1. Lecture 1  Historical Timeline in Nuclear Medicine  Mathematics Review  Image of the week

2. Mathematical Review Graphs Graphs Continous vs. Discrete Functions Continous vs. Discrete Functions Geometry Geometry Exponential Functions Exponential Functions Trigonometry Trigonometry Scientific Notation Scientific Notation Significant Figures Significant Figures

3. Graphical Operations Number Line  one dimensional  infinite in + and - directions

4. Number Line  A ruler is a number line  measuring height of individuals

5. 2-Dimensional Coordinates Y = e x

6. Sudden earthquake activity

7. Expanded Earthquake Graph

8. Another 2D Example e -x**2

9. 3Dimensional Coordinates

10. Three Dimensional Graph e -(x**2 + y**2)

11. Three dimensional display

12. Continuous Function

13. Discrete Function

14. Discrete function “Approximates” Continuous Function

15. Nuclear Medicine Example

16. DATA x=time y1 = x** 2 y2 = x** 3 111 248 3927 41664 525125 636216 749343 864512 981729 101001000 111211331

17. Y = X 2 AND Y = X 3

18. Geometry Area of rectangle

19. Volume of box V = l x w x h

20. Right triangle Area = ?

21. The circle Area = pi x r 2

22. Trigonometry The Navigation Problem

23. Graphing Data Another Way Sine and Cosine waves

24. Periodic function

25. Model of Shape of Electromagnetic Radiation “Wave Function”

26. Periodic Wave Function

27. Exponential Functions 1 Functions of the type e x, e -x, e -ax, e iΘ, e -x2, e x2 have many applications in science. Functions of the type e x, e -x, e -ax, e iΘ, e -x2, e x2 have many applications in science. We will study e -ax, e -x2 and equations derived from these. We will study e -ax, e -x2 and equations derived from these. Some physical processes exhibit “exponential” behavior. Some physical processes exhibit “exponential” behavior. Some examples are attenuation of photon radiation, and radioactive decay. Some examples are attenuation of photon radiation, and radioactive decay.

28. Exponential Functions e -x

29. Normal Distribution e -x**2

30. Scientific Notation Used with constants such as velocity of light: 3.0 x 10 10 cm/sec Used with constants such as velocity of light: 3.0 x 10 10 cm/sec Simplifies writing numbers: 3.0 x 10 10 = 3 x 10000000000 = 30000000000. Simplifies writing numbers: 3.0 x 10 10 = 3 x 10000000000 = 30000000000.

31. Rule for scientific notation: x n = n zeros x -n = n – 1 zeros

32. Proportions Direct Proportion: Y = k * X Direct Proportion: Y = k * X If k = 1, X = Y If k = 1, X = Y Inverse Proportion: Y = k/X Inverse Proportion: Y = k/X If k = 1, Y = 1/X If k = 1, Y = 1/X * means multiplication * means multiplication

33. Examples  Attenuation and Dose Calculations  Inverse square law  Effective half life  Discrete image representation

34. 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. 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.

35. The equation which describes this process is: I = I 0 x e -ux The equation which describes this process is: I = I 0 x e -ux Where, Where, I = Intensity remaining I = Intensity remaining I 0 = initial photon intensity I 0 = initial photon intensity x = thickness of absorber x = thickness of absorber u = constant that determines the attenuation of the photons, and, therefore, the shape of the exponential function. u = constant that determines the attenuation of the photons, and, therefore, the shape of the exponential function.

36. 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. 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 μ is called the linear attenuation coefficient and is a parameter which is a “constant” of attenuation for a given HVL

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

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

39. Variables The value of HVL depends on the energy of the photons, and type of absorber. 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 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. For a given photon energy, the higher the atomic number of the absorber, the higher the HVL.

40. Example 1 The HVL of lead for 140 KeV photons is: 0.3mm The HVL of lead for 140 KeV photons is: 0.3mm What is u? What is u? 0.693/0.3 = 2.31 cm -1 0.693/0.3 = 2.31 cm -1

41. 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? 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 = I 0 x e -ux, with I 0 = 100, u = 2.31 cm -1, x= 0.65, and solving for I, 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% I = 22%

42. 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.

43. 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. 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.

44. 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: First, identify the terms: A = ? A = ? t.= 5 t.= 5 Ao=60 Ao=60 T/12 = 1.8 hrs T/12 = 1.8 hrs We see that A is the unknown. We see that A is the unknown. Then, inserting the values into the equation, we have: Then, inserting the values into the equation, we have: A = 60 x exp(0.693/1.8) x 5) A = 60 x exp(0.693/1.8) x 5) A = 8.8 mCi A = 8.8 mCi So at 1PM you have 8.8mCi/1.3ml. 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. 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 This amounts to 2.27 x 1.3 = 3ml

45. 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. There are really two ways that we can solve this. The first, and easier way: The first, and easier way: Write: Ao = Ax(exp(λt ) Write: Ao = Ax(exp(λt ) Notice we have a positive exponent. Notice we have a positive exponent. In other words, instead of using the law of decay, use the law of growth In other words, instead of using the law of decay, use the law of growth Once again, identify the terms, and the unknown: Once again, identify the terms, and the unknown: A = 500 A = 500 T= 6.5 T= 6.5 Ao= ? Ao= ? T/12 = 1.8 hrs T/12 = 1.8 hrs Exchange Ao and A Exchange Ao and A Ao = A x (exp(λt ) Ao = A x (exp(λt ) A = 500 x (exp(0.693/1.8 x 6.5) A = 500 x (exp(0.693/1.8 x 6.5) A = 6106 mCi at 8:30AM. = 6.106 Ci A = 6106 mCi at 8:30AM. = 6.106 Ci

46. The second way: The second way: A = Ao x exp(-λt ) A = Ao x exp(-λt ) = 500 x (exp(-0.693/1.8 x 6.5) = 500 x (exp(-0.693/1.8 x 6.5) 500 = Ao x exp(-(2.5025) 500 = Ao x exp(-(2.5025) 500 = Ao = 6106 mCi = 6.106 Ci 0.08

47. Effective Half Life 1/Te = 1/Tb + 1/Tp 1/Te = 1/Tb + 1/Tp Where Te = Effective Half Life Where Te = Effective Half Life Tb = biological half life Tb = biological half life Tp = physical half life Tp = physical half life

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

49. Image of the Week

50. This digital image is a 2 dimensional graph. Why?

52. Lecture 2, January 18 Licenses and Regulatory Authorities Licenses and Regulatory Authorities Authorized Users Authorized Users Radiation Safety Officer Radiation Safety Officer Emergency Contacts Emergency Contacts