Lesson11d

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

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the 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 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 – 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 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 = Z E max 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 = 20 MeV ( ignore δ rays.) Solution: For 20 MeV electrons dT ρdx c = 1 . 654 MeV.cm 2 .g- 1 4 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....
View Full Document

This note was uploaded on 10/30/2010 for the course MP 200 taught by Professor Guna during the Fall '10 term at Duke.

Page1 / 16

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

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online