4/2003 rev 2 ii.2.5 – slide 1 of 44 session ii.2.5 iaea post graduate educational course radiation...

4/2003 Rev 24/2003 Rev 2 II.2.5 – slide II.2.5 – slide 11 of 44 of 44

Session II.2.5Session II.2.5

IAEA Post Graduate Educational CourseIAEA Post Graduate Educational CourseRadiation Protection and Safe Use of Radiation SourcesRadiation Protection and Safe Use of Radiation Sources

Part IIPart II Quantities and MeasurementsQuantities and Measurements

Module 2Module 2 Dosimetric Calculations and Dosimetric Calculations and MeasurementsMeasurements

Session 5Session 5 Point, Line, Plane and Volume Point, Line, Plane and Volume SourcesSources


We will discuss how radiation is affected by We will discuss how radiation is affected by distance from a:distance from a:

point sourcepoint source line sourceline source plane (area) sourceplane (area) source volume sourcevolume source

OverviewOverview


4 4 r r22

Point SourcePoint Source

The surface area The surface area of a sphere is:of a sphere is:


For a 1 MBq photon emitter located at the For a 1 MBq photon emitter located at the center of a 1 meter (100 cm) radius sphere center of a 1 meter (100 cm) radius sphere

88photonsphotons

cmcm22 - sec - Bq - sec - Bq44 x 10x 1044 cm cm22

101066 photonsphotons

sec - Bqsec - Bq==



1 cm1 cm

1 cm1 cm

1 m1 m

88photonsphotons

sec sec

1 MBq1 MBq



Same number Same number of photons of photons strike the strike the surface areas surface areas of these two of these two spheres but spheres but the “density” the “density” or “intensity” or “intensity” is greater for is greater for the smaller the smaller sphere (more sphere (more photons per photons per square cm)square cm)



I =I =NN

4 r4 r22 N = I (4 rN = I (4 r22) = 4 (Ir) = 4 (Ir22))

= 4 (I= 4 (Ioorroo22)) 4 (Ir4 (Ir22))

If we increase (or decrease) the radius of the sphere to rIf we increase (or decrease) the radius of the sphere to roo,,

the intensity will change to Ithe intensity will change to Ioo, but N does not change!, but N does not change!

N = 4 (IN = 4 (Ioorroo22))

= (I= (Ioorroo22))(Ir(Ir22))

I = II = Ioo

rroo22

rr22


oror


A measurement made at 1 m yields 1 mGy/hr. What A measurement made at 1 m yields 1 mGy/hr. What would the exposure rate be at 4.5 m?would the exposure rate be at 4.5 m?

where I is unknown, Iwhere I is unknown, Ioo = 1, r = 1, roo = 1 and r = 4.5 so = 1 and r = 4.5 so


I x 4.5I x 4.522 = 1 x 1 = 1 x 122

I =I =1 x 11 x 122

4.54.522= = 1 mGy/hr x 1 m1 mGy/hr x 1 m22

20.25 m20.25 m22= 0.0494 mGy/hr= 0.0494 mGy/hr

Simple ExampleSimple Example


A measurement yields 1 mGy/hr. A second A measurement yields 1 mGy/hr. A second measurement 5 m farther from the source yieldsmeasurement 5 m farther from the source yields0.44 mGy/hr. What is the distance from the source?0.44 mGy/hr. What is the distance from the source?

We know I and IWe know I and Ioo. We don’t know “r” nor “r. We don’t know “r” nor “roo”.”.

We do know that rWe do know that roo = r + 5 m. Solve for “r”. = r + 5 m. Solve for “r”.

(To make the solution general, let “d” represent the (To make the solution general, let “d” represent the distance 5 m)distance 5 m)


More Complex ExampleMore Complex Example


If you have “squares” on If you have “squares” on both sides of an equation, both sides of an equation, DON’T EXPAND. Instead, DON’T EXPAND. Instead, get rid of the “squares”.get rid of the “squares”.

= (I= (Ioorroo22))(Ir(Ir22)) = I= Io o (r+d)(r+d)22

II

IIoo

rr22 = (r+d) = (r+d)22

II

IIoo

rr22 = (r+d) = (r+d)22

II

IIoo

r = (r+d)r = (r+d)

r - r = dr - r = dII

IIoo

- 1 r = d- 1 r = dII

IIoo



Now let’s substitute some real numbersNow let’s substitute some real numbersI = 1, II = 1, Ioo = 0.44 and d = 5 m = 0.44 and d = 5 m

Since we know I, ISince we know I, Ioo and d, let’s solve for r and d, let’s solve for r

r = r = dd

- 1 - 1 II

IIoo

= = 55

- 1 - 1 11

0.440.44

r = r = 5 m5 m

- 1 - 1 2.2732.273= =

5 m5 m

- 1 - 1 1.511.51== = 9.8 m= 9.8 m

5 m5 m

0.510.51



So “r” is 9.8 m and “rSo “r” is 9.8 m and “roo” = r + 5 m = 14.8 m” = r + 5 m = 14.8 m

Remember, the intensity and the distance are Remember, the intensity and the distance are inversely relatedinversely related. As one gets bigger, the other . As one gets bigger, the other gets smaller so that the ratio stays the same.gets smaller so that the ratio stays the same.


11 x (9.8 m)x (9.8 m)22 = 0.44 = 0.44mGymGy

hrhrmGymGy

hrhrx (14.8 m)x (14.8 m)22

(the two sides are only approximately equal due to roundoff)(the two sides are only approximately equal due to roundoff)


According to the equation:According to the equation:


Remember also that there are two point source Remember also that there are two point source equations in which the inverse square is used:equations in which the inverse square is used:

DDraterate = = AA

rr22==

mGymGy

hrhr

DDdosedose = = A tA t

rr22= mGy= mGy



What is the exposure received by an individual What is the exposure received by an individual who spends one minute at 3 m from an who spends one minute at 3 m from an unshielded 3.7 x 10unshielded 3.7 x 101212 Bq Bq 192192Ir source?Ir source?

Sample Problem 1Sample Problem 1


An individual is involved in an exposure An individual is involved in an exposure incident. She states that she was about 4 m incident. She states that she was about 4 m from an unshielded 3.7 x 10from an unshielded 3.7 x 101212 Bq Bq 137137Cs source Cs source for about 10 min. The dosimeter (TLD) worn for about 10 min. The dosimeter (TLD) worn by the individual is processed and registers by the individual is processed and registers 22.4 mSv. What is your conclusion about the 22.4 mSv. What is your conclusion about the individual's statement?individual's statement?



Tech A measures 5 mGy/hr at 6 m from a Tech A measures 5 mGy/hr at 6 m from a point source. Tech B measures 45 mGy/hr. point source. Tech B measures 45 mGy/hr. How far from the source is Tech B?How far from the source is Tech B?



A radiation source is located in a room. The A radiation source is located in a room. The licensee has previously made a measurement licensee has previously made a measurement which indicates 25 mGy/hr. You measurewhich indicates 25 mGy/hr. You measure20 mGy/hr. What are some possible 20 mGy/hr. What are some possible explanations for this discrepancy?explanations for this discrepancy?



You are told that the dose rate at 5 m from an You are told that the dose rate at 5 m from an unshielded unshielded 192192Ir source is 10 mGy/hr. What is Ir source is 10 mGy/hr. What is your estimate of the activity of the source?your estimate of the activity of the source?



LLLL + L + LRR = L = L

L = length of sourceL = length of source

h = perpendicular h = perpendicular distance from sourcedistance from source

r = actual distance from r = actual distance from any point on the sourceany point on the source

““L” & “h” are constantsL” & “h” are constants

““r” is a variabler” is a variable

LLLL LLRR

hh

rrmaxmax

rrmaxmax

LLRR

PP

Line SourceLine Source


D = D = AA

rr22

A “Line” is just a series of pointsA “Line” is just a series of points

You are here !

rr



where A = Cwhere A = CLL x L = ( x L = (Ci/cm) x cm = Ci/cm) x cm = CiCi

cos cos = h/r so that r = (h/cos = h/r so that r = (h/cos ) = h sec ) = h sec [where 1/(cos [where 1/(cos ) = sec ) = sec ]]

also, tan also, tan ii = L = Lii/h so that L/h so that Lii = h tan = h tan ii

and and L = h secL = h sec22

Since tan Since tan = L/h then = L/h then = arctan (L/h) = inverse tan (L/h) = arctan (L/h) = inverse tan (L/h)

D = D = CCLL LL

rr22

D = D = CCLL h sec h sec22

hh2 2 secsec2 2 CCLL

h h ==

LLLL LLRR

hh

rrmaxmax

rrmaxmax

LL

RR

PP

D = D = AArr22

CCLL L Lrr22==



DDraterate = = CCLL

hh 00

LL--RR

00

RR

LL

CCLL

hh++

DDraterate = = CCLL

hh 00

RR00

CCLL

hhLL ++

-arctan-arctanLLLL

hh

arctanarctanLLRR

hh

DDraterate = = CCLL

hh[0 - -arctan[0 - -arctan

LLLL

hh]] + [arctan+ [arctan

LLRR

hh- 0]- 0]


LLLL LLRR

hh

rrmaxmax

rrmaxmax

LLRR

PP


DDraterate = = CCLL

hharctanarctan

LLLL

hh+ arctan+ arctan

LLRR

hh

DDraterate = = CCLL

hhRRLL ++( )( )OROR

If L goes to If L goes to (the line is so long it is appears infinite), (the line is so long it is appears infinite), then arctan (Lthen arctan (Lii/h) goes to 1.57 radians (/h) goes to 1.57 radians ( = = /2) /2)

so that, for a so that, for a VERY LONGVERY LONG line ! line !DDraterate = = CCLL

hh


LLLL LLRR

hh

rrmaxmax

rrmaxmax

LLRR

PP


= 3.14= 3.14

360° 360° = 2= 2 radians radians = 6.28= 6.28180° 180° = = radians radians = 3.14= 3.1490° 90° = = /2 radians /2 radians = 1.57= 1.5745° 45° = = /4 radians /4 radians = 0.79= 0.7930° 30° = = /6 radians /6 radians = 0.52= 0.52

tan 30°tan 30° = 0.58 and arctan 0.58 = 0.52 radians = 30°= 0.58 and arctan 0.58 = 0.52 radians = 30°

Converting fromConverting fromDegrees to RadiansDegrees to Radians


An individual stands 4 m from a pipe carrying An individual stands 4 m from a pipe carrying a Coa Co6060 solution. The concentration is solution. The concentration is3.7 x 103.7 x 1088 Bq/cm Bq/cm33. The pipe penetrates a very . The pipe penetrates a very thick concrete wall at either end. The walls thick concrete wall at either end. The walls are 6 m apart. The individual is 2 m from one are 6 m apart. The individual is 2 m from one of the walls. The radius of the pipe is 10 cm. of the walls. The radius of the pipe is 10 cm. What is the dose rate?What is the dose rate?



What would the dose rate be in Problem 1 if What would the dose rate be in Problem 1 if all the activity in the pipe were concentrated all the activity in the pipe were concentrated at a point 4 m from the individual?at a point 4 m from the individual?



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



How would the dose rates from the point How would the dose rates from the point source in Problem 2 and the line source in source in Problem 2 and the line source in Problem 1 compare if the individual was 8 m Problem 1 compare if the individual was 8 m away?away?



Ratio of distance from Ratio of distance from line to length of line (h/L)line to length of line (h/L)

% difference between point % difference between point & line equations& line equations

{(Point-Line)/Line}x100{(Point-Line)/Line}x100

0.10.10.50.51.01.01.51.52.02.02.52.53.03.03.53.54.04.04.54.55.05.0

264.060264.06027.32427.324

7.8417.8413.6003.6002.0502.0501.3191.3190.9190.9190.6770.6770.5190.5190.4100.4100.3320.332

10% error10% error

1% error1% error

Using the Point Source Using the Point Source Equation for a Line SourceEquation for a Line Source


Using the Point Source Using the Point Source Equation for a Line SourceEquation for a Line Source

LL1L1L

8%8%2%2%0.9%0.9%

2L2L3L3L

Distance from LineDistance from Line

Percent Error Introduced Using the Percent Error Introduced Using the Point Source Equation Instead of Point Source Equation Instead of

the Line Source Equationsthe Line Source Equations


An area source isAn area source issimilar to asimilar to around pizzaround pizza

Each point on Each point on the surface of the surface of

the plane the plane source source

contributes to contributes to the dose via the the dose via the

point source point source equationequation

rr

side viewside view

Plane SourcePlane Source


RR11

Rather than repeat the point Rather than repeat the point source equation thousands of source equation thousands of times, we slice the pizza into times, we slice the pizza into rings (instead of wedges) and rings (instead of wedges) and take a small piece (point) on the take a small piece (point) on the outermost ring Routermost ring R11. We integrate . We integrate

(add up) all the doses contributed (add up) all the doses contributed by each point on that ring.by each point on that ring.

We then move in towards the center to the next ring (RWe then move in towards the center to the next ring (R22) )

and repeat the process. Effectively we are integrating and repeat the process. Effectively we are integrating (adding up) the contribution from each ring as the radius (adding up) the contribution from each ring as the radius varies from R (which is the radius of the whole pizza) to 0.varies from R (which is the radius of the whole pizza) to 0.

RR22

RR33



If the point (P) at which we are If the point (P) at which we are calculating the dose is centered calculating the dose is centered on the circles, then the dose from on the circles, then the dose from each point on a given circle is the each point on a given circle is the same.same.

The dose contribution increases The dose contribution increases as we approach the center of the as we approach the center of the circles since the center is closest circles since the center is closest to the point of interest (“h”).to the point of interest (“h”).

RR

side viewside view

rr

hh


PP


The result of the double integral is:The result of the double integral is:

CCAA = = AA RR22

D = D = rr22

AASimilar toSimilar to

DDraterate = = CCAA ln ln RR22 + h + h22

hh22

RR22 + h + h22

hh22DDraterate = = A lnA ln

RR22


where A is the activity of where A is the activity of the distributed source,the distributed source,R is the radius of the R is the radius of the source and h is the source and h is the perpendicular distance perpendicular distance from the sourcefrom the source

for a point sourcefor a point source


Ratio of the distance from Ratio of the distance from the surface to the the surface to the

diameter of surface (h/2R)diameter of surface (h/2R)

% difference between point % difference between point & plane equations& plane equations

{(Point-Plane)/Plane} x 100{(Point-Plane)/Plane} x 100

0.10.10.50.51.01.01.51.52.02.02.52.53.03.03.53.54.04.04.54.55.05.0

667.319667.31944.27044.27012.03612.036

5.4585.4583.0933.0931.9871.9871.3831.3831.0171.0170.7790.7790.6160.6160.4990.499

10% error10% error

1% error1% error

Using the Point Source Using the Point Source Equation for a Plane SourceEquation for a Plane Source


Using the Point Source Using the Point Source Equation for a Plane SourceEquation for a Plane Source

DD1D1D

12%12%3%3%1.3%1.3%

2D2D3D3D

Distance from LineDistance from Line

Percent Error Introduced Using the Percent Error Introduced Using the Point Source Equation Instead of Point Source Equation Instead of

the Line Source Equationsthe Line Source Equations


A volume source is just like a drum.A volume source is just like a drum.

Each point on the top surface of the Each point on the top surface of the drum is just a plane source and each drum is just a plane source and each point on that surface contributes to the point on that surface contributes to the dose via the point source equation. dose via the point source equation. We can solve for the total dose from We can solve for the total dose from the top of the drum by simply using the top of the drum by simply using the plane source equation.the plane source equation.

Volume SourceVolume Source


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



[ units of T (cm) & (cm-1) cancel ]

where T is the height of the drum, h where T is the height of the drum, h is the distance from the top of the is the distance from the top of the drum to the point of interest & R is drum to the point of interest & R is the radius of the drumthe radius of the drum

CCVV = = AA

RR2 2 TT

DDraterate = =

RR22 + h + h22

hh22CCVV (1 - e (1 - e- T- T) ln) ln

DDraterate = = RR2 2 TT

RR22 + h + h22

hh22A (1 - eA (1 - e- T- T) ln) ln



If T is small such as for a very thin drum (essentially a thick pizza) then:

(this is just the plane source equation)(this is just the plane source equation)

ee- - TT 1 - 1 - T T


RR22 + h + h22

hh22A (1 - [1 - A (1 - [1 - T]) lnT]) ln


RR22 + h + h22

hh22A (A (TT) ln) ln

DDraterate = = RR22

RR22 + h + h22

hh22A lnA ln



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 SourcesGeneric Sources

Plane Source – Circular plane vs actual plane sources which may be square or amorphousin shape such as contamination on a surface

Volume Source – Cylinder vs box, sphere oran amorphous container such as a plastic bag


A small source with some physical dimensions such A small source with some physical dimensions such as a seed or cobalt-60 source can be approximated as a seed or cobalt-60 source can be approximated by the point source equation if the distance from the by the point source equation if the distance from the source is large enough so that the radiationsource is large enough so that the radiationappears to be emanating from a point.appears to be emanating from a point.

A pipe can be approximatedA pipe can be approximatedby a line: by a line:

A square can be approximated by a circle with the A square can be approximated by a circle with the same area:same area:

L x W = L x W = RR22 R =R =


L x WL x W

LL

WW

R


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



Where to Get More InformationWhere to Get More Information

Cember, H., Introduction to Health Physics, 3Cember, H., Introduction to Health Physics, 3rdrd Edition, McGraw-Hill, New York (2000)Edition, McGraw-Hill, New York (2000)

Firestone, R.B., Baglin, C.M., Frank-Chu, S.Y., Eds., Firestone, R.B., Baglin, C.M., Frank-Chu, S.Y., Eds., Table of Isotopes (8Table of Isotopes (8thth Edition, 1999 update), Wiley, Edition, 1999 update), Wiley, New York (1999)New York (1999)

International Atomic Energy Agency, The Safe Use International Atomic Energy Agency, The Safe Use of Radiation Sources, Training Course Series No. 6, of Radiation Sources, Training Course Series No. 6, IAEA, Vienna (1995)IAEA, Vienna (1995)

4/2003 rev 2 ii.2.5 – slide 1 of 44 session ii.2.5 iaea post graduate educational course radiation...

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