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:

xn = 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

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

• If we interposed increasing thickness of absorbers between a source of photons and a detector, we would obtain this graph.

Attenuation vs Thickness of Absorber

0

0.2

0.4

0.6

0.8

1

1 2 3 4 5 6 7 8 9 10

Thickness

Atte

nuat

ion

of P

hoto

ns

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

0

0.2

0.4

0.6

0.8

1

1 2 3 4 5 6 7 8 9 10

Thickness

Att

enu

atio

n o

f P

ho

ton

s

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:Decay Equation:

A = A0 x e-lambda x tA = A0 x e-lambda x tWhere, Where, A = Activity remainingA = Activity remainingA0 = Initial ActivityA0 = Initial Activityt = elapsed timet = elapsed timeu = constant that determines the decay of the u = constant that determines the decay of the radioactive sample, and, therefore, the shape of the radioactive sample, and, therefore, the shape of the exponential function.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 1A dose of FDG is assayed as 60mCi/1.3 ml, at 8AMYou 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?

• 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

• 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, RadiationAs 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?

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