Shielding Radiation 

Alphas, Betas, Gammas and Neutro

7/05/2011 
General

General

Shielding Alpha Emitters

Shielding Alpha Emitters

Shi ldi B t P ti l

Contents

Shielding Beta Particles

GeneralBeta Particle RangeBremsstrahlungExample Calculation

Shielding Positrons

Shielding Positrons
Shielding Gamma Rays

GeneralBasic Equation – First Example CalculationBasic Equation – Second Example CalculationRelaxation LengthCalculating the Required Shielding Thickness

Contents

Calculating the Required Shielding ThicknessCalculating the Required Shielding Thickness – Shielding Multiple Gamma Ray EnergiesShielding Multiple Gamma Ray Energies – ExamLayered (Compound) ShieldingLayered (Compound) Shielding – ExampleHalf Value LayersHalf Value layers – Example
Shielding Neutrons continued

Multipurpose Materials for neutron ShieldsPossible Neutron Shield OptionsNeutron Shielding Calculations – Fast Neutrons

Neutron Shielding Calculations – Alpha-Beryllium 

Neutron Shielding Calculations

Contents

Neutron Shielding Calculations

Gamma and Neutron Shields – General

GeneralRadiation energyShielding MaterialShielding Material - Lead
Shielding Gamma Rays continued

Tenth Value layersBuildupBuildup FactorBuildup Factor ExampleDetermining the Required shield Thickness with 

Contents

Determining the Required shield Thickness with 

Shielding Neutrons

The Three Steps1. Slow the Neutrons2. Absorb the Neutrons3. Absorb the Gamma rays
Gamma and Neutron Shields – General continued

Shielding Material – WaterShielding Material – ConcreteStreamingStreaming – Door DesignStreaming – Labyrinth

Contents

Shine

Appendices

Appendix A – Densities of Common MaterialsAppendix B – How to do a Linear InterpolationAppendix C – Range and Energy of Beta particleAppendix D – Transmission CurvesAppendix F – Buildup Factors
General

August 25, 2009 
• Shielding calculations done “by hand” are often 

approximations

• The most accurate shielding calculations are pe

computersShi ld ft “ d i d” th t th 

General

• Shi

elds are often 

“ove

r

-

designed” so th

a

t they ca

more intense sources of radiation that might be future

• The photon shielding calculations in this handou

apply to monoenergetic photons (e.g., gamma raShielding x-rays, which have a continuous range energies, is a specialized topic not considered h
The required shield thickness depends on three th

1. Radiation Energy. The greater the energy of tradiation (e.g., beta particles, gamma rays, neutthicker the shield must be.

General

2. The shield material. For photons (x-rays, gamthe lower the atomic number of the shield, the thmust be. For neutrons, the general rule is that ththe hydrogen content, the thicker the shield mus3. Radiation intensity. The higher the radiation inand the more it must be reduced, the thicker the 
Shielding Alpha Emitters

August 25, 2009 
• For external alpha emitting sources, the alpha p

themselves are not a radiation hazard. Howeverconsider the possibility that other types of radiatbe present. This possibility must always be conwhen shielding any type of radiation!

Shielding Alpha Emitters

• While we might assume that an alpha emitter do

gamma rays, it might actually do so. Pu-239 for emits a large number of gamma rays. Nevertheloften consider Pu-239 a “pure alpha” emitter becgamma intensity is low .
• Alpha emitters might also emit conversion elect

low energy x-rays. 

• Furthermore, alpha sources often contain trace 

of impurities some of which might emit beta part

Shielding Alpha Emitters

• These kinds of issues are primarily a concern w

quantities of alpha emitters are handled, e.g., in box facilities where curie and megacurie quantitbe encountered.
Shielding Beta Particles

August 25, 2009 
General

• The dose from an external source of beta particl

primarily to the skin. The penetrating power of b- approximately a few mm in tissue. 

Shielding Beta Particles

• Beta particles, like alpha particles, can be comp

stopped by a shield. It rarely requires more than of material to do so.

• The material the shield is made of is not very im

terms of its ability to stop the beta particles. 
Beta Particle Range

• Beta ranges are often expressed as a “density t

which has units of g/cm2. These ranges can be 

converted to centimeters as follows:

Shielding Beta Particles

• The easy way to estimate the range (in mg/cm2) 

particles is to obtain it from the graph on page 1PTP Radiological Health Handbook. The energy on the graph is the maximum beta energy.
Beta Particle Range

Several equations approximate the range of betasthickness of the shield required to stop all the beta• The equation below (Feather's Rule) applies to b

a maximum energy above 0.6 MeV:

Shielding Beta Particles

Range (g/cm2) = 0.542 E – 0.133

E is the maximum β energy (MeV)

• The following equation applies to betas with a m

energy below 0.8 MeV:

Range (g/cm2) = 0.407 E1.38
Bremsstrahlung

• While it is easy to completely stop beta particles, 

act of doing so produces bremsstrahlung, a more penetrating type of radiation. Shielding bremsstrabe more of a problem than shielding gammas.A d hi ldi t i l f b t i i i

Shielding Beta Particles

• A good shielding material for betas minimizes 

bremsstrahlung i.e., low Z material such as plast

• Like beta particles, bremsstrahlung photons have 

energies up to a maximum. The maximum energy bremsstrahlung photons is the same as the maxienergy of the betas. When shielding bremsstrahlucan take the very conservative approach that all tbremsstrahlung photons have this maximum ene
A 1 Ci P-32 source (50 mls aqueous) is in a polyeth

bottle (0.93 g/cm

3). The effective atomic numbers of 

solution and vial wall are assumed to be 7.5. 

a. How thick should the vial be?

Shielding Beta Particles Example calculation

b. What is the exposure rate at 1 meter due to 

bremsstrahlung? 
a. How thick should the vial be?

The maximum beta energy for P-32 is 1.71 MeV. Fcurve (p 162) in the PTP Radiological Health Handbmaximum range of the P-32 betas is 800 mg/cm

2 (0

Shielding Beta Particles Example calculationEssentially the same result is obtained using the eq

Range (g/cm

2) = 0.542 E – 0.133

= 0.542 x 1.71 – 0.133= 0.79 g/cm

2

As such, the vial should be at least 0.86 cm thick ( 0divided by 0.93 g/cm

3).
b. What is the exposure rate at 1 meter due to 

bremsstrahlung? 

The fraction of the beta energy released in the form 

bremsstrahlung from interactions in the solution (anShielding Beta Particles Example calculation

g (

lesser extent in the vial wall) is calculated as follows
b. What is the exposure rate at 1 meter due to 

bremsstrahlung? - continued

The beta energy emission rate from the 1 Ci P-32 s

3 7 1010 d 07M V d

Shielding Beta Particles Example calculation

= 3.7 x 1010 dps x 0.7 MeV per decay

= 2.59 x 1010 MeV/s

The bremsstrahlung energy emission rate is therefo

= 0.0043 x 2.59 x 1010 MeV/s

= 1.11 x 108 MeV/s
b. What is the exposure rate at 1 meter due to 

bremsstrahlung? - continued

The energy fluence rate (R) through a 1 cm2 area at 

is:Shielding Beta Particles Example calculation
b. What is the exposure rate at 1 meter due to 

bremsstrahlung? - continued

The dose rate to air is the energy fluence rate (MeV

times the mass energy absorption coefficient for air 

Shielding Beta Particles Example calculation

Absorbed dose rate = R x uen/D

Mass energy absorption coefficients apply to single energies but bremsstrahlung photons have a range energies. A conservative assumption is that they alenergy of 1.71 MeV (the maximum energy for P-32)The mass energy absorption coefficient for 1.71 Mein air is approximately 0.025 cm2/g. 
b. What is the exposure rate at 1 meter due to 

bremsstrahlung? - continued

Absorbed dose rate = R x u/D

= 883 (MeV cm-2 s-1) x 0.025 

Shielding Beta Particles Example calculation

( ) 

= 22.1 MeV g-1 s-1= 22.1 x 106 eV g-1 s-1

Since one ion pair is produced for every 34 eV absothe air, the ion pair production rate is:

= 22.1 x 106 (eV, g-1, s-1) / 34 (eV/ion 

= 0.65 x 10 6 ion pairs per gram per s
b. What is the exposure rate at 1 meter due to 

bremsstrahlung? - continued

Since the charge produced in air for one member of 

ion pair is 1 6 x 10-19 C the rate of charge productioShielding Beta Particles Example calculationion pair is 1.6 x 10 C, the rate of charge productio

= 0.65 x 10 6 (ion pairs g-1 s-1) x 1.6 x 10

= 1.23 x 10-13 C g-1 s-1= 1.23 x 10-10 C kg-1 s-1= 4.44 x 10-7 C kg-1 hr-1
b. What is the exposure rate at 1 meter due to 

bremsstrahlung? - continued

Finally, the answer:

Shielding Beta Particles Example calculationSince 2.58 x 10-4 C/kg equals 1 roentgen, the expos

in R/hr is 

= 4.44 x 10-7 (C kg-1 hr-1)/2.58 x 10-4 (C/= 1.72 x 10-3 R/hr
Shielding Positrons

August 25, 2009 
• In general, positron energies (and the bremsstrah

produce) tend to be higher than beta energies. Evpositrons themselves are easy to stop. It requires cm or so of plastic.The most important positron emitter, F-18, emits th t f f i l l i 633 5 k V

Shielding Positrons

th

a

t are o

f

f

airly low energy: maximum 633.5

k

e

V

249.8 keV.

• What we don’t want to forget is that each emitted 

will produce two 511 keV annihilation photons thashielded. These annihilation photons are not alwamentioned in tables of decay data.
Shielding Gamma Rays

August 25, 2009 
X0 is the exposure rate in the absence of shielding

Shielding Gamma RaysGeneralSource

X0

d
We want an equation that gives the exposure rate 

shielding of a given thickness (x) is present.

X

Shielding Gamma RaysGeneral

Source

d

x
Basic Equation

The following basic equation assumes a narrow beradiation penetrating a thin shield (a situation referr"good geometry").Shielding Gamma Rays

X is the exposure rate with the shield in place (eXo is the exposure rate without the shield (e.g., R

x is the thickness of the shield (e.g., cm)

u is the linear attenuation coefficient (e.g., cm-1). 

The linear attenuation coefficient (u) is the probabiltype of interaction (PE, CS, PP) per unit path length
Basic Equation

Listings of the linear attenuation coefficient (u) are find. However, listings of a related quantity, the masattenuation coefficient (u/D), are easy to find. Th “ l i ” li ti f tt ti ffi i t Shielding Gamma RaysThe “classic” listing of mass attenuation coefficients 

Hubbell. Hubbell’s tables can be found in the PTP Radiological Health Handbook beginning on page 9The mass attenuation coefficient (u/D) is the probab

interaction (PE, CS, PP) per unit density thickness. are cm2/g. In other words, u/D is the probability of 

interactions per g/cm2 travelled by the photon.
Basic Equation

The value of the mass attenuation coefficient depenshielding material and the energy of the photons. Nevertheless, mass attenuation coefficients for diffematerials are very similar. For example, the followili t tt ti ffi i t f 1 M V Shielding Gamma Rayslists some mass attenuation coefficients for 1 MeV 

The mass attenuation coefficient is the linear attenudivided by the density of the material.

Shield Material u/D (cm2/g)

water 0.0707

concrete 0.0637

air 0.0636

iron 0.0599
Basic Equation

When using the mass attenuation coefficient instealinear attenuation coefficient, the basic shielding eqShielding Gamma Rays

μ/ρ is the mass attenuation coefficient (cm2/g

ρ is the density of the shielding material (g/c

Note, the units in the exponent must cancel out:

cm2/g x g/cm3 x cm
Basic Equation – First example calculation 

The exposure rate at a particular point is 100 R/hr d1332 keV gamma rays from Co-60. What would be resulting exposure rate if a 1 cm lead shield were ebetween the source and the point? Shielding Gamma RaysWarning!In this example, we will use the basic equation whicassumes we have “good” geometry, i.e., a narrow bthin shield. As we will see later, in the “real worldhave good geometry and this equation would result underestimate of the exposure rate.
Basic Equation – First example calculation 

u/

ρ is approximately 0.057 cm

2/g (I eyeballed this)

ρactual is assumed to be the same as the theoretical 

11.35 g/cm

3

Shielding Gamma Rays
Basic Equation – Second example calculation 

The exposure rate at a particular point is 100 R/hr dkeV gamma rays from Cs-137. What would be the exposure rate if a 1 cm lead shield were employed? u/

ρ is approximatel

y 0.11 cm

2/g

(I e

yeballed this

)

Shielding Gamma Rays

ρ pp y g( y )
Relaxation Length

The relaxation length is the thickness of a shielding that will reduce the intensity of the radiation to 1/e (

its original intensity.

Relaxation length (cm) = 1/u

Shielding Gamma Rays

Relaxation length (cm) = 1/u

u is the linear attenuation coefficient (cm-1)
Calculating the required shielding thickness 

The preceding equation can be rewritten as follows calculate the shielding thickness (x) necessary to adesired reduction in the exposure rate:Shielding Gamma Rays
Calculating the required shielding thickness - E

What thickness of a lead shield would be required tthe exposure rate due to Cs-137 from 100 R/hr to 1 Shielding Gamma Rays
Shielding multiple gamma ray energies

Since the mass attenuation coefficient (u/

D) depend

energy of the photons, the preceding shielding calconly work if the exposure rate is due to a single gamenergy. Shielding Gamma RaysThis works fine for Cs-137 which has a single gammHow do we handle the situation where the exposure due to more than one gamma ray energy, e.g., the gamma rays of Co-60 at 1173 keV and 1332 keV? 
Shielding multiple gamma ray energies

We could take a conservative approach and do the calculation only using the highest of the gamma ray This overestimates the exposure rate and the requishielding. Shielding Gamma RaysThis approach might be reasonable for something lwhere the two gamma rays are of similar energy ancontribute nearly equally to the exposure rate. On the other hand, this approach could be a disastesituation where the highest gamma ray energy contonly a little to the exposure rate (e.g., a high activity source combined with a low activity Cs-137 source
Shielding multiple gamma ray energies

The correct approach is to determine the exposure each gamma ray energy. Then we determine the aof each gamma ray energy as an independent calcHere is an example problem of how this can be donShielding Gamma RaysHere is an example problem of how this can be don

The exposure rate at a particular point is 100 R/hr d1173 and 1332 keV Co-60 gamma rays. What wouresulting exposure rate if a 1 cm lead shield were ebetween the source and the point? Once again, w“good” geometry, i.e., a narrow beam and thin shie
Shielding multiple gamma ray energies - Exampl

The 100 R/hr is the sum of the 1173 keV gamma ra

and 1332 keV gamma ray (X1332 ) exposure rates.

X d X b d t i d f th i if

Shielding Gamma Rays

1173 an

d

X1332 can 

be 

d

e

termine

d

from th

eir specif

ray constants: 

'1173 and 

'1332

The specific gamma ray constant for Co-60 (1.3 R m

is the sum of the specific gamma ray constants for th

and 1332 keV gamma rays:
Shielding multiple gamma ray energies - Examp

The specific gamma ray constant for a given gammenergy can be calculated with the equation:Shielding Gamma Rays

' is the specific gamma ray constant in R hr-1

m

2 

E is the gamma ray energy in MeV, I is the gamma ray intensity (abundance)uen/p is the mass energy absorption coefficient
Shielding multiple gamma ray energies - Examp

The specific gamma ray constants for the two gammShielding Gamma Rays
Shielding multiple gamma ray energies - Examp

Shielding Gamma Rays
Shielding multiple gamma ray energies - Examp

First we calculate the effect of 1 cm of lead on the 4due to the 1173 keV photons. The mass attenuatiocoefficient is estimated to be 0.0625 cm

2/g

Shielding Gamma Rays
Shielding multiple gamma ray energies - Examp

Second we calculate the effect of 1 cm of lead on thdue to the 1332 keV photons. The mass attenuatiocoefficient is estimated to be 0.057 cm

2/g

Shielding Gamma Rays
Shielding multiple gamma ray energies - Examp

Finally we add the exposure rates for the 1173 and gamma rays:Shielding Gamma RaysAs expected, the exposure rate (51 R/hr) is not mucdifferent from what we calculated when we assumeexposure rate was only due to the 1332 keV gammR/hr).
Layered (compound) Shielding

It is often the case that the shielding material consisseveral layersShielding Gamma Rays

Source

X

x1 x

2
Layered (compound) Shielding

Shielding Gamma Rays

Source

X

x1 x

2
Layered (compound) Shielding – Example

The exposure rate at a particular point is 100 R/hr dkeV gamma rays. Estimate the resulting exposure shield consisting of 1 cm of lead and 2 cm of iron wemployed between the source and the point? Assu“good

” geometry i e a narrow beam and thin shie

Shielding Gamma Rays

geometry, i.e., 

a narrow beam and thin shie

u/

ρ is approximately 0.11 cm

2/g for lead 

u/

ρ is approximately 0.075 cm

2/g for iron 

ρ for the lead is assumed to be the same as theoretical density of 11.35 g/cm

3

ρ for the iron is assumed to be the same as theoretical density of 7.86 g/cm

3
Layered (compound) Shielding – Example

Shielding Gamma Rays
Half Value Layers

Instead of using attenuation coefficients to perform calculations, we can use half (or tenth) value layersvalue layer is the thickness of material that reduces radiation intensity by one-half. Shielding Gamma RaysThe following equation using HVL's looks very simiradioactive decay equation:
Half Value Layers

A mathematically equivalent form of this equation isShielding Gamma RaysHalf value and tenth value layers for some materialfound in PTP’s Radiological Health Handbook on pDon’t be surprised if you see slightly different valueliterature for the half value layers of different nuclide0.6 cm or 0.65 cm for Cs-137 in lead.
Half Value Layers

The following table gives an idea as to the approximvalue layer thicknesses for several shielding materiShielding Gamma Rays

Energy 

(MeV)

Approximate Half Value Layers in cm (TVL in parent

Uranium Tungsten Lead Iron Concrete 

0.5 0.51 1.0 3.30 1.0 0.76 1.52 4.57 1.5 1.27 1.78 5.84 2.0 1.52 2.03 6.60 

Ir-192 0.28 0.33 0.48 1.27 4.5

Cs-137 0.65 (2.16) 1.6 (5.3) 4.8 (15.7

Co-60 0.69 0.79 1.2 (4.0) 2.1 (6.9) 6.2 (20.6

Ra-226 1.66 (5.5) 2.2 (7.4) 6.9 (23.4
Half Value Layers - Example

The exposure rate is 100 R/hr due to a Cs-137 souwould the exposure rate be if the source were shielcm of lead? Shielding Gamma Rays
Tenth Value Layers

In a similar manner, a tenth value layer (TVL) is defthat thickness which will reduce the radiation intenstenth of its original value. The equations for using Tas follows: Shielding Gamma Rays
Buildup

When the exposure rate at a given distance from a source is calculated using the standard equation, wscattered radiation doesn’t contribute to the exposuOnly those gamma rays originally directed to the pointerest contributeShielding Gamma Raysinterest contribute.

Source
Buildup

Similarly, the shielding equations considered so far assume no contribution from radiation scattered by shield towards the point of interest. This is the so-c“good geometry” only possible with a narrow (collimbeam) and a thin shield.Shielding Gamma Raysbeam) and a thin shield.

Collimated

Source

Shield
Buildup

For a broad beam or thick shield situation like that bradiation scattered in the shield contributes to the erate. This is the real world (“bad geometry”). Scatteradiation towards the point of interest is referred to buildup. Shielding Gamma Rays

Source 

Shield
Buildup

To account for buildup, we can incorporate a “fudge factor” known as the buildup factor into our equations. Shielding Gamma Rays

Source X

Shield
Buildup Factor

The buildup factor is a unitless number greater than 1.Shielding Gamma Rays

The value we assign to the buildup factor depends things:

• Shield material• Energy of the photons• Value of the equation’s exponent (:/D) D x (i.e

Buildup factors can be found on page 88 of PTP’s Radiological Health Handbook.
Buildup Factor

The equation’s exponent, the product ux , is sometreferred to as the number of mean free paths. The mean free path (mfp) is the average distance a travels before an interaction takes place. It is the re

f th li tt ti ffi i t ( )

Shielding Gamma Raysof the linear attenuation coefficient (u).
Buildup Factor - Example

The exposure rate at a particular point is 100 R/hr d1332 keV gamma rays. What would be the resultinexposure rate if a 1 cm lead shield were employed the source and the point? Assume “poor” geometrybroad beam and thick shield.Shielding Gamma Rays

u/

ρ is approximately 0.057 cm

2/g

ρactual is assumed to be 11.35 g/cm

3

B is estimated to be 1.2, based on an energy of a value of 0.647 for ux (number of mean free pat
Buildup Factor - Example

Shielding Gamma Rays
Determining the Required Shield Thickness with 

The following equation gives the shield thickness (xnecessary to reduce the exposure rate from X

0 to X

Shielding Gamma RaysA problem with this equation is that the buildup factdepends on the thickness of the shield (x), which is what we are trying to determine. In other words, we can’t determine B before we kno
Determining the Required Shield Thickness with 

First approach - iterationWe could estimate B by guessing what x is likely to Then we could plug B and x into the following equasee how close we 

get to the re

quired ex

posure rate 

Shielding Gamma Rays

g qp

If X is too high, increase the shield thickness and recalculation. Keep doing this until we find the shield that achieves the desired exposure rate. This apprreferred to as iteration. An unsavory sledgehammeapproach to be sure, but it works.
Determining the Required Shield Thickness with 

Second approach – energy absorption coefficientWe could eliminate the buildup factor in our equatiosubstituting the mass energy absorption coefficient 

the mass attenuation coefficient (

:

/

D) . Then we us

Shielding Gamma Raysfollowing equation for x:Although not technically correct, this approach is si
Determining the Required Shield Thickness with 

Third approach- computersSwallow our pride and use a computer.Com

puter calculations can 

produce ver

y accurate r

Shielding Gamma Rays

p py

complicated shield and source geometries.Computer modeling typically uses the Discrete OrdTechnique (DOT for short) or the Monte Carlo TechThe DOT is a numerical integration technique used solutions of the Boltzmann Transport equation in cesimplified cases. 
Determining the Required Shield Thickness with 

Third approach - computersThe Monte Carlo method uses extensive data tablerandom number generator to mathematically "trackthousands of simulated emissions of radiation throuShielding Gamma Raysshield. Perhaps the best known shielding code is Microshie
Determining the Required Shield Thickness with 

Fourth approach – transmission curvesOne way to determining the required shielding thicreduction in exposure rate due to a given shield thicto use a transmission curve that accounts for builduShielding Gamma RaysTransmission curves are plots of transmission (X/X

shield thickness for various radionuclides. Exampleare on page 82 of PTP’s Radiological Health HandbAlas, transmission graphs are only available for a linumber of radionuclides and shield materials and thhave somewhat limited applicability.
Determining the Required Shield Thickness with 

Fourth approach – transmission curves exampleWhat thickness of lead shielding is required to reduexposure rate due to Co-60 from 100 R/hr to 22 mRShielding Gamma Rays
Determining the Required Shield Thickness with 

Fourth approach – transmission curves exampleFrom the curve on page 83 of PTP’s Radiological HHandbook, approximately 15 cm of lead correspontransmission factor of 2.2 x 10-4Shielding Gamma Rays
Shielding Neutrons
Neutron Shielding

The Three StepsShielding neutrons involves three steps: 

1. Slow the neutrons

2. Absorb the neutrons3. Absorb the gamma rays
Neutron Shielding

1. Slow the neutrons• Neutrons are slowed to thermal energies with 

hydrogenous material: water, paraffin, plastic.

• Water can evaporate or leak, paraffin is flamma

l ti i i

plastic is expensive.

• To slow down very fast neutrons, iron or lead m

used in front of the hydrogenous material.
Neutron Shielding

2. Absorb the neutrons• Hydrogenous materials are also very effective a

absorbing neutrons - the cross section for neucapture by H-1 is 0.33 barns. 

• Unfortunately a difficult to shield 2 2 MeV gamm

Unfortunately, a difficult to shield 2.2 MeV gamm

emitted when H-1 absorbs a neutron.

• Boron might be incorporated into the shield bec

has a large cross section for neutron absorption emits a low energy capture gamma ray.

• To slow down very fast neutrons, iron or lead m

used in front of the hydrogenous material.
Neutron Shielding

3. Absorb the gamma rays• Gamma rays are produced in the neutron shie

neutron (radiative) capture, inelastic scatteringdecay of activation products.
Neutron Shielding

Multipurpose Materials for Neutron Shields• Concrete, especially with barium mixed in, ca

absorb the neutrons, and shield the gamma ra

• Plastic with boron is also a 

good multi

pur

pose 

g pp 

material.
Possible Neutron Shield Options

n

Neutron Shielding

Hydrogenous 

material

Gamma

shield

n

Hi Z moderator for fast 

neutrons

n

Combinatio

borated plas

with 
Neutron Shielding Calculations – fast ne

Neutron shielding calculations are best done by coNevertheless, in some limited situations, it is possemploy a simplistic exponential equation similar to for monoenergetic photons.Neutron ShieldingThe following equation (Schaeffer 1973) describes of a given shielding material (e.g., steel) on fast nerate. It only works if the there is at least 50 cm of wequivalent hydrogenous material) behind the shiel
Neutron Shielding Calculations – fast ne

t

H

O

D

Neutron Shielding

t

H

2

O

D

D is the dose rate with shieldD0 is the dose rate without shieldt is the shield thickness (e.g., cm)ER is the neutron removal cross section (e
The following equation (modified from one in from 

Handbook 63) describes the effect of shield thicknthe neutron dose rate associated with a radioactivNeutron Shielding Calculations – alpha-beryllium sourcesNeutron Shieldingsource (e.g., AmBe).

B is a buildup factor usually assumed to be 
The following values for removal cross sections ar

NBS Handbook 63.

Material Removal Cross Section 

E

R (cm-1

)

water 0 103

Neutron Shielding CalculationsNeutron Shielding

0.103

iron 0.1576

ordinary concrete 0.0942

barytes concrete 0.0945

graphite 0.0785

For what it is worth, the removal cross section is approximately 2/3 to 3/4 of the total cross section
In neutron shielding calculations, we might also us

mass attenuation coefficient symbolized 

E

R

/

D

According to Schaeffer (1973), the mass attenuaticoefficient (

E

R

/

D) for fast neutrons can be approxim

Neutron Shielding CalculationsNeutron Shieldingyour choice of one of the following:

E

R

/

D = 0.19 Z-0.743 cm

2/g (Z 

≤ 8)

= 0.125 Z-0.565 cm

2/g (Z > 8)

E

R

/

D = 0.206 A-1/3 

Z-0.294 ~ 0.206 (A Z)-1/

E

R

/

D = 0.21 A-0.58
When the neutron mass attenuation coefficient (

E

R

employed instead of the neutron removal cross seprevious equation becomes:

Neutron Shielding Calculations Neutron Shielding
Neutron Shielding Calculations

Another simple approach that might be possible is employ the neutron transmission/attenuation curveAllen and Futterer which can be found beginning o12 of PTP’s Radiological Health Handbook.Neutron ShieldingThese graphs provide dose reduction factors for nof specified energies as a function of shield thicknshield is indicated to be polyethylene, but the grapprovide conversion factors for water, concrete and (NTS).
Gamma and Neutron Shields 

General
General

Consider the possible use of temporary shielding (mfixed) rather than, or in addition to, permanent shieConsider the consequences of the shield becoming contaminated or if neutrons are present activatedGamma and Neutron Shields - Gencontaminated or, if neutrons are present, activated

If permanent shielding is used, consider the cost anassociated with dismantling it – this might be neceswhen the facility is decommissioned. Concrete shiesometimes built with an inner (closest to the sourcethat is intended to be removed at the end of the fac
General

If a source emits gamma rays at a variety of energihigh energy gamma rays sometimes determine the shielding requirements. For example, even though 1764 and 2204 keV gamma rays of radium sources normally contribute very little to the exposure rate tGamma and Neutron Shields - Gennormally contribute very little to the exposure rate, t

become the controlling radiation once the shield haeliminated the more predominant low energy photoHVL for Ra-226 is even larger than that for Co-60!
General

Consider the possible use of local shielding inside svaults/areas not normally occupied. For example shmight be installed inside accelerator vaults near thoaccelerator components most prone to activation. Treduces doses during maintenance when the area Gamma and Neutron Shields - Genreduces doses during maintenance when the area 

temporarily occupied.
Radiation Energy

The lower the energy of the gamma rays or neutroneasier it is to shield them (i.e., the thinner the shield to be).Gamma and Neutron Shields - GenIf a source emits gamma rays at a variety of energihigh energy gamma rays sometimes determine the shielding requirements. For example, even though 1764 and 2204 keV gamma rays of radium sources normally contribute very little to the exposure rate, tbecome the controlling radiation once the shield haeliminated the more predominant low energy photoHVL for Ra-226 is even larger than that for Co-60!
Shielding Material 

For gamma rays, the higher the atomic number of tmaterial, the greater the attenuation of the radiationFor neutrons, the lower the atomic number of the s

t i l th t th tt ti f th di ti

Gamma and Neutron Shields - Genmaterial, the greater the attenuation of the radiation

For both radiations, the greater the density of the smaterial the greater the attenuation of the radiationthough water and air have essentially the same effeatomic number, water is a better shield because it denser.
Shielding Material – Lead

For attenuating photons

Advantages: 

C high atomic number (Z=82)

C moderately high density (11.34 

C easily machined and melted

Gamma and Neutron Shields - Gen

C relatively inexpensive

Disadvantages: 

• lack of mechanical strength• chemical toxicity (some facilities require glove

handling) . Oxidized lead (white) is especially 

• “new” lead can be naturally radioactive (Pb-21• recycled lead is often contaminated 
Shielding Material – Lead

Lead is available in a variety of forms including:

• standard bricks (4” x 8” x 2” approx 25-30 lbs)

• interlocking bricks

• sheets (typically 1/64 to 1/8

” thick) including le

Gamma and Neutron Shields - Gen

• sheets (typically 1/64 to 1/8 thick), including le

backed drywall board and plywood

• tape• wire• pipe• shot – for filling voids• wool – for filling voids
Shielding Material – Lead

Leaded hardware/items of various types:

• leaded louvers for vents• leaded hinges and handles for doors• lead covers for electrical switches and covers

Gamma and Neutron Shields - Gen

lead covers for electrical switches and covers

• lead-headed nails for installing lead sheets• lead plugs, caps for screw hole covers.• batten strips for joints and corners• lead glass for windows - it varies, but 4 mm of 

typical lead glass is roughly equivalent to 1 mm lead 
Shielding Material – Tungsten Alloys

For attenuating photons. Tungsten alloys can contairon, copper, cobalt and nickel.

Advantages: 

C tungsten’s high atomic number (

C ver

y hi

gh densit

y

(ca. 14 - 18 

g/c

Gamma and Neutron Shields - Gen

yg y

(

g

C strongC non-toxicC thinner shields possible than wit

Disadvantages: 

• very difficult to machine and fabricate• expensive
Shielding Material – Depleted Uranium

For attenuating photons. 

Advantages: 

C very high atomic number (Z= 92

C very high density (19.1 g/cm

3

)

C good mechanical strength

Gamma and Neutron Shields - Gen

good mechanical strength

C thinner shields possible than wit

Disadvantages: 

• pyrophoric• expensive, but less so than tungsten• radioactive, hence there are regulatory issues 
Shielding Material – Water

For attenuating photons and neutrons

Advantages: 

• completely fills all voids unlike a solid shi• trans

parent

Gamma and Neutron Shields - Gen

• can manipulate source and equipment in • inexpensive• source leaks can be detected by samplin

Disadvantages: 

C it can leak

C it can evaporateC a leaking source can contamin

volumes of the water 
Shielding Material – Concrete

For attenuating photons and neutrons

Advantages:

• can also serve a structural role

Gamma and Neutron Shields - Gen

• can incorporate additives such as barium to i

its characteristics

• can be used to shield both neutrons and gam

Disadvantages: 

• very thick walls can be needed• it can be difficult to modify and dismantle
Streaming

Cracks or voids in the shield can allow narrow beamradiation to stream through the shield. Such openings in a shield might be unavoidable, e.d t ti f i i d t bl tGamma and Neutron Shields - Gendoors, penetrations for piping, ducts, cables ,etc. 

Whenever possible, such openings should be curveangled so as to prevent the radiation from having a straight shot through the shield. Even so, some garays and especially neutrons can scatter their way tthese openings.Quantifying such streaming is best done with a com
Streaming

Penetrations in the shield should generally be at leaseven feet above the floor.Consider filling any voids in such penetrations (e.g.Gamma and Neutron Shields - Genlead shot).Penetrations should not be directed toward occupieareas or areas where there is traffic, e.g., stairwaysDetermine if shields could be employed at ends of tpenetration.
Streaming – Door design

Neutron streaming can be harder to minimize than streaming.Gamma and Neutron Shields - Gen

Wall Wall

Door

Overlap Overlap

Doo

Wall 
Door

Streaming – Door designIt can be very difficult to prevent radiation from goin

through the gap under a door. 

Gamma and Neutron Shields - Gen

Floor

Scattered Radiation

High atomic number inserts under a door can minscattered gamma radiation. 

Insert
Streaming - Labyrinth

Gamma and Neutron Shields - GenA labyrinth (aka maze) might be employed at the ena shielded area.This labyrinth prevents a “line of site view” of the sowhile allowing equipment and material to be moved while allowing equipment and material to be moved 

with relative ease.It avoids the need for complicated door designs like shown in the previous slides.The general rule is that the radiation must scatter atwice to reach the entrance.
SOURCE

Streaming - LabyrinthGamma and Neutron Shields - Gen

LOCATION

Labyrinth/maze at 

entrance to 

minimize streaming 

through door
Shine

Shine is a term that usually refers to radiation that hpenetrated a weak spot in the shielding and scatterOne such weak spot is often the ceiling. Gamma and Neutron Shields - Gen

Sk hi

Source Ground shine

Shield

Sky shine
Shine

Radiation that goes over the shield and scatters in towards the ground is known as sky shine.Radiation that has gone through the floor and scattGamma and Neutron Shields - Genupwards in the ground is known as ground shine. Tlatter is rarely a major concern.The primary dose to the public near a nuclear facilitaccelerator) is often due to sky shine. This is espectrue for neutrons.
Appendices