IAEA Post Graduate Educational Course Radiation Protection and

IAEA Post Graduate Educational Course Radiation Protection and Safe Use of Radiation Sources Part IIQuantities and Measurements Module 2 Radiometric and dosimetric Calculations and Measurements Session 3 Point, Line, Plane and Volume Sources IAEA International Atomic Energy Agency

Overview We will discuss how radiation is affected by distance from a: • • IAEA point source line source plane (area) source volume source 2

Point Source The surface area of a sphere is: 4 IAEA 2 r 3

Point Source For a 1 MBq photon emitter located at the center of a 1 meter (100 cm) radius sphere 106 photons sec - Bq 4 x 104 cm 2 IAEA = 8 photons cm 2 - sec - Bq 4

Point Source 1 MBq 1 m 1 cm IAEA 8 photons sec 5

Point Source 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) IAEA 6

Point Source I= N N = I (4 r 2) = 4 (Ir 2) 4 r 2 If we increase (or decrease) the radius of the sphere to ro, the intensity will change to Io, but N does not change! N = 4 (Ioro 2) 4 (Ir 2) = 4 (Ioro 2) (Ir 2) = (Ioro 2) IAEA or I = Io ro 2 r 2 7

Simple Example A measurement made at 1 m yields 1 m. Gy/hr. What would the exposure rate be at 4. 5 m? (Ir 2) = (Ioro 2) where I is unknown, Io = 1, ro = 1 and r = 4. 5 so I x 4. 52 = 1 x 12 I = IAEA 1 x 12 4. 52 2 1 m. Gy/hr x 1 m = 20. 25 m 2 = 0. 0494 m. Gy/hr 8

More Complex Example A measurement yields 1 m. Gy/hr. A second measurement 5 m farther from the source yields 0. 44 m. Gy/hr. What is the distance from the source? (Ir 2) = (Ioro 2) We know I and Io. We don’t know “r” nor “ro”. We do know that ro = r + 5 m. Solve for “r”. (To make the solution general, let “d” represent the distance 5 m) IAEA 9

More Complex Example (Ir 2) = (Ioro 2) = Io I 2 r = (r+d)2 Io (r+d)2 If you have “squares” on both sides of an equation, DON’T EXPAND. Instead, get rid of the “squares”. I r = (r+d) Io I r - r = d Io I - 1 r = d Io IAEA 10

More Complex Example Since we know I, Io and d, let’s solve for r d r= 5 = I - 1 Io 1 - 1 0. 44 Now let’s substitute some real numbers I = 1, Io = 0. 44 and d = 5 m r= 5 m 2. 273 IAEA = - 1 5 m 1. 51 - 1 5 m = = 9. 8 m 0. 51 11

More Complex Example So “r” is 9. 8 m and “ro” = r + 5 m = 14. 8 m According to the equation: m. Gy 1 x (9. 8 m)2 = 0. 44 hr (Ir 2) = (Ioro 2) m. Gy x (14. 8 m)2 hr (the two sides are only approximately equal due to roundoff) Remember, the intensity and the distance are inversely related. As one gets bigger, the other gets smaller so that the ratio stays the same. IAEA 12

Point Source Remember also that there are two point source equations in which the inverse square is used: IAEA Drate = A r 2 m. Gy = hr Ddose = A t r 2 = m. Gy 13

Sample Problem 1 What is the exposure received by an individual who spends one minute at 3 m from an unshielded 3. 7 x 1012 Bq 192 Ir source? IAEA 14

Sample Problem 2 An individual is involved in an exposure incident. She states that she was about 4 m from an unshielded 3. 7 x 1012 Bq 137 Cs source for about 10 min. The dosimeter (TLD) worn by the individual is processed and registers 22. 4 m. Sv. What is your conclusion about the individual's statement? IAEA 15

Sample Problem 3 Tech A measures 5 m. Gy/hr at 6 m from a point source. Tech B measures 45 m. Gy/hr. How far from the source is Tech B? IAEA 16

Sample Problem 4 A radiation source is located in a room. The licensee has previously made a measurement which indicates 25 m. Gy/hr. You measure 20 m. Gy/hr. What are some possible explanations for this discrepancy? IAEA 17

Sample Problem 5 You are told that the dose rate at 5 m from an unshielded 192 Ir source is 10 m. Gy/hr. What is your estimate of the activity of the source? IAEA 18

Line Source LL LR LL + L R = L L = length of source h = perpendicular distance from source h rmax L R P IAEA r = actual distance from any point on the source “L” & “h” are constants “r” is a variable 19

Line Source A “Line” is just a series of points r A D= 2 r You are here ! IAEA 20

Line Source LL LR D= h rmax R L P D = = CL L r 2 D = C L L r 2 where A = CL x L = ( Ci/cm) x cm = Ci rmax cos = h/r so that r = (h/cos ) = h sec [where 1/(cos ) = sec ] also, tan i = Li/h so that Li = h tan i and L = h sec 2 CL h sec 2 h 2 sec 2 Since tan = L/h then IAEA A r 2 C L = h = arctan (L/h) = inverse tan (L/h) 21

Line Source LL Drate = IAEA CL h 0 - L CL h L + 0 -arctan LL h [0 - -arctan L rmax R R 0 h rmax L R P CL + h LL ] + [arctan h LR arctan 0 LR h R LR - 0] h 22

Line Source CL h Drate = OR arctan CL h LL + arctan h LR h LL rmax ( L + R) LR h rmax L R P If L goes to (the line is so long it is appears infinite), then arctan (Li/h) goes to 1. 57 radians ( = /2) so that, IAEA CL Drate = for a VERY LONG line ! h 23

Converting from Degrees to Radians = 3. 14 360° 180° 90° 45° 30° = 2 radians = /2 radians = /4 radians = /6 radians = 6. 28 = 3. 14 = 1. 57 = 0. 79 = 0. 52 tan 30° = 0. 58 and arctan 0. 58 = 0. 52 radians = 30° IAEA 24

Sample Problem 1 An individual stands 4 m from a pipe carrying a Co 60 solution. The concentration is 3. 7 x 108 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? IAEA 25

Sample Problem 2 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? IAEA 26

Sample Problem 3 What would the dose rate be in Problem 1 if the pipe was very long (“infinite”)? IAEA 27

Sample Problem 4 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? IAEA 28

Using the Point Source Equation for a Line Source Ratio of distance from line to length of line (h/L) 10% error 1% error IAEA 0. 1 0. 5 1. 0 1. 5 2. 0 2. 5 3. 0 3. 5 4. 0 4. 5 5. 0 % difference between point & line equations {(Point-Line)/Line}x 100 264. 060 27. 324 7. 841 3. 600 2. 050 1. 319 0. 919 0. 677 0. 519 0. 410 0. 332 29

Using the Point Source Equation for a Line Source Percent Error Introduced Using the Point Source Equation Instead of the Line Source Equations 0. 9% 2% 8% 3 L 2 L 1 L L Distance from Line IAEA 30

Plane Source r An area source is similar to a round pizza IAEA side view Each point on the surface of the plane source contributes to the dose via the point source equation 31

Plane Source R 1 R 2 R 3 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. IAEA 32

Plane Source 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. R r h P The dose contribution increases as we approach the center of the circles since the center is closest to the point of interest (“h”). side view IAEA 33

Plane Source The result of the double integral is: Drate = CA ln Drate = A ln R 2 Similar to D= for a point source IAEA R 2 + h 2 h 2 A r 2 CA = A R 2 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 34

Using the Point Source Equation for a Plane Source Ratio of the distance from the surface to the diameter of surface (h/2 R) 10% error 1% error IAEA 0. 1 0. 5 1. 0 1. 5 2. 0 2. 5 3. 0 3. 5 4. 0 4. 5 5. 0 % difference between point & plane equations {(Point-Plane)/Plane} x 100 667. 319 44. 270 12. 036 5. 458 3. 093 1. 987 1. 383 1. 017 0. 779 0. 616 0. 499 35

Using the Point Source Equation for a Plane Source Percent Error Introduced Using the Point Source Equation Instead of the Line Source Equations 1. 3% 3% 12% 3 D 2 D 1 D D Distance from Line IAEA 36

Volume Source 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. IAEA 37

Volume Source 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. IAEA 38

Volume Source Drate = CV (1 CV = A R 2 T Drate = e- T) R 2 + h 2 ln 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 (1 A R 2 T e- T ) ln R 2 + h 2 [ units of T (cm) & (cm-1) cancel ] IAEA 39

Volume Source If T is small such as for a very thin drum (essentially a thick pizza) then: e - T 1 - T Drate = (1 - [1 - T]) ln A R 2 T ( T) ln A R 2 T A ln R 2 + h 2 h 2 R 2 + h 2 (this is just the plane source equation) IAEA 40

Generic Sources Ø Point Source – True point vs actual source which has some dimension such as a “seed” Ø Line Source – True line vs an actual source such as a wire, pipe etc Ø Plane Source – Circular plane vs actual plane sources which may be square or amorphous in shape such as contamination on a surface Ø Volume Source – Cylinder vs box, sphere or an amorphous container such as a plastic bag IAEA 41

Generic Sources Ø 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. Ø A pipe can be approximated by a line: Ø A square can be approximated by a circle with the same area: L x W = R 2 IAEA R= L x W L R W 42

Generic Sources Ø 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 Ø 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 Ø Such equations are available in texts but are much beyond the scope of this training IAEA 43