Measurement of radiation Chapter 7 Measurement of Radiation: • Description of radiation beam • Calculation of the absorbed dose Dosimetry – Bragg-Grey cavity theory – Practical ion chambers Radiation Dosimetry I – Determination of absorbed dose for energies above 3 MeV Text: H.E Johns and J.R. Cunningham, The – Dosimetry of radio-nuclides physics of radiology, 4th ed. http://www.utoledo.edu/med/depts/radther

Description of radiation beam Energy transfer

• Photon interaction involves two stages: (a) energy is • Fluence transferred to charged particles and (b) charged particles dN   transfer energy directly through excitations and ionizations da • The initial interaction can be described by kerma (kinetic • Energy fluence energy released in medium): dN hv   dEtr da K  • Fluence rate dm d • Kerma is related to photon fluence   dt    K    Etr   

Absorbed dose Electronic equilibrium

• Absorbed dose originates in the second interaction • Transfer of energy to charged particles (kerma) stage, describing the energy retained by the does not take place at the same location as the medium absorption of energy deposited by charged dE D  ab particles (dose) dm • Kerma can be directly related to the fluence, but • Units: 1Gy (gray)=1 J/kg dose can be calculated only in the assumption of – Older unit: 1 rad=10-2 Gy=1 cGy the electronic equilibrium: in any volume as many • This absorbed energy causes ionizations along the electrons are stopped as set in motion charged particle track • Under this condition dose is equal to kerma

1 Electronic equilibrium Example 1

No photon • In reality dose deposition attenuation at any point is the result of • Calculate the kerma given the photon flux kerma upstream 1016/m2, photon energy 10 MeV, linear • In case of electronic attenuation coefficient 0.028 cm2/g and equilibrium: energy transfer attenuation coefficient 0.022

   cm2/g. D    E  K(1 g) tr /    ab K   /  Etr   /  E     /  A. 5 J/kg 1 104 m2 • g- fraction of energy lost to K 1016 0.022 10MeV  bremsstrahlung B. 15 J/kg m2 103 kg • Typically approximate C. 25 J/kg 2.21014 MeV / kg  electronic equilibrium is 14 13 D. 35 J/kg 2.210 1.610 J / kg  35J / kg assumed

Bragg-Gray cavity theory Bragg-Gray cavity theory

• Most dose measurements are based on a • Can relate the dose in gas to the dose in the measurement of charge produced through surrounding medium (“wall”) through the ratio of mean stopping powers in gas and wall gas ionization: Q • Bragg-Gray formula relates ionization in the gas Dgas  W mgas cavity to absorbed dose in the medium Q • W – is the average energy wall Dwall  W  S gas required to cause one m gas ionization in the gas wall • S gas designate averaging over both photon and • In air W=33.85 eV/ion pair electron spectra

Bragg-Gray cavity theory Absolute ion chamber • An ionization chamber made of a known material and having a cavity of a known volume • Have three materials involved • The wall thickness has to be greater than the range of electrons to separate the wall from the medium • From measurement can find the dose to the wall • Knowing the ratio of average energy absorption coefficients in med    wall and medium  ab  arrive at:    • Using restricted stopping powers gives more accurate result  wall • The ratio is not very sensitive to the choice of D med  ab  • The cavity is always assumed so small that it does not affect Dmed  Dwall   the beam spectrum   wall

2 Determination of absorbed dose Correction factor • Both the air cavity and wall introduce perturbations to the beam • In order to account for the finite size of both the air cavity and wall, need to introduce attenuation correction factor kc

• Values of kc are determined approximately

med Q wall  ab  Dmed  W Sair   kc m   wall

Effect of temperature and pressure Example 2

• Since the volume of an ion chamber is fixed, need • Find the ratio of barometer readings taken at heights to correct for change in gas mass due to change in X and X+500 meters. Molar mass of Earth's air is temperature and pressure 0.029 kg/mol, universal gas constant • Correction factor relative to conditions of 0oC and R=8.31 N·m/(mol·K). 101.3 kPa (760 mm Hg): Barometric formula: gh  273.2  t 101.3 A. 0.98  k    RT TP    P  P0e  273.2  p B. 0.96   gDh 0.0299.8500   C. 0.94 RT 8.31295 o P / P  e  e  • If the instrument is calibrated for 22 C – adjust the D. 0.92 500 temperature in denominator e0.058 1 0.058  0.94

Exposure Standard air chamber

• Exposure is a measure of the ability of radiation to ionize the air; defined as dQ X  dm • Defined only for photons, and only for energies below 3 MeV • Exposure can only be measured directly by standard air • Roentgen is defined as: 1R  2.58104 C / kg of air chamber • It has to be large since the sensitive volume is defined by – 1 C/kg=3876 R the range of electrons set in motion: 3MeV photons produce – Equivalent ionization: 3.335x10-10 C/cm3 of air electron tracks ~1.5 m long

3 Example 3 Practical ion chambers

• Air kerma is 5 mGy. What is the exposure? Q Q K  W; X  A. 0.3 R m m 3 J  J  • Assume that even for a volume of air small compared to the X  K /W  510 /33.4   B. 0.6 R kg  C  range of electrons the ionization is produced by electrons C within the volume C. 0.9 R 0.15103  0.15103 3.9103 R  0.58R kg • Adding air-equivalent wall and two electrodes – obtain a D. 1.2 R practical device for measurement of exposure • It has to be calibrated against the standard chamber to produce energy dependent calibration factor Nx: get X=MNx

Effective atomic number Effective atomic number • Air-equivalent material has to have appropriate Z to represent photoelectric effect interaction coefficient at low energies (30 to 80 keV) • The effective atomic number of a mixture (typically take m=3.5):

m m m Z  m a Z  a Z ...  a Z 1 1 2 2 n n

• For high energies only electron density is important since Compton interaction is dominant

Absorbed dose determination Fluence and exposure above 3 MeV • Ion chamber is still used as the basis for measurements • Energy fluence per • A set of correction factors is employed to convert the roentgen: raw measurement to the dose  0.00873 J • AAPM task group protocols for clinical dosimetry of  X  /  kg R high-energy photon and electron beams: ab – Older TG-21 (1983) is based on exposure (or air-kerma) • Fluence per roentgen standard and calibration factor (Nx) – New TG-51 (1999) is based on an absorbed-dose to water  0.00873 J standard and calibration factor (ND,w)  X hv  /  kg R – Parameters are published for ion chambers from different  ab  manufacturers C J 1R is equivalent to 2.58104 33.85  0.00873J kg  0.873cGy in air kg C

4 Exposure rate from g-emitters Exposure rate from g-emitters

• The exposure rate constant is the exposure rate in R/hr at a point 1 meter away from a source having • Exposure rate activity of 1 Ci constant in air for a • From the inverse square law the exposure rate at source emitting 1 any point distance d away from a source with photon of energy hv activity A: per disintegration: X  A  2 t d 2 1 1  194.5hvab / air Rm hr Ci

R m2  • Units of      hr Ci 

Example 4 Example 5

• Four 30 mCi 1-125 seeds are arranged at the • The exposure rate constant for a radionuclide is corners of a 1 cm square. Neglecting tissue 12.9 R cm2/(mCi h). How many half-value layers attenuation, the exposure rate in tissue at the (HVLs) of shielding are required to reduce the center of the square is: (Exposure rate constant = exposure rate from a 19.5 mCi source at 2 m to 1.46 Rcm2/mCi-hr) less than 1 mR/h? A. 1 A. 3 15 R/hr X   A 12.919.5 B. 2    B. 376 R/h X 4 A 41.4630 2 2    C. 3 t d 200 C. 264 R/hr 2 2 2 t d ( 20.5 ) D. 4 0.00629 R/h  6.3 mR/h D. 192 R/hr E. 6 6.3 ln6.3 E. 350 R/hr 350.4 R/hr 1  n   2.65  n  3 2n ln2

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