage, Compton dominates and bone has slightly fewer electrons per gram than soft tissue, so its dose per gram is slightly lower. Thin soft tissue inside bone, such as marrow and osteocytes, is an intermediate cavity and receives less than the bulk bone figure.

11 · Clinic layer

In the planning room

You will never compute a Spencer-Attix ratio in clinic. You will hear these ideas every time a plan is reported in dose to medium, a detector is chosen for skin dose, or a chamber reading is turned into a calibration.

Ballpark numbers to carry in your head

QuantityRough numberWhat it tells you
W/e in dry air33.97 J/CCharge to energy; the anchor of every ion chamber
Air kerma per roentgen0.876 cGy/RExposure to air collision kerma
f-factor, water, MV≈ 0.97 cGy/RWater absorbs about 11 percent more than air
s̄water,air, Co-60≈ 1.13Air dose to water dose in a chamber
Farmer chamber sensitivity≈ 20 nC/Gy0.6 cm³ of air; small chambers collect proportionally less
Dose to water versus dose to medium in cortical bone, MV≈ 10% higher DwRatio of water to bone stopping powers; small in soft tissue

Dose to water or dose to medium

Monte Carlo and grid-based Boltzmann solvers (Acuros XB) compute dose in the actual tissue, Dm. Convolution algorithms and every clinical trial before them report dose to water, Dw. Converting one to the other is a cavity theory problem: a small water cavity in bone is Bragg-Gray, so Dw = Dm × s̄w,m. In soft tissue the difference is about 1 to 2 percent; in cortical bone it is around 10 percent. A spine SBRT plan or a mandible constraint can therefore look different with no change in the delivered beam, only in what is being reported.

Detectors where CPE fails

Skin and buildup-region dose cannot be measured with a detector that needs CPE. Extrapolation chambers and thin-window parallel-plate chambers are the reference tools; OSLDs, TLDs and diodes used in vivo need either buildup caps, which measure dose at dmax rather than at the skin, or calibration under the same conditions as the measurement.

Say this to physics or dosimetryWhy it lands
"Is this plan reported as dose to water or dose to medium?"Explains a 5 to 10 percent change in bone or in a prosthesis region between algorithms
"How was the skin dose measured, and with what buildup?"Shows you know a capped diode reports dmax dose, not surface dose
"Is the calc point in a region with electronic equilibrium?"Points at buildup, interfaces, penumbra and small fields in lung, where kerma-based checks disagree
"Was PTP applied on this output check?"Ties a percent-level output drift to air density in a vented chamber
"Which detector did we use for these small fields?"Detector size versus electron range decides whether volume averaging and fluence perturbation matter (Module 17)
In clinic

Postmastectomy chest wall with bolus: the bolus supplies the upstream electrons so that the skin sits beyond the buildup, where dose approaches the kerma-limited value. Without bolus, the skin is in the region where D is far below Kcol. That is skin sparing, explained as a CPE problem.

12 · Rapid review

Board pearls and traps

Kerma: energy transferred, at the interaction

Indirectly ionizing radiation only. Highest at the surface.

Dose: energy imparted, where it stays

Any radiation, any medium. 1 Gy = 1 J/kg = 100 rad.

K = Ψ·μtr/ρ; Kcol = Ψ·μen/ρ

Kcol = K(1 − g).

CPE: D = Kcol

Not total kerma. TCPE beyond dmax: D = β·Kcol, β slightly above 1.

Buildup: D < Kcol

Curves cross near dmax; beyond it D > Kcol.

Exposure: photons, air, under about 3 MeV

1 R = 2.58 × 10−4 C/kg.

W/e = 33.97 J/C

1 R gives 0.876 cGy air kerma.

f-factor = 0.876 × μen ratio

Water ≈ 0.97 cGy/R at MV. Bone ≫ tissue at kV, slightly below at MV.

Bragg-Gray: stopping power ratio

Cavity small, does not perturb fluence, no photon interactions in it. CPE not required.

Spencer-Attix: restricted stopping powers

Cutoff Δ ≈ 10 keV accounts for delta rays.

Large cavity: μen ratio

Needs CPE in the detector. Burlin interpolates between the two.

s̄water,air ≈ 1.13 at Co-60

Falls slowly with energy.

Common traps
  • Setting dose equal to total kerma under CPE. It equals collision kerma.
  • Saying kerma peaks at dmax. Kerma peaks at the surface; dose peaks at dmax.
  • Listing CPE as a Bragg-Gray condition.
  • Using the μen ratio for an air-filled ion chamber in a megavoltage beam.
  • Applying exposure to electron beams or to photons above about 3 MeV.
  • Forgetting that "small" and "large" cavity are judged against electron range, so they change with beam energy.
  • Dividing by the stopping power ratio when going from air to water. Water dose is higher.
13 · Test yourself

Question bank

Exam-style single best answer. Pick one to see the explanation.

0 / 0 answered
14 · Keep this

Formula card and sources

Fluence
Φ = dN/da   Ψ = dE/da = Φ·hν
Kerma
K = Ψ · μtr/ρ
Kcol = Ψ · μen/ρ = K(1 − g)
Equilibrium
CPE: D = Kcol
TCPE: D = β · Kcol, β > 1
Exposure
Kcol,air = X · W/e
1 R → 0.876 cGy; f = 0.876 · (μen/ρ)med/air
Ion chamber
Dair = (Q / m) · W/e
m = ρair(T,P) · V
Cavity theory
Small: Dmed/Dgas = s̄med,gas
Large: Dmed/Ddet = (μen/ρ)med/det
Burlin: d·s̄ + (1 − d)·μen ratio

Sources

  1. American Board of Radiology. Study Guide for Medical Physics for Radiation Oncology (2023 domain ranges; domain II.A, radiation quantities and units).
  2. Attix FH. Introduction to Radiological Physics and Radiation Dosimetry. Wiley, 1986. Chapters on kerma, CPE and cavity theory.
  3. Khan FM, Gibbons JP. The Physics of Radiation Therapy, 5th ed. Chapters on measurement of ionizing radiation and quantification of absorbed dose.
  4. Podgorsak EB, ed. Radiation Oncology Physics: A Handbook for Teachers and Students. IAEA, 2005. Chapters 1 to 3.
  5. ICRU Report 85. Fundamental Quantities and Units for Ionizing Radiation. J ICRU 2011;11(1).
  6. ICRU Report 90. Key Data for Ionizing-Radiation Dosimetry: Measurement Standards and Applications. J ICRU 2014;14(1). Wair = 33.97 eV, uncertainty 0.35%.
  7. Almond PR et al. AAPM TG-51 protocol for clinical reference dosimetry of high-energy photon and electron beams. Med Phys 1999;26:1847.
  8. IAEA Technical Reports Series 398. Absorbed Dose Determination in External Beam Radiotherapy. Vienna, 2000.
  9. ICRP Publication 103. The 2007 Recommendations of the International Commission on Radiological Protection. Ann ICRP 2007;37.

The buildup panel uses a two-exponential teaching model and the cavity bench applies bare Bragg-Gray without perturbation factors. Both teach direction and magnitude. Clinical numbers come from your physicist's calibration and your machine's commissioning data.