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X-Ray Physics. Assumptions: Matter is composed of discrete particles (i.e. electrons, nucleus) Distance between particles >> particle size X-ray photons are small particles Interact with body in binomial process Pass through body with probability p

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## X-Ray Physics

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**X-Ray Physics**Assumptions: Matter is composed of discrete particles (i.e. electrons, nucleus) Distance between particles >> particle size X-ray photons are small particles Interact with body in binomial process Pass through body with probability p Interact with body with probability 1-p (Absorption or scatter) No scatter photons for now (i.e. receive photons at original energy or not at all.**N |∆x|N + ∆N**The number of interactions (removals) number of x-ray photons and ∆x ∆N = -µN∆x µ = linear attenuation coefficient (units cm-1)**Id (x,y) = ∫ I0 () exp [ -∫ u (x,y,z,) dz] d **Integrate over and depth. If a single energy I0() = I0 ( - o), If homogeneous material, then µ (x,y,z, 0) = µ0 Id (x,y) = I0 e -µ0l**Notation** I0 I = I0 e - l l Often to simplify discussion in the book or problems on homework, the intensity transmission, t, will be given for an object instead of the attenuation coefficient t = I/Io = e-µl**X-ray Source**Accelerate electrons towards anode. Three types of events can happen. • Collision events -> heat • Photoelectric effect • Braking of electron by nucleus creates an x-ray (Bremstrahlung effect) Typically Tungsten Target High melting point High atomic number Andrew Webb, Introduction to Biomedical Imaging, 2003, Wiley-Interscience.**Thin Target X-ray Formation**There are different interactions creating X-ray photons between the accelerated electrons and the target. Maximum energy is created when an electron gives all of its energy, 0 , to one photon. Or, the electron can produce n photons, each with energy 0/n. Or it can produce a number of events in between. Interestingly, this process creates a relatively uniform spectrum. Power output is proportional to 02 Intensity = nh 0 Photon energy spectrum**Thick Target X-ray Formation**We can model target as a series of thin targets. Electrons successively loses energy as they moves deeper into the target. Gun X-rays Relative Intensity 0 Each layer produces a flat energy spectrum with decreasing peak energy level.**Thick Target X-ray Formation**In the limit as the thin target planes get thinner, a wedge shaped energy profile is generated. Relative Intensity 0 Again, 0 is the energy of the accelerated electrons.**Thick Target X-ray Formation**Andrew Webb, Introduction to Biomedical Imaging, 2003, Wiley-Interscience. ( Lower energy photons are absorbed with aluminum to block radiation that will be absorbed by surface of body and won’t contribute to image. The photoelectric effect(details coming in attenuation section) will create significant spikes of energy when accelerated electrons collide with tightly bound electrons, usually in the K shell.**How do we describe attenuation of X-rays by body?**µ = f(Z, ) Attenuation a function of atomic number Z and energy Solving the differential equation suggested by the second slide of this lecture, dN = -µNdx Ninx Nout µ Noutx ∫ dN/N = -µ ∫ dx Nin 0 ln (Nout/Nin) = -µx Nout = Nin e-µx**If material attenuation varies in x, we can write**attenuation as u(x) Nout = Nin e -∫µ(x) dx Io photons/cm2 (µ (x,y,z)) Id (x,y) = I0 exp [ -∫ µ(x,y,z) dz] Assume: perfectly collimated beam ( for now), perfect detector no loss of resolution Actually recall that attenuation is also a function of energy , µ = µ(x,y,z, ). We will often assume a single energy source, I0 = I0(). After analyzing a single energy, we can add the effects of other energies by superposition. Detector Plane Id (x,y)**Diagnostic Range**50 keV < E < 150 keV ≈ 0.5% Rotate anode to prevent melting What parameters do we have to play with? • Current • Units are in mA • Time • Units · sec 3. Energy ( keV)**Mass Attenuation Coefficient**Since mass is providing the attenuation, we will consider the linear attenuation coefficient, µ, as normalized to the density of the object first. This is termed the mass attenuation coefficient. µ/p cm2/gm We simply remultiply by the density to return to the linear attenuation coefficient. For example: t = e- (µ/p)pl Mixture µ/p = (µ1/p1) w1 + (µ2/p2) w2 + … w0 = fraction weight of each element**Mechanisms of Interaction**1. Coherent scatter or Rayleigh (Small significance) 2. Photoelectric absorption 3. Compton Scattering – Most serious significance**Physical Basis of**Attenuation Coefficient Coherent Scattering - Rayleigh • • Coherent scattering varies over diagnostic energy range as: • µ/p 1/2 •• •• • • • **Photoelectric Effect**Andrew Webb, Introduction to Biomedical Imaging, 2003, Wiley-Interscience. Photoelectric effect varies over diagnostic energy range as: log /r 1 p 3 log ( Photon energy) K-edge**Photoelectric Effect**Longest photoelectron range 0.03 cm Fluorescent radiation example: Calcium 4 keV Too low to be of interest. Quickly absorbed Items introduced to the body: Ba, Iodine have K-lines close to region of diagnostic interest.**We can use K-edge to dramatically increase absorption in**areas where material is injected, ingested, etc. Photoelectric linear attenuation varies by Z4/ 3 ln /r Log () Photon energy K edge µ/r 1/ 3**Compton Scatter**Interaction of photons and electrons produce scattered photons of reduced energy. When will this be a problem? Is reduced energy a problem? Is change in direction a problem? E’ photon E a Outer Shell electron v Electron (“recoil”)**Satisfy Conservation of Energy and Momentum**(m-mo = electron mass : relativistic effects) Conservation of Momentum 2) 3)**Energy of recoil or Compton electron**can be rewritten as h = 6.63 x 10-34 Jsec eV = 1.62 x 10-19 J mo = 9.31 x 10-31 kg ∆ = h/ moc (1 - cos ) = 0.0241 A0 (1 - cos ) ∆ at = π = 0.048 Angstroms Energy of Compton photon**Greatest effect ∆/ occurs at high energy**At 50 kev, x-ray wavelength is .2 Angstroms Low energy small change in energy High energy higher change in energy**µ/r electron mass density**Unfortunately, almost all elements have electron mass density ≈ 3 x 10 23 electrons/gram Hydrogen (exception) ≈ 6.0 x 1023 electrons/gram Mass attenuation coefficient (µ/r) for Compton scattering is Z independent Compton Linear Attenuation Coefficient µ p Avg atomic number for Bone ~ 20 Avg atomic number for body 7 or 8**Rayleigh, Compton, Photoelectric are independent sources of**attenuation t = I/I0 = e-µl = exp [ -(uR + up + uc)l] µ () ≈ pNg { f() + CR (Z2/ 1.9) + Cp (Z3.8/ 3.2)} Compton Rayleigh Photoelectric Ng electrons/gram ( electron mass density) So rNg is electrons/cm3 Ng = NA (Z/A) ≈ NA /2 (all but H) A = atomic mass f() = 0.597 x 10-24 exp [ -0.0028 (-30)] for 50 keV to 200 keV in keV**Attenuation Mechanisms**Andrew Webb, Introduction to Biomedical Imaging, 2003, Wiley-Interscience. Curve on left shows how photoelectric effects dominates at lower energies and how Compton effect dominates at higher energies. Curve on right shows that mass attenuation coefficient varies little over 100 kev. Ideally, we would image at lower energies to create contrast.**Photoelectric vs. Compton Effect**Macovski, Medical Imaging Systems, Prentice-Hall The curve above shows that the Compton effect dominates at higher energy values as a function of atomic number. Ideally, we would like to use lower energies to use the higher contrast available with The photoelectric effect. Higher energies are needed however as the body gets thicker.

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