Lesson11d - Lesson 11d Calculation of absorbed dose MP200...

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Lesson 11d Calculation of absorbed dose MP200 Radiation Physics - 2010 Duke Medical Physics Graduate Program 1
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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 ρdx 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 δ -rays is negligible: If t >> range of δ -rays, all the secondary electrons lose en- ergy in the material. 2
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There exists a CPE of δ 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 θ Energy deposited in the slab = ρt cos θ dT ρdx c Particle fluence is defined by, Φ = dN da = (track lengths in volume) Volume Note: Another definition of Φ: 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. Φ = t cos θ t.dA = 1 cos θ.dA Energy absorbed, E = Φ ρt dT ρdx c dA Dose, D = E m = Φ ρt dT ρdx c dA ρtdA D = Φ dT ρdx c Dose is independent of thickness t and θ . Therefore, the same equation holds for particles traveling in all directions, and an arbitrary volume. 3
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When electrons have a spectrum of energies: Assume, dN = number of electrons between E and E + dE crossing an area dA . Then, dN = d Φ dE .dA.dE Here, d Φ dE is called the fluence spectrum. Φ T = d Φ dE Units: of Φ T , electrons.MeV - 1 .cm - 2 . D = E max 0 Φ T dT ρdx c dE Dose Units: D = Φ dT ρdx c ( Units : MeV g - 1 ) But, 1 MeV.g - 1 = 1 . 602 × 10 - 10 Gy D = 1 . 602 × 10 - 10 Φ dT ρdx 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 0 = 20 MeV ( ignore δ rays.) Solution: For 20 MeV electrons dT ρdx c = 1 . 654 MeV.cm 2 .g - 1 4
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D = 1 . 602 × 10 - 10 Φ dT ρdx c D = 1 . 602 × 10 - 10 × 3 × 10 11 × 1 . 654 = 79 . 5 Gy Estimating δ ray energy losses In the case where the foil is comparable in thickness to the range of the δ -rays produced in it, assumption c may not be satisfied unless the target foil is sandwiched between ”buffer” foils of the same material.
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