Planar X-ray Imaging
Measure the integeral of the linear attenuation coefficient over the
beam path through the object.
P(x,y) = I e ∫
µdz
o
µ
where µ = µ (x,y,z)
has two main contributions
1. Photoelectric effect - this removes photons from the beam and
has the properties we want for imaging
2. Compton Scattering - scatters photons which may end up in the
detector. Can lead to noise.
Calculation of Attenuation Coefficient
To the extent that Compton scattered photons do not reach the detector,
they also contribute to the signal, but contracts is low.
A reasonable (approximate) analytical expression for the attenuation
coefficient is:

Z 
µ = ρN  f (E ) + C

E 
m
g
p
n
Z
A
Z = atomic number
N =N
g
A
A = atomic mass
N = Avogadro' s number
A
ρ = density
f (E ) = 0.597 ×10 e
−24
−0.0028 ( E − 30 )
= Compton scattering part for low E
C = 9.8 ×10
−24
p
m = 3.8,n = 3.2
Calculation of Effective Z
This is quite approximate but does permit simple computation provided
that the energy is high enough, the simple scattering is not an issue, and
less than, or equal to, 200 keV.
For practical problems, still need to calculate an effective Z.
(
Z = ∑α Z
eff
i
i
)
1
m
i
m
where α is the electron fraction of the i
N
α =
∑N
i
th
element.
gi
i
gj
j
Z 
N = N w 
A 
i
gi
A
i
i
fraction by weight of the element
Sample Effective Z Calculations
Calculate Z
for :
eff
1. water, H O
2
2. oil, CH (CH
3
2
) CH
4
3
→ ρ = 0.66, mw = 86.2
3. calcium carbonate, CaCO → ρ = 2.930, mw = 100.09
3
Plot µ and µ
p
c
for each
Atomic # (Z)
Atomic Weight
(A)
H
1
1.008
C
6
12.011
O
8
15.994
Ca
20
40.08
Sample Effective Z Calculations (Water)
16  8 
; N
= N   
=N
18 16 
= 0.444 N
;
= 0.111N
0.444
α =
= 0.8
; α = 0.2
0.555
N
go
A
gH
A
o
Z
eff
A
eff
A
A
H
[
= 0.8(8) + 0.2(1)
3.8
3.8
]
1
3.8
= 7.54
is just a normal weighted sum.
 2  1 
 
18 1.008 
Sample Effective Z Calculations (Hexane)
 72  6 
 14 
; N
N =N 
=N 
 
(1)
 86.2 12 
 86.2 
= 0.42N ;
= 0.16N
0.42
0.16
α =
= 0.724 ; α =
= 0.275
0.58
0.58
go
A
gH
A
A
o
Z
eff
A
eff
A
H
[
= 0.724(6) + 0.275(1)
= 5.51
= 12
3.8
3.8
]
1
3.8
Sample Effective Z Calculations (Calcium Carbonate)
N
gCa
N
gC
N
gO
α
Z
Ca
eff
 20  20 
=N 
 
100  40 
 12  6 
=N 
 
100 12 
 48  8 
=N 
 
100 16 
0.1
=
= 0.25
; α
0.4
A
A
A
A
A
[

= 0.10N  

= 0.06N ∑ = 0.40


= 0.24 N 

0.06
0.24
=
= 0.15 ; α =
= 0.6
0.4
0.4
A
C
O
= 0.25(20) + 0.15(6) + 0.6(8)
3.8
3.8
= [21,971+135 +1,621
]
1
A
eff
= 14.2
= 21.4
3.8
3.8
]
1
3.8
Determining The Signal
We would like to know what the signal is at the detector, but this depends
on the geometry since Compton scattering is important.
E
E
e
'
θ
α
−
v
e
−
Conservation of energy:
E = E + (m − m )c
'
2
o
where c is the velocity of light
and m is the rest mass of electron
m
m =
= mass of moving electrons
1− v
c
2
o
o
( )
2
Determining The Signal
Conservation of momentum:
E E
= cos(θ ) + mv cos(α )
c c
E
0 = sin(θ ) − mv sin(α )
c
'
'
Solving these together yield:
EE
E
−
E
=
1− cos(θ ))
(
)
(
mc
'
'
2
o
Angular Dependence of Compton Scattering (Low Energy)
At low energies the scatter angle distribution is approximately isotropic.
Plot ∆E vs angle for various energies
Note: ∆τ = 0.0241(1 - cos(θ)) where ∆τ is in Angstroms. To convert to keV,
recall that 50keV is about 0.2Å.
E = hν =
hc
(
;
λ
1
λ = • 50keV • 0.2 A
E
o
in
E =
out
1
λ
out
)
1
∆E =
× 50keV • 0.2 A
0.241(1− cos(θ ))
;
(50keV • 0.2 A)
(
o
)
1
λ =
50keV • 0.2 A − 0.0241(1− cos(θ ))
E
out
in
o
o
o
∆E = E − E = E −
in
out
in
(
50keV • 0.2 A
)
1
50keV • 0.2 A − 0.0241(1− cos(θ ))
E
in
o
Angular Dependence of Compton Scattering (High Energy)
At high energies, the photon energy changes significantly with scatter
angle. This results in most scatter being forward-directed and thus highenergy X-ray is extremely challenging since we can not distinguish
scattered from transmitted radiation.
Photon electron (transmitted radiation) reaches the detector with the
original beam geometry. Compton reaches as the solid angle subtended
by the detector.
Compton-based Imaging
Can try Compton-based imaging with energy detection
θ
θ
Detector at energy E’ < E,
thus scattering angle is θ.
I, E
{
o
o
monoergetic
This specifies a cone that the radiation can come form.
This can be reconstructed but it is difficult and only used in cosmology
where the original source is at infinity.
Compton-based Imaging
θ
θ
This specifies a cone that the radiation can come form.
This can be reconstructed but it is difficult and only used in cosmology
where the original source is at infinity.