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 cm/sec Simplifies writing numbers: 3.0 x = 3 x =

Rule for scientific notation: x n = 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 = I 0 x e -ux Where, I = Intensity remaining I 0 = 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.

The line through the data points is a mathematical determination which best describes the measured points. The equation describes an exponential process

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 cm are placed between the source and detector? Solution: using I = I 0 x e -ux, with I 0 = 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? 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. = 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 = Ao = 6106 mCi = 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, 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 lawRoentgens radsrems

Inverse Square Law The intensity of Radiation from a point source is inversely proportional to the square of the distance I1/I2 = D2 2 / D1 2

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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