Download BME 412 Biomedical Signal Processing Class 1

BME 560
Medical Imaging: X-ray, CT, and
Nuclear Methods
Radiation Physics Part 2
Today
• EM Radiation interactions with matter
–
–
–
–
Rayleigh scatter
Photoelectric effect
Compton scatter
Pair production
• Attenuation of EM Radiation
X-ray interactions with matter
Possible outcomes for an x-ray or gamma ray when traveling through matter are:
– No effect (a)
(it travels along the original
path with original energy)
(a) (b)
(c)
– Absorption (b)
(it is absorbed in the matter)
– Scatter (c)
(It changes its direction of
travel and/or energy)
Imaging detector
EM Radiation
• Four possible interactions with matter
–
–
–
–
Rayleigh scatter
Photoelectric effect
Compton scatter
Pair production
• Each has its own physical properties
Rayleigh
Scattering
• Elastic (coherent, classical) scattering scatter with direction change but no energy loss.
huincident = huscattered ; lincident = lscattered
• Characteristics
– an interaction of the x-ray with the entire atom via a
refractory mechanism (wavelength ~ atomic radius)
– occurs for longer wavelengths (low x-ray energies in
the mammography range - 15 - 30keV)
– reduces imaging quality in x-ray imaging.
E(keV)
P(%)
~30
<12
>70
<5
P: probability
Compton scattering
1. It is also referred as inelastic or
non-classical scattering
2. an interaction of incident x-ray
with outer shell electrons;
3. almost atomic (proton) number
(Z) independent;
4. proportional to re;
5. weakly energy dependent until
energy is high (1MeV);
6. a disturbance to x-ray image
quality.
Compton Scatter Equation
Planck’s constant
l
scattered
l
original
h

(1  cos  )
m0 c
Scatter angle
wavelengths
Electron rest mass
Compton Scattering energy and angular dependence
hu0
Energy and
momentum
conservation
hu sc 
f

husc
hu 0
1
Ee-
hu 0
(1 cos )
2
m0c
hu0
(1  cos  )
2
m0 c
Ee   hu0
hu0
1
(1  cos  )
2
m0 c
1 v 2 /c 2
cos 
(hu 0  hu sc  cos  )
m0v  c
f= 00 &  = 1800 Compton scattering
(Max energy transfer to electron)
Ee
hu 0
hu 0
(1
cos180)
2
m 0c 2
m 0c 2
 hu 0
 hu 0
hu 0
hu 0
1
(1
cos180)
1
2
m 0c 2
m 0c 2
hu sc,min  hu 0
1
hu 0
1 2
m 0c 2
Ee
hu sc
 180,hu 0 100keV
min
 180,hu 0 100keV
 100keV
Ee
max
hu sc
 180,hu 0 10MeV
min
 180,hu 0 100keV
2  0.2
 28keV
1 2  0.2

f

Ee-
husc
(hu 0  10MeV ;
hu 0
(hu 0  100keV;
 0.2)
2
m0c
max
hu0
 100keV  28keV  72keV
hu 0
 20)
m0c 2
 10MeV
2  20
 9.8MeV
1 2  20
 10MeV  9.8MeV  0.2MeV
 = 900 Compton scattering
Ee
hu0
hu 0
hu 0
(1
cos90)
m 0c 2
m 0c 2
 hu 0
 hu 0
hu 0
hu 0
1
(1
cos90)
1
m 0c 2
m 0c 2
hu sc,min  hu 0
1
hu 0
1
m 0c 2
f

husc
(hu 0  100keV;
Ee
Ee-
hu 0
 0.2)
2
m0c
 100keV
0.2
 17keV
1 0.2
0
hu 0
(hu 0  10MeV ;
 20)
min
2
h
u
m0c
sc  90,hu 100keV  100keV 17keV  83keV
0
20
Ee
 10MeV
 9.5MeV
 90,hu 0 10MeV
1 20
hu sc
min
hu sc
min
 90,hu 0 10MeV
 90
 90,hu 100keV
 10MeV  9.5MeV  0.5MeV
 m0c 2  511keV

Compton scatter angular distribution
hu0
f

Ee-
husc
The lower the incident x-ray energy, the more isotropic (angular distribution) the
(compton) scattered x-ray is.
(Isotropic - each point of compton interaction becomes an x-ray source - bad for imaging)
Photoelectric Effect (PE)
1)
An total absorption event (not a scatter): all or none;
2)
An interaction of incident x-ray with inner shell electron;
3)
Highly dependent on atomic (proton) number (Z) - Z3;
4)
Highly energy dependent - E-3;
Photoelectric effect
• Incident electron deposits all of its energy into
a shell electron
– Must overcome electron binding energy
• The shell electron (photoelectron) escapes
– Ionization event
– Energetic electron interacts with tissue (short
range)
• The photon must start with energy greater than
the electron binding energy.
Photoelectric Effect
• “K-edges”
– Photoelectric
attenuation as a
function of energy is
discontinuous at the
shell binding
energies.
– We make use of this
property in several
ways.
From Sprawls, The Physical Basis of Medical Imaging
Photoelectric effect energy and Z dependence
High Z dependence (Z3) of PE is the basic reason why diagnostic x-ray is
suited to medical imaging:
– PE completely removes the incident x-ray (no scatter or partial absorption)
– PE differentiates different atomic compositions in tissue (bone and fat);
– PE is behind the use of high Z image contrast materials;
– PE decreases drastically (E-3) with increasing x-ray energy
(not good for imaging but good for therapy)
X-ray imaging contrast
ZBa=56; r =3.5g/cm3
Zwater, effective=7.25; r =1g/cm3
(ZBa/Zwater)3=512
Pair Production
• occurs under strong electric field of nucleus (=>Z dependent);
• only occurs if the incident x-ray energy is > 1.022MeV (enough to
create the rest masses of an e- and a e+);
• completely removes the incident x-ray (no scatter or partial absorption)
• increases with increasing x-ray energy
• generates an electron and a positron
Positron annihilation
2nd part of fig 3-12, redbook
• Positron annihilation gamma rays are ~1800 apart;
• each gamma ray has 511keV;
• principle of PET scanner (a measure of PET agent
distribution in patient).
EM Radiation Interactions
• Rayleigh scatter: only significant at low energies
• Compton scatter: dominates at imaging energies
• Photoelectric effect: significant at imaging energies
• Pair production: cannot occur at imaging energies

Total linear attenuation coefficient
N x  N 0e   x
  Rayleigh   photoelectric  compton   pair
Soft Tissue
productin
X-ray attenuation
N
Absorbed
&
scattered
n
transmitted
N-n

Dx
n    N  Dx
DN
   Dx
N
ln N  x
N x  N 0e
N t  N 0e l t
(l : decay constan t)
 x

Linear attenuation coefficient (cm-1):
0.10cm-1 ==> 10% per cm.
Total attenuation coefficients of water and lead
Characteristic x-ray peak (K-shell)
Table of interaction vs. x-ray energy
(Khan?)
Relative importance of each x-ray interaction in water
Mass attenuation coefficient (/r)
linear attenuation coefficient () normalized to the density (r) of the
irradiated material. Therefore it is independent of density.
/r (cm2/g) =  (cm-1) /r (g/cm3)
Nx = N0 e- (/r)rx
Exercise
Given:
CT x-ray energy is120kVp. The patient
thickness is 20 cm and 25 cm in the AP and
lateral direction, respectively. If the x-rays
attenuate 20% per cm of tissue, what is the xray intensity difference as measured by the
detector in the AP and lateral positions?
xap
patient
xat
Solution:
 0.20 cm-1
N x  N 0e x
X-ray detector
N ap
e   20
    25  e  (2520)
N lateral e
 e 0.2 5  e  2.72
Half Value Layer (HVL)
N HVL  N 0 /2  N 0e(  / r ) r  HVL
HVL 
(T1/ 2 
ln 2

ln 2
l

0.693

100
(a)(b)
(c)
)
HVL
• HVL or is a function of BOTH the
x-ray (E) and the matter (Z,r).

• T1/2 is a function of radionucleus only.
50 = 100/2
Narrow-beam and broad-beam HVL
• which HVL is larger, broad-beam or narrow-beam?
Polyenergetic x-ray beam
Beam Hardening:
HVL (x)
• Beam hardening - mean energy of a poly energetic x-ray beam
increases as it travels through the irradiated material.
• Reason - lower energy x-rays have larger than higher energy x-rays
and thus attenuate quicker.
• HVL(x1) < HVL (x2) when x1 < x2 (for polyenergetic x-ray beams)
Review of interactions
Particles (electrons)
• Collisional
– Ionization, heat
• Radiative
– X-rays (characteristic,
Bremsstrahlung)
EM
• Rayleigh scatter
– Change direction, no energy
loss
• Compton scatter
– Change direction, energy loss
• Photoelectric
– Absorption, photoelectron
released
• Pair production
– Absorption, electron-positron
pair released