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Interaction of Radiation with Matter - 4

Buildup and Shielding

Day 2 – Lecture 4

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To discuss shielding for photon beams and the increase in photon transmission through a shield resulting from buildup

Objective

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HVL and TVL

• The amount of shielding required to reduce the incident radiation levels by ½ is called the “half-value layer” or HVL

• The HVL is dependent on the energy of the photon and the type of material.

• Similarly, the amount of shielding required to reduce the incident radiation levels by 1/10 is called the “tenth-value layer” or TVL.

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HVL and TVL

HVL (cm) TVL (cm)

Isotope Photon E (MeV)

Concrete Steel Lead Concrete Steel Lead

137Cs 0.66 4.8 1.6 0.65 15.7 5.3 2.1

60Co 1.17, 1.33

6.2 2.1 1.2 20.6 6.9 4

198Au 0.41 4.1 0.33 13.5 1.1

192Ir 0.13 to 1.06

4.3 1.3 0.6 14.7 4.3 2

226Ra 0.047 to 2.4

6.9 2.2 1.66 23.4 7.4 5.5

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HVL and TVL

The half value layer (HVL) and tenth value layer (TVL) are mathematically related as follows:

HVL = ln(2)

TVL = ln(10)

=TVLHVL

ln(10)

ln(2)

= ln(10)ln(2)

=2.3030.693

= 3.323

TVL = 3.323 x HVL

and

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Shielding

Shielding is intended to reduce the radiation level at a specific location

The amount of shielding (thickness) depends on:

the energy of the radiation the shielding material the distance from the source

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Inverse Square Law

If the radiation emanates from a point source, the radiation follows what is commonly known as the “Inverse Square Law” or ISL.

Most real sources which are considered to be “point” sources are actually not “point” sources. Most sources such as a 60Co teletherapy source have finite dimensions (a few cm in each direction). These sources appear to behave like point sources at some distance away but as one gets closer to the source, the physical dimensions of the source result in a breakdown of the ISL.

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Inverse Square Law

If the source of the radiation is not a point but is a line, a flat surface (plane) or a finite volume, the ISL does not apply.

However, if one gets far enough away from a finite line, plane or volume, they appear to be a point and the ISL applies with some acceptable error.

For the remaining discussion let’s assume we have a point source. The ISL predicts that the intensity of the radiation will decrease as distance from the source increases even without any shielding.

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These photons should not strike the individual. But due to scatter, they do, so the calculated value is too low. It needs to be increased by the buildup factor.

Scatter

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Photon Attenuation and Absorption

• Absorption refers to the total number of photons absorbed by the material (dark blue arrows)

Attenuation refers to total number of photons removed from incident beam (absorbed + scattered) (dark blue and light blue arrows)

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Photon Attenuation

Ix = Io e-x = Io ewhere:

Ix = photon intensity after traversing x cm of some material

Io = initial or incident photon intensity

x = thickness of material (cm)

= linear attenuation coefficient (cm-1)

= density (g/cm3)

/ = mass attenuation coefficient (cm2/g)

(x)-

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Attenuation and Buildup

Io IxBIo I

where B1

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I = Io B e(-x)

primary photons + scattered photonsprimary photonsB =

If there are no scattered photons, then B = 1

If there are scattered photons, then B > 1

Buildup

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What amount of lead shielding is needed to reduce the dose rate beyond the shield from 1 mSv/hr to 0.02 mSv/hr for a 1 MeV photon beam?

Sample Buildup

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Sample Buildup

PhotonEnergy Material

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Sample Buildup

The mass attenuation coefficient (/) for a1 MeV photon incident on a lead shield is:

(/) = 0.0708 cm2/g

The density of lead = 11.35 g/cm3

The linear attenuation coefficient is:

(/) x = 0.0708 cm2/g x 11.35 g/cm3 = 0.804 cm-1

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I = Io B e(-x)

I = 1 mSv/hr

Io = 0.02 mSv/hr

B = 1 (assumed)

= 0.804 cm-1

Solve for “x”

0.02 mSv/hr = (1 mSv/hr) (1) e(-x)

Sample Buildup

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Sample Buildup

ln(0.02) = ln[e(-x)]-3.91 = -x

Although it is not required:x = -3.91/-0.804 = 4.86 cmThis would be the calculated value of the lead shield if scatter was not considered.

0.02 mSv/hr

1 mSv/hr= e(-x)

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Sample Buildup

x = 3.91

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Sample Buildup

x = 3.91 Let’s interpolate:when x = 4, B = 2.26when x = 2, B = 1.69

1.27 = 0.09/(2.26 - x)(2.26 - x) = 0.09/1.27 = 0.0712.26 – 0.071 = x = 2.19

(4-2)

(2.26-1.69)

(4-3.91)(2.26 - x)

=

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Solve for x using B = 2.2

I = I0 B e(-x)

0.02 mSv/hr = (1 mSv/hr) (2.2) e(-x)

ln(0.02/2.2) = ln[e(-x)]

-4.71 = - x

so x = -4.71/-0.804 = 5.86 cm

The thickness of the shield increased from 4.86 cm to 5.86 cm (20%) due to buildup

Sample Buildup

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Skyshine

Photons can scatter or “bounce” off atoms in materials such as the ceiling of a room or even air molecules !

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Where to Get More Information

Cember, H., Johnson, T. E, Introduction to Health Physics, 4th Edition, McGraw-Hill, New York (2009)

International Atomic Energy Agency, Postgraduate Educational Course in Radiation Protection and the Safety of Radiation Sources (PGEC), Training Course Series 18, IAEA, Vienna (2002)