3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 11 of 42 of 42IAEA Post Graduate Educational CourseIAEA Post Graduate Educational Course

Radiation 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 8Session 8 Doses from Neutron SourcesDoses from Neutron Sources

Session II.2.8Session II.2.8

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 22 of 42 of 42

OverviewOverview

Principles of neutron dose calculation will be Principles of neutron dose calculation will be discusseddiscussed

Students will learn about fast and thermal Students will learn about fast and thermal neutron interactions with tissue; dose neutron interactions with tissue; dose calculation from neutron beams, neutron calculation from neutron beams, neutron point sources, fast neutrons, thermal point sources, fast neutrons, thermal neutronsneutrons

Example problems illustrating neutron dose Example problems illustrating neutron dose calculation will be workedcalculation will be worked

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 33 of 42 of 42

ContentContent

Fast and thermal neutron interactions with tissueFast and thermal neutron interactions with tissue Dose from fast neutronsDose from fast neutrons Dose from thermal neutronsDose from thermal neutrons Dose from neutron charged particle and gamma Dose from neutron charged particle and gamma

ray emissionsray emissions Example problems calculating dose from Example problems calculating dose from

neutron beamsneutron beams Example problems calculating dose from Example problems calculating dose from

neutron point sources, including calibration neutron point sources, including calibration sources, criticalities, and acceleratorssources, criticalities, and accelerators

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 44 of 42 of 42

Tissue Dose From NeutronsTissue Dose From Neutrons

Dose from fast neutrons is primarily due to Dose from fast neutrons is primarily due to elastic scattering with hydrogenelastic scattering with hydrogen

Dose from thermal neutrons is due to two Dose from thermal neutrons is due to two reactions, primarily:reactions, primarily:

1414N(n,p)N(n,p)1414C C

11H(n,H(n,))22HH

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 55 of 42 of 42

Dose from Fast NeutronsDose from Fast Neutrons

In case of elastic scattering, scattered nuclei In case of elastic scattering, scattered nuclei dissipate their energy in the immediate vicinity of dissipate their energy in the immediate vicinity of the primary neutron interactionthe primary neutron interaction

This is called the “first collision dose” and is This is called the “first collision dose” and is entirely determined by the primary neutron fluxentirely determined by the primary neutron flux

The scattered neutron is not considered after the The scattered neutron is not considered after the primary interactionprimary interaction

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 66 of 42 of 42

Dose from Fast NeutronsDose from Fast Neutrons

For fast neutrons, the first collision dose is For fast neutrons, the first collision dose is given by:given by:

DDnn(E) =(E) =

where:where:

= neutron flux, n/cm= neutron flux, n/cm22/sec/sec

E = neutron energy, ergsE = neutron energy, ergs

EENNiiiiffii

101044 ergs ergsggGyGy

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 77 of 42 of 42

Dose from Fast NeutronsDose from Fast Neutrons

NNii = atoms per gram of the i = atoms per gram of the ithth element element

i i = scattering cross section of the i= scattering cross section of the ithth

element for neutrons of energy E in barns element for neutrons of energy E in barns (x10(x10-24-24 cm cm22))

ffii = mean fractional energy transfer to = mean fractional energy transfer to scattered atoms of mass M atomic mass scattered atoms of mass M atomic mass units during collision with a neutron of m units during collision with a neutron of m atomic mass unitsatomic mass units

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 88 of 42 of 42

Mean Fractional Energy TransferMean Fractional Energy Transfer

For isotropic scattering of fast neutrons, the For isotropic scattering of fast neutrons, the average fraction of energy lost in an elastic average fraction of energy lost in an elastic collision with a nucleus of atomic mass collision with a nucleus of atomic mass number M is given by:number M is given by:

f =f =(M + 1)(M + 1)22

2M2M

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 99 of 42 of 42

f Values for Tissuef Values for Tissue

ElementElement N (atoms/g)N (atoms/g) ff

HH 6.70 x 106.70 x 102222 0.5000.500

CC 4.18 x 104.18 x 102121 0.1420.142

NN 1.17 x 101.17 x 102121 0.1240.124

OO 2.93 x 102.93 x 102222 0.1110.111

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 1010 of 42 of 42

Sample Problem 1Sample Problem 1

What is the absorbed dose rate to soft tissue What is the absorbed dose rate to soft tissue from a beam of 5 MeV neutrons whose from a beam of 5 MeV neutrons whose intensity is 2000 n/cmintensity is 2000 n/cm22sec?sec?

Given: scattering cross sections of each of Given: scattering cross sections of each of the tissue elements for 5 MeV neutrons are the tissue elements for 5 MeV neutrons are shown in table in next slideshown in table in next slide

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 1111 of 42 of 42

Sample Problem No. 1Sample Problem No. 1

ElementElement (cm(cm22))

HH 1.5 x 101.5 x 10-24-24

CC 1.65 x 101.65 x 10-24-24

NN 1.00 x 101.00 x 10-24-24

OO 1.55 x 101.55 x 10-24-24

Scattering Cross Scattering Cross Sections in Tissue Sections in Tissue for 5 MeV Neutronsfor 5 MeV Neutrons

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 1212 of 42 of 42

Solution toSolution toSample Problem No. 1Sample Problem No. 1

Substituting the values for hydrogen into the Substituting the values for hydrogen into the equation, we have:equation, we have:

8.05 x 108.05 x 10-8-8 Gy Gysecsec hrhr

2.9 x 102.9 x 10-4-4 Gy Gy== ==

1.6x101.6x10-6-6 erg erg

MeVMeV

5 MeV5 MeVnn

6.7x106.7x102222 atoms atomsgg

1.5x101.5x10-24-24 cm cm22

atomatomcmcm2 2 -sec-sec2000 n2000 n

101044 ergs ergsgg

GyGy

xx x 0.5 x 0.5 xxxxxx

DDHH = =

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 1313 of 42 of 42

Solution to Sample Problem No. 1Solution to Sample Problem No. 1

In a similar manner, the dose would be calculated In a similar manner, the dose would be calculated for 5 MeV neutron interactions with C, N, and Ofor 5 MeV neutron interactions with C, N, and O

The total dose is found to be 3.2 x 10The total dose is found to be 3.2 x 10-4-4 Gy/hr Gy/hr

Thus, 5 MeV neutron interactions with H contributed Thus, 5 MeV neutron interactions with H contributed about 89% of the the total tissue dose!about 89% of the the total tissue dose!

It should be noted that this calculation was It should be noted that this calculation was performed for a monoenergetic neutron beamperformed for a monoenergetic neutron beam

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 1414 of 42 of 42

Dose from Thermal NeutronsDose from Thermal Neutrons

The dose rate from the The dose rate from the 1414N(n,p)N(n,p)1414C reaction can be C reaction can be calculated as follows:calculated as follows:

DD(n,p)(n,p) = = NNQ x 1.6 x 10Q x 1.6 x 10-6-6 erg/MeV erg/MeV

where:where:

= thermal flux, n/cm= thermal flux, n/cm22/sec/secN = nitrogen atoms per gram tissue (1.17x10N = nitrogen atoms per gram tissue (1.17x102121)) = nitrogen absorption cross section (1.75x10= nitrogen absorption cross section (1.75x10-24-24 cm cm22))Q = energy released by the reaction (0.63 MeV)Q = energy released by the reaction (0.63 MeV)

101044 erg/g/Gy erg/g/Gy

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 1515 of 42 of 42

Sample Problem No. 2Sample Problem No. 2

Calculate the dose rate from the Calculate the dose rate from the 1414N(n,p)N(n,p)1414C reaction from an average total C reaction from an average total body exposure rate of 10,000 thermal body exposure rate of 10,000 thermal neutrons per cmneutrons per cm2 2 per second.per second.

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 1616 of 42 of 42

Solution toSolution toSample Problem No. 2Sample Problem No. 2

= 2.1 x 10= 2.1 x 10-9-9 Gy/sec = 7.56 x 10 Gy/sec = 7.56 x 10-6-6 Gy/hr Gy/hr

1.6x101.6x10-6-6 erg erg

MeVMeV

0.63 MeV0.63 MeVnn

1.17x101.17x102121 atoms atomsgg

1.75x101.75x10-24-24 cm cm22

atomatomcmcm2 2 -sec-sec101044 n n

101044 ergs ergsgg

GyGy

xx xxxxxx

DDn,pn,p = =

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 1717 of 42 of 42

Dose from Thermal NeutronsDose from Thermal Neutrons

The dose rate from the The dose rate from the 11H(n,H(n,))22H reaction can H reaction can be calculated based on the following:be calculated based on the following:

This reaction is equivalent to having a This reaction is equivalent to having a gamma-emitting isotope uniformly gamma-emitting isotope uniformly distributed throughout the bodydistributed throughout the body

It results in what is called an “autointegral It results in what is called an “autointegral gamma-ray dose”gamma-ray dose”

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 1818 of 42 of 42

Dose from Thermal NeutronsDose from Thermal Neutrons

The specific activity of this distributed gamma source, The specific activity of this distributed gamma source, the number of reactions per second per gram of tissue, the number of reactions per second per gram of tissue, is given by:is given by:

A = A = NNwhere:where:

= thermal neutron flux, n/cm= thermal neutron flux, n/cm22/sec/secN = No. of hydrogen atoms per gram of tissue (6.7x10N = No. of hydrogen atoms per gram of tissue (6.7x102222)) = hydrogen absorption cross section (0.33x10= hydrogen absorption cross section (0.33x10-24-24 cm cm22))

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 1919 of 42 of 42

Autointegral Gamma-Ray Dose RateAutointegral Gamma-Ray Dose Rate

For a gamma-ray emitter uniformly distributed in the For a gamma-ray emitter uniformly distributed in the body, the autointegral gamma-ray dose rate may be body, the autointegral gamma-ray dose rate may be calculated by the following equation:calculated by the following equation:

DD = C = Cg Gy/hrg Gy/hr

where:where:

C = the concentration of the gamma emitter (Bq/cmC = the concentration of the gamma emitter (Bq/cm33)) = source strength (Gy/hr per Bq at 1 cm)= source strength (Gy/hr per Bq at 1 cm)g = average geometry factor (cm)g = average geometry factor (cm)

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 2020 of 42 of 42

Sample Problem No. 3Sample Problem No. 3

Calculate the absorbed dose rate to the standard man Calculate the absorbed dose rate to the standard man from the from the 11H(n,H(n,))22H reaction due to an average total body H reaction due to an average total body exposure rate of 10,000 thermal neutrons/cmexposure rate of 10,000 thermal neutrons/cm22/sec./sec.

Given:Given:

g (70 kg, 170 cm) = 126g (70 kg, 170 cm) = 126

(2.23 MeV photons ) = 2.70 x 10(2.23 MeV photons ) = 2.70 x 10-9-9 Gy-cm Gy-cm22/Bq-hr/Bq-hr

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 2121 of 42 of 42

Solution toSolution toSample Problem No. 3Sample Problem No. 3

Calculate the “specific activity” from the Calculate the “specific activity” from the 11H(n,H(n,))22H reaction in tissue:H reaction in tissue:

101044 n ncmcm22 - sec - sec

6.7 x 106.7 x 102222 atoms atomsgg

3.3 x 103.3 x 10-25-25 cm cm22

atomatom

1 g1 g

cmcm33C =C = xx xx xx

= 221 photons = 221 Bq/cm= 221 photons = 221 Bq/cm33

sec – cmsec – cm33

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 2222 of 42 of 42

Solution toSolution toSample Problem No. 3Sample Problem No. 3

Thus, the absorbed dose rate from gamma Thus, the absorbed dose rate from gamma emission in tissue due to the emission in tissue due to the 11H(n,H(n,))22H H reaction is given by:reaction is given by:

DD = C = Cg Gy/hrg Gy/hr

= 7.5 x 10= 7.5 x 10-5-5 Gy/hr Gy/hr

221 Bq221 Bq

cmcm33

2.70 x 102.70 x 10-9-9 Gy- cm Gy- cm22

Bq-hrBq-hrDD = 126 cm = 126 cmxxxx

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 2323 of 42 of 42

Sample Problem No. 4Sample Problem No. 4

1616N calibration sources are commonly used N calibration sources are commonly used at nuclear power stations. The source at nuclear power stations. The source generates generates 1616N via the (N via the (,p) reaction, using ,p) reaction, using 5.92 x 105.92 x 1099 Bq of Bq of 244244Cm and Cm and 1313C.C.

Calculate the total neutron dose equivalent Calculate the total neutron dose equivalent rate at 1 m from the source.rate at 1 m from the source.

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 2424 of 42 of 42

Sample Problem No. 4Sample Problem No. 4

Given:Given:

Neutron source strength = 2.0 x 10Neutron source strength = 2.0 x 1055 n/sec n/sec

Neutron energy = 2.5 MeVNeutron energy = 2.5 MeV

Neutron flux to dose equivalent conversion Neutron flux to dose equivalent conversion factor (k) = 20 n/cmfactor (k) = 20 n/cm22/sec = 2.5 x 10/sec = 2.5 x 10-5-5 Sv/hr Sv/hr

Source is a point sourceSource is a point source

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 2525 of 42 of 42

Solution toSolution toSample Problem No. 4Sample Problem No. 4

The total neutron dose rate at 1 m is given by:The total neutron dose rate at 1 m is given by:

DDnn = k = k

where:where:

SSn n = source strength= source strength

r = distance from source to receptorr = distance from source to receptork = neutron flux to dose equivalent rate conversion k = neutron flux to dose equivalent rate conversion

factorfactor

SSnn

44rr22

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 2626 of 42 of 42

Solution toSolution toSample Problem No. 4Sample Problem No. 4

Solving the equation, we have:Solving the equation, we have:

44 (100 cm) (100 cm)22

2.0 x 102.0 x 1055 n/sec n/secDDnn = = xx 2.5 x 102.5 x 10-5-5 Sv/hr Sv/hr20 n/cm20 n/cm22/sec/sec

= 2.0 x 10= 2.0 x 10-6-6 Sv/hr Sv/hr

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 2727 of 42 of 42

Sample Problem No. 5Sample Problem No. 5

You are the health physics manager at a You are the health physics manager at a uranium fuel reprocessing facility. Building A uranium fuel reprocessing facility. Building A includes a tank farm used to process highly includes a tank farm used to process highly enriched uranium.enriched uranium.

During a batch-processing operation in During a batch-processing operation in Building A, a technician violates standard Building A, a technician violates standard operating procedures, which leads to a critical operating procedures, which leads to a critical geometry in a small tank. A criticality occurs geometry in a small tank. A criticality occurs which releases 1.0 x 10which releases 1.0 x 101616 fissions. fissions.

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 2828 of 42 of 42

Sample Problem No. 5Sample Problem No. 5

The plant manager is standing behind a 30.5 cm The plant manager is standing behind a 30.5 cm polyethylene shield and he is 3 meters from the polyethylene shield and he is 3 meters from the center of the tank.center of the tank.

Calculate the neutron dose equivalent in Sv to Calculate the neutron dose equivalent in Sv to the plant manager during the Building A the plant manager during the Building A criticality.criticality.

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 2929 of 42 of 42

Sample Problem No. 5Sample Problem No. 5

Given:Given:

Density of the polyethylene shield is 1.4 g/cmDensity of the polyethylene shield is 1.4 g/cm33

Each fission event produces 3 neutronsEach fission event produces 3 neutrons

The neutron spectrum for the criticality is The neutron spectrum for the criticality is represented by a dose conversion factor ofrepresented by a dose conversion factor of

2.5 x 102.5 x 10-5-5 Sv/hr per 20 neutrons/cm Sv/hr per 20 neutrons/cm22-sec-sec

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 3030 of 42 of 42

Sample Problem No. 5Sample Problem No. 5

The mean neutron energy of the spectrum is The mean neutron energy of the spectrum is 2.5 MeV2.5 MeV

The neutron dose attenuation factor forThe neutron dose attenuation factor for2.5 MeV neutrons through 30.5 cm of 2.5 MeV neutrons through 30.5 cm of polyethylene is 0.005 and the buildup factor is polyethylene is 0.005 and the buildup factor is oneone

The criticality is adequately represented by a The criticality is adequately represented by a point source and it lasts one hourpoint source and it lasts one hour

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 3131 of 42 of 42

Solution toSolution toSample Problem No. 5Sample Problem No. 5

The general equation for the neutron dose The general equation for the neutron dose equivalent in this example is:equivalent in this example is:

NNfissionsfissions k e k e--tt B B

44rr22DDnn = =

where:where:NNfissionsfissions = 1.0 x 10 = 1.0 x 101616

k = 2.5 x 10k = 2.5 x 10-5-5 Sv/hr per 20 neutrons/cm Sv/hr per 20 neutrons/cm22-sec-secB = 1B = 1ee--tt B = 0.005 B = 0.005R = 3 m = 300 cmR = 3 m = 300 cm = number of neutrons per fission = 3= number of neutrons per fission = 3

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 3232 of 42 of 42

Solution toSolution toSample Problem No. 5Sample Problem No. 5

= 4.5 x 10= 4.5 x 10-2-2 Sv Sv1 hr1 hr

3600 sec3600 sec0.005 x0.005 x

(1.0x10(1.0x101616 fissions)x(3 ) fissions)x(3 )

(4)x(3.14)x(300(4)x(3.14)x(30022 cm cm22))DDnn = =

2.5 x 102.5 x 10-5-5 Sv Svhrhr

20 neutrons20 neutronscmcm22-sec-sec

neutronsneutronsfissionfission

xx xx

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 3333 of 42 of 42

Sample Problem No. 6Sample Problem No. 6

A linear accelerator (LINAC) bombards a tritium A linear accelerator (LINAC) bombards a tritium target with a 25 target with a 25 A beam of 2.5 MeV protons. A beam of 2.5 MeV protons. This produces 1.2 MeV neutrons via the This produces 1.2 MeV neutrons via the 33H(p,n)H(p,n)33He reaction.He reaction.

Calculate the neutron dose equivalent rate in Calculate the neutron dose equivalent rate in Sv/hr at a point 40 cm away from the target Sv/hr at a point 40 cm away from the target along the beam centerline.along the beam centerline.

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 3434 of 42 of 42

Sample Problem No. 6Sample Problem No. 6

Given:Given:

Production rate = 1.8 x 10Production rate = 1.8 x 10-6-6 neutrons/proton neutrons/proton

6.24 x 106.24 x 101818 protons/amp-sec protons/amp-sec

Dose conversion factor = 3.5 x 10Dose conversion factor = 3.5 x 10-10-10 Sv cm Sv cm22 n n-1-1

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 3535 of 42 of 42

Solution toSolution toSample Problem No. 6Sample Problem No. 6

Assuming that the distribution of neutrons Assuming that the distribution of neutrons is isotropic, the dose equivalent rate can be is isotropic, the dose equivalent rate can be written as:written as:

DDnn = =IKkIKk11P(DCF)P(DCF)

(4(4rr22 ) )

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 3636 of 42 of 42

Solution toSolution toSample Problem No. 6Sample Problem No. 6

where:where:

I = proton beam current = 25.0 x 10I = proton beam current = 25.0 x 10-6-6 A A

k = charge/proton = 1.602 x 10k = charge/proton = 1.602 x 10-19 -19 coulomb/protoncoulomb/proton

K = 1/k = 6.24 x 10K = 1/k = 6.24 x 101818 protons/A-sec protons/A-sec

kk11 = time conversion factor = 3600 sec/hr = time conversion factor = 3600 sec/hr

P = neutron production rate = 1.8 x 10P = neutron production rate = 1.8 x 10-6 -6 neutrons/protonneutrons/proton

DCF = dose conversion factor = 3.5 x 10DCF = dose conversion factor = 3.5 x 10-10-10 Sv cm Sv cm22 n n-1-1

r = distance from the target = 40 cmr = distance from the target = 40 cm

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 3737 of 42 of 42

Solution toSolution toSample Problem No. 6Sample Problem No. 6

Therefore, Therefore,

= 1.8 x 10= 1.8 x 10-2-2 Sv/hr Sv/hr

(25.0x10(25.0x10-6-6 A) x A) x

(4(4 40 4022 cm cm22))

3600 sec3600 sechrhr

1.8 x 101.8 x 10-6 -6 n nprotonproton

3.5x103.5x10-10-10 Sv cm Sv cm22

nn

6.24x106.24x1018 18 protonsprotonsA-secA-sec

xx xx

DDnn = =

4.9x104.9x10-6-6 Sv x Sv xsecsec

==

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 3838 of 42 of 42

SummarySummary

Principles of neutron dose calculation Principles of neutron dose calculation were discussedwere discussed

Students learned about fast and thermal Students learned about fast and thermal neutron interactions; dose calculation neutron interactions; dose calculation from neutron beams, neutron point from neutron beams, neutron point sources, fast neutrons, thermal neutrons; sources, fast neutrons, thermal neutrons; and worked example problemsand worked example problems

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 3939 of 42 of 42

Knoll, G.T., Knoll, G.T., Radiation Detection and Radiation Detection and MeasurementMeasurement, 3, 3rdrd Edition, Wiley, New York (2000) Edition, Wiley, New York (2000)

Attix, F.H., Attix, F.H., Introduction to Radiological Physics Introduction to Radiological Physics and Radiation Dosimetryand Radiation Dosimetry, Wiley, New York (1986), Wiley, New York (1986)

International Atomic Energy Agency, International Atomic Energy Agency, Determination of Absorbed Dose in Photon and Determination of Absorbed Dose in Photon and Electron Beams, 2Electron Beams, 2ndnd Edition, Technical Reports Edition, Technical Reports Series No. 277, IAEA, Vienna (1997)Series No. 277, IAEA, Vienna (1997)

Where to Get More InformationWhere to Get More Information

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 4040 of 42 of 42

International Commission on Radiation Units and International Commission on Radiation Units and Measurements, Quantities and Units in Radiation Measurements, Quantities and Units in Radiation Protection Dosimetry, Report No. 51, ICRU, Protection Dosimetry, Report No. 51, ICRU, Bethesda (1993)Bethesda (1993)

International Commission on Radiation Units and International Commission on Radiation Units and Measurements, Fundamental Quantities and Measurements, Fundamental Quantities and Units for Ionizing Radiation, Report No. 60, ICRU, Units for Ionizing Radiation, Report No. 60, ICRU, Bethesda (1998)Bethesda (1998)

Where to Get More InformationWhere to Get More Information

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 4141 of 42 of 42

Hine, G. J. and Brownell, G. L., (Ed. ), Hine, G. J. and Brownell, G. L., (Ed. ), Radiation Radiation DosimetryDosimetry, Academic Press (New York, 1956), Academic Press (New York, 1956)

Bevelacqua, Joseph J., Bevelacqua, Joseph J., Contemporary Health Contemporary Health PhysicsPhysics, John Wiley & Sons, Inc. (New York, 1995), John Wiley & Sons, Inc. (New York, 1995)

International Commission on Radiological International Commission on Radiological Protection, Data for Protection Against Ionizing Protection, Data for Protection Against Ionizing Radiation from External Sources: Supplement to Radiation from External Sources: Supplement to ICRP Publication 15. A Report of ICRP Committee 3, ICRP Publication 15. A Report of ICRP Committee 3, ICRP Publication 21, Pergamon Press (Oxford, 1973)ICRP Publication 21, Pergamon Press (Oxford, 1973)

Where to Get More InformationWhere to Get More Information

3/2003 Rev 13/2003 Rev 1 II.2.8 – slide II.2.8 – slide 4242 of 42 of 42

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)