Transcript Slide 1

Lecture 1
 Historical Timeline in Nuclear Medicine
Mathematics Review
 Image of the week
Mathematical Review
• Graphs
• Continous vs. Discrete Functions
• Geometry
• Exponential Functions
• Trigonometry
• Scientific Notation
• Significant Figures
Graphical Operations
Number Line
 one dimensional
 infinite in + and - directions
Number Line
 A ruler is a number line
 measuring height of individuals
2-Dimensional Coordinates
Y = ex
9000
8000
7000
6000
5000
4000
3000
2000
1000
0
1
2
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5
6
7
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19
20
Sudden earthquake activity
Expanded Earthquake Graph
Another 2D Example e-x**2
1
0.8
Y axis: Amount
0.6
0.4
0.2
0
-0.2
X axis: time
3Dimensional Coordinates
Three Dimensional Graph
e-(x**2 + y**2)
Three dimensional display
Continuous Function
120
0.0625
0.125
100
0.1875
0.25
80
0.3125
0.375
60
0.4375
0.5
0.5625
40
0.625
0.6875
20
0.75
0.8125
0.875
0
0.06 0.13 0.19 0.25 0.31 0.38 0.44 0.5 0.56 0.63 0.69 0.75 0.81 0.88 0.94
0.9375
Discrete Function
120
0.0625
0.125
100
0.1875
0.25
0.3125
80
0.375
0.4375
60
0.5
0.5625
40
0.625
0.6875
0.75
20
0.8125
0.875
0
0.06 0.13 0.19 0.25 0.31 0.38 0.44 0.5
0.9375
0.56 0.63 0.69 0.75 0.81 0.88 0.94
Discrete function “Approximates”
Continuous Function
120
100
80
60
Series1
40
20
0
1
2
3
4
5
6
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9
10
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Nuclear Medicine Example
DATA
x=time
1
2
y1 = x**2
1
4
y2 = x**3
1
8
3
4
9
16
27
64
5
6
7
8
9
10
11
25
36
49
64
81
100
121
125
216
343
512
729
1000
1331
Y = X2 AND Y = X3
9000
8000
7000
6000
5000
4000
3000
2000
1000
0
1
2
3
4
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6
7
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20
Geometry
Area of rectangle
Volume of box
V=lxwxh
Right triangle
Area = ?
The circle
Area = pi x r2
Trigonometry
The Navigation Problem
Graphing Data Another Way
Sine and Cosine waves
Periodic function
Model of Shape of Electromagnetic
Radiation “Wave Function”
Periodic Wave Function
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0
-0.2
0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 3
-0.4
-0.6
Tim e
Exponential Functions 1
• Functions of the type ex, e-x, e-ax, eiΘ, e-x2, ex2
have many applications in science.
• We will study e-ax , e-x2 and equations
derived from these.
• Some physical processes exhibit
“exponential” behavior.
• Some examples are attenuation of photon
radiation, and radioactive decay.
Exponential Functions
e-x
1
0.8
0.6
0.4
0.2
0
Normal Distribution e-x**2
1
0.8
Y axis: Amount
0.6
0.4
0.2
0
-0.2
X axis: time
Scientific Notation
• Used with constants such as velocity of
light: 3.0 x 1010 cm/sec
• Simplifies writing numbers: 3.0 x 1010
10000000000 = 30000000000.
=
3x
Rule for scientific notation:
n
x =
n zeros
x-n = n – 1 zeros
Proportions
• Direct Proportion: Y = k * X
• If k = 1, X = Y
• Inverse Proportion: Y = k/X
• If k = 1, Y = 1/X
• * means multiplication
Examples
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Attenuation and Dose Calculations
Inverse square law
Effective half life
Discrete image representation
The Attenuation Equation
• Given a beam containing a large flux of
monoenergetic photons, and a uniform
absorber, the removal (attenuation) of
photons from the beam can be described as
an exponential process.
• The equation which describes this process is:
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I = I0 x e-ux
Where,
I = Intensity remaining
I0 = initial photon intensity
x = thickness of absorber
u = constant that determines the attenuation of
the photons, and, therefore, the shape of the
exponential function.
• Experimental data demonstrates that
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μ = 0.693/ HVL,
where HVL stands for Half Value Layer and
represents that thickness of absorber material
which reduces I to one/half its value.
μ is called the linear attenuation coefficient and is
a parameter which is a “constant” of attenuation
for a given HVL
Derivation
• If we interposed
1
Attenuation of Photons
increasing thickness of
absorbers between a
source of photons and
a detector, we would
obtain this graph.
Attenuation vs Thickness of Absorber
0.8
0.6
0.4
0.2
0
1
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5
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Thickness
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10
The line through the data points is a mathematical
determination which best describes the measured
points. The equation describes an exponential
process
Attenuation vs Thickness of Absorber
Attenuation of Photons
1
0.8
0.6
0.4
0.2
0
1
2
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Thickness
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9
10
Variables
• The value of HVL depends on the energy of
the photons, and type of absorber.
• For a given absorber, the higher the photon
energy, the lower the HVL.
• For a given photon energy, the higher the
atomic number of the absorber, the higher
the HVL.
Example 1
• The HVL of lead for 140 KeV photons is:
0.3mm
• What is u?
• 0.693/0.3 = 2.31 cm-1
Example 2
Given the data in Example 1, what % of photons
are detected after a thickness of 0.65 mm are
placed between the source and detector?
Solution: using I = I0 x e-ux, with I0 = 100,
u = 2.31 cm-1, x= 0.65, and solving for I,
I = 22%
Decay Equation:
A = A0 x e-lambda x t
Where,
A = Activity remaining
A0 = Initial Activity
t = elapsed time
u = constant that determines the decay of the
radioactive sample, and, therefore, the shape of the
exponential function.
• Experimental data demonstrates that
lambda = 0.693/ Half Life
where Half Life represents the time it takes
for a sample to decay to 50% of it’s value.
Example 1
A dose of FDG is assayed as 60mCi/1.3 ml, at 8AM
You need to administer a dose of 20mCi at 1PM.
How much volume should you draw into the syringe?
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First, identify the terms:
A
=?
t.
=5
Ao =60
T/12
= 1.8 hrs
We see that A is the unknown.
Then, inserting the values into the equation, we have:
A = 60 x exp(0.693/1.8) x 5)
A = 8.8 mCi
So at 1PM you have 8.8mCi/1.3ml.
You need to draw up 20/8.8 = 2.27 times the volume
needed 5 hours ago.
• This amounts to 2.27 x 1.3 = 3ml
Example 2: A cyclotron operator needs to irradiate enough H2O to be
able to supply the radiochemist with 500mCi/ml F-18 at 3PM. The
operator runs the cyclotron at 8:30AM. How much activity/ml is needed
at that time?
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There are really two ways that we can solve this.
The first, and easier way:
Write: Ao = Ax(exp(λt )
Notice we have a positive exponent.
In other words, instead of using the law of decay, use the law of growth
Once again, identify the terms, and the unknown:
A
= 500
T
= 6.5
Ao = ?
T/12 = 1.8 hrs
Exchange Ao and A
Ao = A x (exp(λt )
A = 500 x (exp(0.693/1.8 x 6.5)
A = 6106 mCi at 8:30AM. = 6.106 Ci
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The second way:
A = Ao x exp(-λt )
= 500 x (exp(-0.693/1.8 x 6.5)
500 = Ao x exp(-(2.5025)
500 = Ao = 6106 mCi = 6.106 Ci
0.08
Effective Half Life
• 1/Te = 1/Tb + 1/Tp
• Where Te = Effective Half Life
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Tb = biological half life
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Tp = physical half life
Inverse Square Law
• The intensity of Radiation from a
point source is inversely
proportional to the square of the
distance
• I1/I2 = D22 / D12
Image of the Week
This digital image is a
2 dimensional graph. Why?
Lecture 2, January 18
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