Transcript Mathematics Review 2

Lecture 2, January 19
• Conclusion of Mathematics Review with
Examples
• Historical Timeline in Nuclear Medicine
• Radiation Safety Introduction
• Image of the Week
Scientific Notation
• Used with constants such as velocity of
light: 3.0 x 1010 cm/sec
• Simplifies writing numbers: 3.0 x 1010 = 3 x
10000000000 = 30000000000.
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:
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
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
Attenuation vs Thickness of Absorber
1
Attenuation of Photons
• If we interposed
increasing thickness of
absorbers between a
source of photons and
a detector, we would
obtain this graph.
0.8
0.6
0.4
0.2
0
1
2
3
4
5
6
Thickness
7
8
9
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
3
4
5
6
Thickness
7
8
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?
Example 2
Given the data in Example 1, what % of photons
are detected after a thickness of 0.065 cm
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.
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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?
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 = (60/1.3 = 46.2) x exp(0.693/1.8) x 5)
A = 6.7 mCi
So at 1PM you have 6.7mCi/ml.
You need to draw up 20/6.7 = 2.98 ml.
Draw up 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:
Write: Ao = A x (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
• The second way:
• 500 = Ao x exp(-λt )
• = 500 x (exp(-0.693/1.8 x 6.5)
• 500 = Ao x exp(-(2.5025) = Ao x 0.082
500 = Ao = 6106 mCi = 6.106 Ci
0.08
Effective Half Life
• 1/Te = 1/Tb + 1/Tp
• Where Te = Effective Half Life
•
Tb = biological half life
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Tp = physical half life
Effective T1/2 Example
• A compound of Nitrogen-13, a positron
emitter used in some research applications
involving protein metabolism, has a 10
minute biological half life.
• What is the effective half life?
• Info: physical T1/2 = 9.97 minutes.
Inverse Square Law, Radiation
As one of the fields which obey the general inverse square law, a point radiation source can
be characterized by the relationship below whether you are talking about Roentgens , rads,
or rems . All measures of exposure will drop off by inverse square law.
Inverse Square Law
• The intensity of Radiation from a
point source is inversely
proportional to the square of the
distance
• I1/I2 = D22 / D12
Inverse Square Law Example 1
• The exposure rate at one meter from a
source of O-15 is 36R/min. What is the
exposure rate at 5 meters from the
source?
Inverse Square Law Example 2
• The exposure rate at 20 cm to a body
from a source of C-11 is 22R/hr. By
what factor would you increase the
distance to decrease the exposure to
8R?
Radiation Safety 1
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