Lesson4_1

# Lesson4_1 - Lesson 04 Exponential Attenuation MP200...

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 Document

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lesson 04 Exponential Attenuation MP200 Radiation Physics - 2010 Duke Medical Physics Graduate Program 1 Introduction This concept is relevant primarily to uncharged ( photons, neutrons) radiation. Charged particles undergo many small collisions and lose their kinetic energy gradually. Simple Exponential Attenuation Indirectly ionizing radiation incidents on a slab of thickness L . Then, Figure 1: Simple Exponential Attenuation N L = N e- μL Scattered and secondary particles may be produced, but are not counted in N L μ = Linear attenuation coefficient Units of μ is cm Definition of μ μ = ( n N ) dx 2 Probability of uncharged particle interactions per unit distance N L N = e- μL ( if μL < . 05 ) e- μL = 1- μL + ( μL ) 2 2!- · = 1- μL Mean free path or Relaxation Length Relaxation length = 1 μ cm It is the average distance a single particle travels through a given medium before interacting. It is also the depth to which a fraction of 1 e (37%) of a large homogeneous beam can penetrate. Exponential Attenuation for Plural Modes of Absorption Assume that each event by each process is totally absorbing producing no scatterings. μ = μ 1 + μ 2 + μ 3 + ··· N L = N e- μL = N e- ( μ 1 + μ 2 ) L + ··· or N L = N ( e- μ 1 L )( e- μ 2 L ) ··· Total number of interactions by all types of processes, Δ N = N- N L = N- N e- μL No. of interactions by a single process, Δ N x = ( N- N L ) μ x μ = N (1- N e- μL ) μ x μ μ x μ = fraction of interactions for process x 3 Example μ 1 = 0 . 02 cm- 1 and μ 2 = 0 . 04 cm- 1 are partial linear attenuation coeffi- cients. If L = 3 cm and N = 10 6 particles, what is N L and how many absorbed in each process in slab? Solution: N L = N e- ( μ 1 + μ 2 ) L = 10 6 e- (0 . 02+0 . 04)5 = 7 . 408 × 10 5 Total number of particles absorbed is, N- N L = (10 6- 7 . 408 × 10 5 ) = 2 . 592 × 10 5 The number absorbed by process 1, Δ N 1 = ( N- N L ) μ 1 μ = 2 . 592 × 10 5 × . 02 . 06 = 8 . 64 × 10 4 Similarly, by process 2, Δ N 2 = ( N- N L ) μ 2 μ = 2 . 592 × 10 5 × . 04 . 06 = 1 . 728 × 10 5 ”Narrow Beam” Attenuation of Uncharged Radiation Consider a beam on uncharged particles on a slab. The scattered and secondary uncharged particles can be counted in N L or not. Figure 2: Narrow-beam geometry. If they are counted, the attenuation equation is invalid and the case is said to be ” broad beam attenuation ”. 4 If scattered or secondary uncharged radiation reaches the detector, but only the pri- maries are counted in N L , one has Broad beam geometry but narrow beam attenuation . Then, the attenuation remains valid under these conditions even for real beams of uncharged primary radiation....
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 / 17

Lesson4_1 - Lesson 04 Exponential Attenuation MP200...

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

View Full Document
Ask a homework question - tutors are online