Simulating the Light
Distribution in Sodium Iodide
Detectors
Lisi Goodlett
Dr. Ingo Wiedenhöver
STRIPSI Detector System
Contains twenty-four sodium iodide detectors, in
two concentric rings around the experiment.
Sodium Iodide Detectors
Used to detect gamma rays
Composed of a sodium iodide crystal
Windows at either end
Photomultiplier tubes affixed to each
window
Current Method of Calibration
E1
Based on an
exponential attenuation
of light.
E gamma e
E2 Egamma e
E
m
m
L 2
L 2
E1 E 2
x log E 1 E 2
x
x
Data taken from a calibration
run
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●
This data was analyzed
using the previous
formulas.
Notice that the energy is
not constant as one
would expect.
The current means for
calibration do not
completely describe the
the behavior of the light
in the detector.
Energy vs. Asymmetry
Objective
Computationally simulate the light
distribution within the sodium iodide
detector.
Basic Assumptions
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Attenuation of light can be described through
intensity losses due to reflections within the
cavity walls.
About half of the scintillation photons
produced propogate to either end of the
crystal.
Amount of light detected at either end can be
calculated by varying the point of interaction
and integrating over possible angles.
A few Uncertainties:
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●
Our detectors were given to us (third or
fourth hand), and had been specially made.
In consequence,
we do not know all of the materials used to make the detectors.
The index of refraction for sodium iodide is
n=1.8. In general, glass has a value of n=1.5.
Critical angle:
thetac arcsin ni nt
approx: 56 degrees
n (of window)=1.5
detector 1
1.00
0.90
0.80
0.60
energy
energy
0.70
0.50
0.40
0.30
0.20
0.10
0.00
-0.60
-0.40
-0.20
0.00
0.20
asymmetry
0.40
0.60
160
150
140
130
120
110
100
90
80
70
60
50
40
30
20
10
0
30
40
50
60
70
asymmetry
80
90
Results:
Index of refraction of glass must be greater than
1.8.
An ideal detector would transmit no light, assume a
large value of n for coating.
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●
Tested many combinations of the two variables,
graphing for each:
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Asymmetry vs. energy
Position vs. asymmetry
– Where asymmetry is defined as: A E2 E1 E2 E1
Graphs from data:
detector 1
detector 1
90
85
80
70
65
energy
asymmetry
75
60
55
50
45
40
35
30
-20.0
-15.0
-10.0
160
150
140
130
120
110
100
90
80
70
60
50
40
30
20
10
0
-5.0
30
40
50
position
60
70
80
90
asymmetry
Graphs from simulation:
n (window)=2.0 n (coating)=3.0
1.00
1.00
.80
.90
.60
.80
.40
.70
energy
asymmetry
n (window)=2.0 n (coating)=3.0
.20
.00
.60
.50
-.20
.40
-.40
.30
-.60
.20
-.80
.10
-1.00
.00
-200.00
-100.00
position
-1.00 -.75
-.50
-.25
.00
.25
asymmetry
.50
.75
1.00
n (window)=3.0 n (coating)=4.0
1.00
.90
.70
energy
Increasing the values
gives a closer
approximation in the
graph of energy vs.
asymmetry, but
leads to a far worse
approximation in the
graph of position vs.
asymmetry.
.60
.50
.40
.30
.20
.10
.00
-.80
-.60
-.40
-.20
.00
.20
.40
.60
.80
asymmetry
n (window)=3.0 n (coating)=4.0
1.00
0.80
0.60
asymmetry
●
.80
0.40
0.20
0.00
-0.20
-0.40
-0.60
-0.80
-1.00
-200.0
-150.0
-100.0
position
-50.0
What is causing this
discrepancy?
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Possibility that coating on cavity walls is a
type of titanium powder sometimes used in
making detectors.
How would this change a simulation?
–
Absorption and emission of photons.
Work in
progress...
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I have since begun incorporating absorption
into my current program.
I will continue this work into the fall and hope
to find my results improved.
Counting Reflections
d
2 R tan thetai
reflections D x
x
2 R
R tan thetai
positionon window
reflections back
D x d
d
Calculating for intensity loss
reflectance R
R
n i cos thetai
n t cos thetat
transmittance T
T
t
n i cos thetai
I t cos thetat
n t cos thetat
2 ni cos thetai
Ir Ii
I i cos thetai
n i cos thetai
n i cos thetai
n t cos thetat
t
2
n t cos thetat