1. 4/2003 Rev 2 II.2.5 – slide 1 of 44 Session II.2.5 IAEA Post Graduate Educational Course Radiation Protection and Safe Use of Radiation Sources Part IIQuantities and Measurements Module 2Dosimetric Calculations and Measurements Session 5Point, Line, Plane and Volume Sources

2. 4/2003 Rev 2 II.2.5 – slide 2 of 44 We will discuss how radiation is affected by distance from a:  point source  line source  plane (area) source  volume source Overview

3. 4/2003 Rev 2 II.2.5 – slide 3 of 44 4  r 2 Point Source The surface area of a sphere is:

4. 4/2003 Rev 2 II.2.5 – slide 4 of 44 For a 1 MBq photon emitter located at the center of a 1 meter (100 cm) radius sphere 8photons cm 2 - sec - Bq 4  x 10 4 cm 2 10 6 photons sec - Bq = Point Source

5. 4/2003 Rev 2 II.2.5 – slide 5 of 44 1 cm 1 m 8photons sec sec 1 MBq Point Source

6. 4/2003 Rev 2 II.2.5 – slide 6 of 44 Same number of photons strike the surface areas of these two spheres but the “density” or “intensity” is greater for the smaller sphere (more photons per square cm) Point Source

7. 4/2003 Rev 2 II.2.5 – slide 7 of 44 I = N 4 r 2  N = I (4 r 2 ) = 4 (Ir 2 )  = 4 (I o r o 2 )  4 (Ir 2 )  If we increase (or decrease) the radius of the sphere to r o, the intensity will change to I o, but N does not change! N = 4 (I o r o 2 )  = (I o r o 2 ) (Ir 2 ) I = I o ro2ro2ro2ro2 r2r2r2r2 Point Source or

8. 4/2003 Rev 2 II.2.5 – slide 8 of 44 A measurement made at 1 m yields 1 mGy/hr. What would the exposure rate be at 4.5 m? where I is unknown, I o = 1, r o = 1 and r = 4.5 so = (I o r o 2 ) (Ir 2 ) I x 4.5 2 = 1 x 1 2 I = 1 x 1 2 4.5 2 = 1 mGy/hr x 1 m 2 20.25 m 2 = 0.0494 mGy/hr Simple Example

9. 4/2003 Rev 2 II.2.5 – slide 9 of 44 A measurement yields 1 mGy/hr. A second measurement 5 m farther from the source yields 0.44 mGy/hr. What is the distance from the source? We know I and I o. We don’t know “r” nor “r o ”. We do know that r o = r + 5 m. Solve for “r”. (To make the solution general, let “d” represent the distance 5 m) = (I o r o 2 ) (Ir 2 ) More Complex Example

10. 4/2003 Rev 2 II.2.5 – slide 10 of 44 If you have “squares” on both sides of an equation, DON’T EXPAND. Instead, get rid of the “squares”. = (I o r o 2 ) (Ir 2 ) = I o (r+d) 2 I IoIoIoIo r 2 = (r+d) 2 I IoIoIoIo I IoIoIoIo r = (r+d) r - r = d I IoIoIoIo - 1 r = d I IoIoIoIo More Complex Example

11. 4/2003 Rev 2 II.2.5 – slide 11 of 44 Now let’s substitute some real numbers I = 1, I o = 0.44 and d = 5 m Since we know I, I o and d, let’s solve for r r = d - 1 I IoIoIoIo =5 1 0.44 r = 5 m - 1 2.273 = 5 m - 1 1.51 = = 9.8 m 5 m 0.51 More Complex Example

12. 4/2003 Rev 2 II.2.5 – slide 12 of 44 So “r” is 9.8 m and “r o ” = r + 5 m = 14.8 m Remember, the intensity and the distance are inversely related. As one gets bigger, the other gets smaller so that the ratio stays the same. = (I o r o 2 ) (Ir 2 ) 1 x (9.8 m) 2 = 0.44 mGyhrmGyhr x (14.8 m) 2 (the two sides are only approximately equal due to roundoff) More Complex Example According to the equation:

13. 4/2003 Rev 2 II.2.5 – slide 13 of 44 Remember also that there are two point source equations in which the inverse square is used: D rate = A r2r2r2r2 = mGy hr D dose =  A t r2r2r2r2 = mGy Point Source

14. 4/2003 Rev 2 II.2.5 – slide 14 of 44 What is the exposure received by an individual who spends one minute at 3 m from an unshielded 3.7 x 10 12 Bq 192 Ir source? Sample Problem 1

15. 4/2003 Rev 2 II.2.5 – slide 15 of 44 An individual is involved in an exposure incident. She states that she was about 4 m from an unshielded 3.7 x 10 12 Bq 137 Cs source for about 10 min. The dosimeter (TLD) worn by the individual is processed and registers 22.4 mSv. What is your conclusion about the individual's statement? Sample Problem 2

16. 4/2003 Rev 2 II.2.5 – slide 16 of 44 Tech A measures 5 mGy/hr at 6 m from a point source. Tech B measures 45 mGy/hr. How far from the source is Tech B? Sample Problem 3

17. 4/2003 Rev 2 II.2.5 – slide 17 of 44 A radiation source is located in a room. The licensee has previously made a measurement which indicates 25 mGy/hr. You measure 20 mGy/hr. What are some possible explanations for this discrepancy? Sample Problem 4

18. 4/2003 Rev 2 II.2.5 – slide 18 of 44 You are told that the dose rate at 5 m from an unshielded 192 Ir source is 10 mGy/hr. What is your estimate of the activity of the source? Sample Problem 5

19. 4/2003 Rev 2 II.2.5 – slide 19 of 44 L L + L R = L L = length of source h = perpendicular distance from source r = actual distance from any point on the source “L” & “h” are constants “r” is a variable LLLLLLLL LRLRLRLR h r max LLLL RRRR P Line Source

20. 4/2003 Rev 2 II.2.5 – slide 20 of 44 D =  A A A A r2r2r2r2 A “Line” is just a series of points You are here ! r Line Source

21. 4/2003 Rev 2 II.2.5 – slide 21 of 44 where A = C L x L = (  Ci/cm) x cm =  Ci cos  = h/r so that r = (h/cos  ) = h sec  [where 1/(cos  ) = sec  ] also, tan  i = L i /h so that L i = h tan  i and  L = h sec 2   Since tan  = L/h then  = arctan (L/h) = inverse tan (L/h)  D =  CL L CL L CL L CL L r2r2r2r2  C L h sec 2   h 2 sec 2   CL  CL  CL  CL h = LLLLLLLL LRLRLRLR h r max LLLL RRRR P D =  A A A A r2r2r2r2  CL L CL L CL L CL L r2r2r2r2 = Line Source

22. 4/2003 Rev 2 II.2.5 – slide 22 of 44 D rate =  CL CL CL CL h0 LLLL -  R 0 RRRR  L  CL CL CL CL h + D rate =  CL CL CL CL h  0  R  0  CL CL CL CL h  L + -arctan LLLLLLLL h arctan LRLRLRLR h D rate =  CL CL CL CLh [0 - -arctan LLLLLLLL h ] + [arctan LRLRLRLR h - 0] Line Source LLLLLLLL LRLRLRLR h r max LLLL RRRR P

23. 4/2003 Rev 2 II.2.5 – slide 23 of 44 D rate =  CL CL CL CLh arctan LLLLLLLL h + arctan LRLRLRLR h D rate =  CL CL CL CLh RRRR LLLL + ( ) OR If L goes to  (the line is so long it is appears infinite), then arctan (L i /h) goes to 1.57 radians (  =  /2) so that, for a VERY LONG line ! D rate =   CL  CL  CL  CL h Line Source LLLLLLLL LRLRLRLR h r max LLLL RRRR P

24. 4/2003 Rev 2 II.2.5 – slide 24 of 44  = 3.14 360° = 2  radians = 6.28 180° =  radians = 3.14 90° =  /2 radians = 1.57 45° =  /4 radians = 0.79 30° =  /6 radians = 0.52 tan 30° = 0.58 and arctan 0.58 = 0.52 radians = 30° Converting from Degrees to Radians

25. 4/2003 Rev 2 II.2.5 – slide 25 of 44 An individual stands 4 m from a pipe carrying a Co 60 solution. The concentration is 3.7 x 10 8 Bq/cm 3. The pipe penetrates a very thick concrete wall at either end. The walls are 6 m apart. The individual is 2 m from one of the walls. The radius of the pipe is 10 cm. What is the dose rate? Sample Problem 1

26. 4/2003 Rev 2 II.2.5 – slide 26 of 44 What would the dose rate be in Problem 1 if all the activity in the pipe were concentrated at a point 4 m from the individual? Sample Problem 2

27. 4/2003 Rev 2 II.2.5 – slide 27 of 44 What would the dose rate be in Problem 1 if the pipe was very long (“infinite”)? Sample Problem 3

28. 4/2003 Rev 2 II.2.5 – slide 28 of 44 How would the dose rates from the point source in Problem 2 and the line source in Problem 1 compare if the individual was 8 m away? Sample Problem 4

29. 4/2003 Rev 2 II.2.5 – slide 29 of 44 Ratio of distance from line to length of line (h/L) % difference between point & line equations {(Point-Line)/Line}x100 0.10.51.01.52.02.53.03.54.04.55.0264.06027.3247.8413.6002.0501.3190.9190.6770.5190.4100.332 10% error 1% error Using the Point Source Equation for a Line Source

30. 4/2003 Rev 2 II.2.5 – slide 30 of 44 Using the Point Source Equation for a Line Source L 1L 8% 2%0.9% 2L3L Distance from Line Percent Error Introduced Using the Point Source Equation Instead of the Line Source Equations

31. 4/2003 Rev 2 II.2.5 – slide 31 of 44 An area source is similar to a round pizza Each point on the surface of the plane source contributes to the dose via the point source equation r side view Plane Source

32. 4/2003 Rev 2 II.2.5 – slide 32 of 44 R1R1R1R1 Rather than repeat the point source equation thousands of times, we slice the pizza into rings (instead of wedges) and take a small piece (point) on the outermost ring R 1. We integrate (add up) all the doses contributed by each point on that ring. We then move in towards the center to the next ring (R 2 ) and repeat the process. Effectively we are integrating (adding up) the contribution from each ring as the radius varies from R (which is the radius of the whole pizza) to 0. R2R2R2R2 R3R3R3R3 Plane Source

33. 4/2003 Rev 2 II.2.5 – slide 33 of 44 If the point (P) at which we are calculating the dose is centered on the circles, then the dose from each point on a given circle is the same. The dose contribution increases as we approach the center of the circles since the center is closest to the point of interest (“h”). R side view r h Plane Source P

34. 4/2003 Rev 2 II.2.5 – slide 34 of 44 The result of the double integral is: CA =CA =CA =CA =A  R2 R2 R2 R2 D = r2r2r2r2  A A A A Similar to  D rate =  C A ln R 2 + h 2 h2h2h2h2 h2h2h2h2 D rate =  A ln R2R2R2R2 Plane Source where A is the activity of the distributed source, R is the radius of the source and h is the perpendicular distance from the source for a point source

35. 4/2003 Rev 2 II.2.5 – slide 35 of 44 Ratio of the distance from the surface to the diameter of surface (h/2R) % difference between point & plane equations {(Point-Plane)/Plane} x 100 0.10.51.01.52.02.53.03.54.04.55.0667.31944.27012.0365.4583.0931.9871.3831.0170.7790.6160.499 10% error 1% error Using the Point Source Equation for a Plane Source

36. 4/2003 Rev 2 II.2.5 – slide 36 of 44 Using the Point Source Equation for a Plane Source D 1D 12% 3%1.3% 2D3D Distance from Line Percent Error Introduced Using the Point Source Equation Instead of the Line Source Equations

37. 4/2003 Rev 2 II.2.5 – slide 37 of 44 A volume source is just like a drum. Each point on the top surface of the drum is just a plane source and each point on that surface contributes to the dose via the point source equation. We can solve for the total dose from the top of the drum by simply using the plane source equation. Volume Source

38. 4/2003 Rev 2 II.2.5 – slide 38 of 44 Now we can slice the drum into many stacked plane sources (like a stack of pizzas). We can calculate the dose from each pizza using the plane source equation. The only difference is that the photons emitted from all the pizzas except the top one have to penetrate the slices above them. This results in a correction factor that includes the exponential attenuation equation. Volume Source

39. 4/2003 Rev 2 II.2.5 – slide 39 of 44 [ units of T (cm) &  (cm -1 ) cancel ] where T is the height of the drum, h is the distance from the top of the drum to the point of interest & R is the radius of the drum CV =CV =CV =CV =A  R 2 T  D rate =   R 2 + h 2 h2h2h2h2 C V (1 - e - T ) ln  D rate =   R 2 T R 2 + h 2 h2h2h2h2 A (1 - e - T ) ln  Volume Source

40. 4/2003 Rev 2 II.2.5 – slide 40 of 44 If  T is small such as for a very thin drum (essentially a thick pizza) then: (this is just the plane source equation) e -  T  1 -  T D rate =  R 2 T  R 2 T R 2 + h 2 h2h2h2h2 A (1 - [1 - T]) ln A (1 - [1 -  T]) ln D rate =   R 2 T R 2 + h 2 h2h2h2h2 A (  T) ln D rate =  R2R2R2R2 R 2 + h 2 h2h2h2h2 A ln Volume Source

41. 4/2003 Rev 2 II.2.5 – slide 41 of 44   Point Source – True point vs actual source which has some dimension such as a “seed”   Line Source – True line vs an actual source such as a wire, pipe etc Generic Sources   Plane Source – Circular plane vs actual plane sources which may be square or amorphous in shape such as contamination on a surface   Volume Source – Cylinder vs box, sphere or an amorphous container such as a plastic bag

42. 4/2003 Rev 2 II.2.5 – slide 42 of 44  A small source with some physical dimensions such as a seed or cobalt-60 source can be approximated by the point source equation if the distance from the source is large enough so that the radiation appears to be emanating from a point.  A pipe can be approximated by a line:  A square can be approximated by a circle with the same area: L x W =  R 2 R = Generic Sources L x W   L W R

43. 4/2003 Rev 2 II.2.5 – slide 43 of 44   The equation provided for a volume source is for the specific case of a drum containing radioactive material and the dose rate is determined at some distance from one end   A general equation applicable to a point along any surface of the drum is much more complex and a still more general equation for any solid shape such as a rectangular box would be even more complex   Such equations are available in texts but are much beyond the scope of this training Generic Sources

44. 4/2003 Rev 2 II.2.5 – slide 44 of 44 Where to Get More Information  Cember, H., Introduction to Health Physics, 3 rd Edition, McGraw-Hill, New York (2000)  Firestone, R.B., Baglin, C.M., Frank-Chu, S.Y., Eds., Table of Isotopes (8 th Edition, 1999 update), Wiley, New York (1999)  International Atomic Energy Agency, The Safe Use of Radiation Sources, Training Course Series No. 6, IAEA, Vienna (1995)