This preview shows pages 1–6. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Lesson 11d Calculation of absorbed dose MP200 Radiation Physics  2010 Duke Medical Physics Graduate Program 1 Calculation of absorbed dose In this section, we calculate doses in thin foils and thick foils due to heavy ( monoenergetic and monodirectional) charged particles, and electrons. Simplest case: Dose in thin foils Consider a passage of monoenergetic charged particles, all traveling in the same direction incident on a thin slab of thickness t as shown. dA is the cross sectional area. Assume that the foil is thin enough so that: the collision stopping power dT dx c is nearly constant through the slab. every particle passes straight through the foil. i.e. scattering is negligible. Assume the energy carried out of slab by rays is negligible: If t >> range of rays, all the secondary electrons lose en ergy in the material. 2 There exists a CPE of rays with matter adjacent to slab. For heavy charged particles, all of these assumptions are satisfied. For electrons, the second assumption is weaker. From the figure, path length through slab = t cos Energy deposited in the slab = t cos dT dx ! c Particle fluence is defined by, = dN da = (track lengths in volume) Volume Note: Another definition of : The fluence at a point is numerically equal to the expectation value of the sum of the particle track lengths ( assumed to be straight) that occur in an infinitesimal volume dV at P , divided by dV . i.e. = t cos t.dA = 1 cos .dA Energy absorbed, E = t dT dx ! c dA Dose, D = E m = t dT dx c dA tdA D = dT dx ! c Dose is independent of thickness t and . Therefore, the same equation holds for particles traveling in all directions, and an arbitrary volume. 3 When electrons have a spectrum of energies: Assume, dN = number of electrons between E and E + dE crossing an area dA . Then, dN = d dE ! .dA.dE Here, d dE is called the fluence spectrum. T = d dE ! Units: of T , electrons.MeV 1 .cm 2 . D = Z E max T dT dx ! c dE Dose Units: D = dT dx ! c ( Units : MeV g 1 ) But, 1 MeV.g 1 = 1 . 602 10 10 Gy D = 1 . 602 10 10 dT dx ! c Gy Example: What is the dose ( Gy ) in a thin LiF dosimeter struck by a fluence of 3 10 11 electrons.cm 2 with T = 20 MeV ( ignore rays.) Solution: For 20 MeV electrons dT dx c = 1 . 654 MeV.cm 2 .g 1 4 D = 1 . 602 10 10 dT dx ! c D = 1 . 602 10 10 3 10 11 1 . 654 = 79 . 5 Gy Estimating ray energy losses In the case where the foil is comparable in thickness to the range of the rays produced in it, assumption c may not be satisfied unless the target foil is sandwiched between buffer foils of the same material....
View Full
Document
 Fall '10
 Guna

Click to edit the document details