Measurement of radiation. Chapter 7 Measurement of Radiation: Dosimetry. Description of radiation beam. Energy transfer.

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1 Chapter 7 Measureent of Radiation: osietry Radiation osietry I Text: H.E Johns and J.R. Cunningha, The physics of radiology, 4 th ed. Measureent of radiation escription of radiation bea Calculation of the absorbed dose Bragg-Grey cavity theory Practical ion chabers eterination of absorbed dose for energies above 3 MeV osietry of radio-nuclides escription of radiation bea Fluence Energy fluence Fluence rate dn da dn hv da d dt Energy transfer Photon interaction involves two stages: (a) energy is transferred to charged particles and (b) charged particles transfer energy directly through excitations and ionizations The initial interaction can be described by kera (kinetic energy released in ediu): detr K d Kera is related to photon fluence K E tr Absorbed dose Absorbed dose originates in the second interaction stage, describing the energy retained by the ediu deab d Units: 1Gy (gray)=1 J/ Older unit: 1 rad=10 - Gy=1 cgy This absorbed energy causes ionizations along the charged particle track Electronic equilibriu Transfer of energy to charged particles (kera) does not take place at the sae location as the absorption of energy deposited by charged particles (dose) Kera can be directly related to the fluence, but dose can be calculated only in the assuption of the electronic equilibriu: in any volue as any electrons are stopped as set in otion Under this condition dose is equal to kera 1

2 Electronic equilibriu No photon attenuation In reality dose deposition at any point is the result of kera upstrea In case of electronic equilibriu: Eab K( 1 g) g- fraction of energy lost to bresstrahlung Typically approxiate electronic equilibriu is assued Exaple 1 Calculate the kera given the photon flux /, photon energy 10 MeV, linear attenuation coefficient 0.08 c /g and energy transfer attenuation coefficient 0.0 c /g. tr / K / Etr / E / 4 A. 5 J/ K MeV 3 B. 15 J/ 10 C. 5 J/. 35 J/ MeV / J / 35J / Most dose easureents are based on a easureent of charge produced through ionization: W W is the average energy required to cause one ionization in the In air W=33.85 ev/ion pair Can relate the dose in to the dose in the surrounding ediu ( ) through the ratio of ean stopping powers in and Bragg-Gray forula relates ionization in the cavity to absorbed dose in the ediu S designate averaging over both photon and electron spectra W S Using restricted stopping powers gives ore accurate result The ratio is not very sensitive to the choice of The cavity is always assued so sall that it does not affect the bea spectru Absolute ion chaber An ionization chaber ade of a known aterial and having a cavity of a known volue Have three aterials involved The thickness has to be greater than the range of electrons to separate the fro the ediu Fro easureent can find the dose to the Knowing the ratio of average energy absorption coefficients in ed and ediu ab arrive at: ed ab ed

3 eterination of absorbed dose Correction factor Both the air cavity and introduce perturbations to the bea In order to account for the finite size of both the air cavity and, need to introduce attenuation correction factor k c Values of k c are deterined approxiately ed W S air ab ed k c Effect of teperature and pressure Since the volue of an ion chaber is fixed, need to correct for change in ass due to change in teperature and pressure Correction factor relative to conditions of 0 o C and kpa (760 Hg): 73. t k TP 73. p If the instruent is calibrated for o C adjust the teperature in denoinator Exaple Find the ratio of baroeter readings taken at heights X and X+500 eters. Molar ass of Earth's air is 0.09 /ol, universal constant R=8.31 N /(ol K). A B C P P e P / P e Baroetric forula: gh RT 0 e gh RT e Exposure Exposure is a easure of the ability of radiation to ionize the air; defined as d X d efined only for photons, and only for energies below 3 MeV Roentgen is defined as: 1 C/=3876 R 4 1R.5810 C / of air Equivalent ionization: 3.335x10-10 C/c 3 of air Standard air chaber Exposure can only be easured directly by standard air chaber It has to be large since the sensitive volue is defined by the range of electrons set in otion: 3MeV photons produce electron tracks ~1.5 long 3

4 Exaple 3 Practical ion chabers Air kera is 5 Gy. What is the exposure? A. 0.3 R B. 0.6 R C. 0.9 R. 1. R K W; X K / W C X 3 J J / 33.4 C R 0.58R Assue that even for a volue of air sall copared to the range of electrons the ionization is produced by electrons within the volue Adding air-equivalent and two electrodes obtain a practical device for easureent of exposure It has to be calibrated against the standard chaber to produce energy dependent calibration factor N x : get X=MN x Effective atoic nuber Air-equivalent aterial has to have appropriate Z to represent photoelectric effect interaction coefficient at low energies (30 to 80 kev) The effective atoic nuber of a ixture (typically take =3.5): Z a1z1 az... a Z n n Effective atoic nuber For high energies only electron density is iportant since Copton interaction is doinant Absorbed dose deterination above 3 MeV Ion chaber is still used as the basis for easureents A set of correction factors is eployed to convert the raw easureent to the dose AAPM task group protocols for clinical dosietry of high-energy photon and electron beas: Older TG-1 (1983) is based on exposure (or air-kera) standard and calibration factor (N x ) New TG-51 (1999) is based on an absorbed-dose to water standard and calibration factor (N,w ) Paraeters are published for ion chabers fro different anufacturers Fluence and exposure Energy fluence per roentgen: X / J R ab Fluence per roentgen X 4 C J 1R is equivalent to J 0.873cGy in air C hv / J R ab 4

5 Exposure rate fro g-eitters The exposure rate constant is the exposure rate in R/hr at a point 1 eter away fro a source having activity of 1 Ci Fro the inverse square law the exposure rate at any point distance d away fro a source with activity A: Units of X t R hr Ci d A Exposure rate fro g-eitters Exposure rate constant in air for a source eitting 1 photon of energy hv per disintegration: 194.5hv 1 / R hr 1 Ci ab air Exaple 4 Four 30 Ci 1-15 seeds are arranged at the corners of a 1 c square. Neglecting tissue attenuation, the exposure rate in tissue at the center of the square is: (Exposure rate constant = 1.46 Rc/Ci-hr) A R/hr B. 376 R/h C. 64 R/hr. 19 R/hr E. 350 R/hr X 4 A t d ( 0.5 ) R/hr Exaple 5 The exposure rate constant for a radionuclide is 1.9 R c /(Ci h). How any half-value layers (HVLs) of shielding are required to reduce the exposure rate fro a 19.5 Ci source at to less than 1 R/h? A. 1 B. C E. 6 X A t d R/h 6.3 R/h n ln6.3 n.65 ln n 3 5