Assumptions: 1) Matter is composed of discrete particles (i.e. electrons, nucleus) 2) Distance between particles >> particle size 3) 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) 4) 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)  photons and ∆x number of x-ray ∆N = -µN∆x µ = linear attenuation coefficient (units cm -1 )

I d (x,y) = ∫ I 0 (  ) exp [ -∫ u (x,y,z,  ) dz] d  Integrate over  and depth.

If a single energy I 0 (  ) = I 0  (   o ), If homogeneous material, then µ (x,y,z,  0 ) = µ 0 I d (x,y) = I 0 e -µ0l

Notation

I 0  I = I 0 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/I o = e -µl

Accelerate electrons towards anode. Three types of events can happen. 1. Collision events -> heat 2. Photoelectric effect 3. 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, proportional to  0 2  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 Intensity = nh  Photon energy spectrum  0

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 Again,  0  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 N in  x   N out µ Nout x ∫ dN/N = -µ ∫ dx Nin 0 ln (N out /N in ) = -µx N out = N in e -µx

If material attenuation varies in x, we can write attenuation as u(x) N out = N in e -∫µ(x) dx I o photons/cm 2 (µ (x,y,z)) I d (x,y) = I 0 exp [ -∫ µ(x,y,z) dz] Assume: perfectly collimated beam ( for now), perfect detector no loss of resolution I d (x,y) Detector Plane Actually recall that attenuation is also a function of energy  , µ = µ(x,y,z,  ). We will often assume a single energy source, I 0 = I 0 (  ). After analyzing a single energy, we can add the effects of other energies by superposition.

Diagnostic Range 50 keV < E < 150 keV  ≈ 0.5% Rotate anode to prevent melting What parameters do we have to play with? 1. Current Units are in mA 2. Time Units · sec 3. Energy ( keV)

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 cm 2 /gm We simply remultiply by the density to return to the linear attenuation coefficient. For example: t = e - (µ/p)pl Mixture µ/p = (µ 1 /p 1 ) w 1 + (µ 2 /p 2 ) w 2 + … w 0 = fraction weight of each element

1. Coherent scatter or Rayleigh (Small significance) 2. Photoelectric absorption 3. Compton Scattering – Most serious significance

Coherent Scattering - Rayleigh • • • Coherent scattering varies over diagnostic energy range as: µ/p  1/  2 • •  • 

Andrew Webb, Introduction to Biomedical Imaging, 2003, Wiley-Interscience.

log  / r K-edge log  Photoelectric effect varies over diagnostic energy range as: ( Photon energy)   1 p  3

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 Z 4 /  3 ln  / r K edge µ/ r  1/  3 Log (  ) Photon energy

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 Outer Shell electron   v Electron (“recoil”)

Satisfy Conservation of Energy and Momentum

1)

E



E

  (

m



m

0 )

c

2 (m-m o = electron mass : relativistic effects)

m



m

0 / 1  (

v

/

c

) 2

Conservation of Momentum

2) 3)

E c

 0 

E

 cos(  )

E c

 sin

c

  

mv

cos(  )

mv

sin 

Energy of recoil or Compton electron



E



E



E

 can be rewritten as

E

  1 

E

   h = 6.63 x 10 -34 Jsec eV = 1.62 x 10 -19 J m o = 9.31 x 10 -31 kg ∆  ∆  = h/ m o c (1 - cos  ) = 0.0241 A 0 at  = π = 0.048 Angstroms (1 - cos  )

h

Energy of Compton photon

   1  ( 1 

h

 cos  )

hv m

0

c

2

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 10 23 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/I 0 = e -µl = exp [ -(u R + u p + u c )l] µ (  ) ≈ pN g { f(  ) + C R (Z 2 /  1.9

) + C p (Z 3.8

/  3.2

)} Compton Rayleigh Photoelectric N g electrons/gram ( electron mass density) So r N g is electrons/cm 3 N g = N A (Z/A) ≈ N A /2 (all but H) f(  ) = 0.597 x 10 -24 exp [ -0.0028 (  -30)] for 50 keV to 200 keV A = atomic mass  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.