4/2003 Rev 2 I.3.6 – slide 1 of 23 Session I.3.6 Part I Review of Fundamentals Module 3Interaction of Radiation with Matter Session 6Buildup and Shielding IAEA Post Graduate Educational Course Radiation Protection and Safety of Radiation Sources

4/2003 Rev 2 I.3.6 – slide 2 of 23  In this session we will discuss shielding for photon beams  We will also discuss the increase in photon transmission through a shield resulting from buildup Overview

4/2003 Rev 2 I.3.6 – slide 3 of 23 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.

4/2003 Rev 2 I.3.6 – slide 4 of 23 HVL and TVL HVL (cm) TVL (cm) Isotope Photon E (MeV) ConcreteSteelLeadConcreteSteelLead 137 Cs Co 1.17, Au Ir 0.13 to Ra to

4/2003 Rev 2 I.3.6 – slide 5 of 23 HVL and TVL The half value layer (HVL) and tenth value layer (TVL) are mathematically related as follows: HVL = ln(2) ln(2) TVL = ln(10) ln(10) = TVL HVL  ln(2) ln(2)  = ln(10) ln(10)ln(2) = = TVL = x HVL and

4/2003 Rev 2 I.3.6 – slide 6 of 23 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

4/2003 Rev 2 I.3.6 – slide 7 of 23 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 60 Co 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.

4/2003 Rev 2 I.3.6 – slide 8 of 23 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 (to be discussed in Session II.2.5).  However, if one is far enough away from a finite line, plane or volume, they appear to be a point and the ISL can be applied, with an acceptable error band.  For the remaining discussion, a point source will be assumed. The ISL predicts that the intensity of the radiation will decrease as the distance from the source increases even without any shielding.

4/2003 Rev 2 I.3.6 – slide 9 of 23 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

4/2003 Rev 2 I.3.6 – slide 10 of 23 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)

4/2003 Rev 2 I.3.6 – slide 11 of 23 Photon Attenuation I x = I o e -  x = I o e where: I x = photon intensity after traversing x cm of some material I o = initial or incident photon intensity x = thickness of material (cm)  = linear attenuation coefficient (cm -1 )  = density (g/cm 3 )  /  = mass attenuation coefficient (cm 2 /g)  (  x) -

4/2003 Rev 2 I.3.6 – slide 12 of 23 Attenuation and Buildup IoIoIoIo IxB IoIoIoIo I where B  1

4/2003 Rev 2 I.3.6 – slide 13 of 23 I = I o B e (-  x) primary photons + scattered photons primary photons B = If there are no scattered photons, then B = 1 If there are scattered photons, then B > 1 Buildup

4/2003 Rev 2 I.3.6 – slide 14 of 23 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

4/2003 Rev 2 I.3.6 – slide 15 of 23 Sample Buildup PhotonEnergy Material

4/2003 Rev 2 I.3.6 – slide 16 of 23 Sample Buildup The mass attenuation coefficient (  /  ) for a 1 MeV photon incident on a lead shield is: (  /  ) = cm 2 /g The density of lead = g/cm 3 The linear attenuation coefficient is: (  /  ) x  = cm 2 /g x g/cm 3 = cm -1

4/2003 Rev 2 I.3.6 – slide 17 of 23 I = I o B e (-  x) I = 1 mSv/hr I o = 0.02 mSv/hr B = 1 (assumed)  = cm -1 Solve for “  x” 0.02 mSv/hr = (1 mSv/hr) (1) e (-  x) Sample Buildup

4/2003 Rev 2 I.3.6 – slide 18 of 23 Sample Buildup ln(0.02) = ln[e (-  x) ] = -  x Although it is not required: x = -3.91/ = 4.86 cm This would be the calculated value of the lead shield if scatter was not considered mSv/hr 1 mSv/hr = e (-  x)

4/2003 Rev 2 I.3.6 – slide 19 of 23 Sample Buildup  x = 3.91

4/2003 Rev 2 I.3.6 – slide 20 of 23 Sample Buildup  x = 3.91Let’s interpolate: when  x = 4, B = 2.26 when  x = 2, B = = 0.09/( x) ( x) = 0.09/1.27 = – = x = 2.19 (4 - 2) ( ) ( ) ( x) =

4/2003 Rev 2 I.3.6 – slide 21 of 23 Solve for  x using B = 2.2 I = I 0 B e (-  x) 0.02 mSv/hr = (1 mSv/hr) (2.2) e (-  x) ln(0.02/2.2) = ln[e (-  x) ] = -  x so x = / = 5.86 cm The thickness of the shield increased from 4.86 cm to 5.86 cm (20%) due to buildup Sample Buildup

4/2003 Rev 2 I.3.6 – slide 22 of 23 Skyshine Photons can scatter or “bounce” off atoms in materials such as the ceiling of a room or even air molecules !

4/2003 Rev 2 I.3.6 – slide 23 of 23 Where to Get More Information  Cember, H., Johnson, T. E., Introduction to Health Physics, 4th Edition, McGraw-Hill, New York (2008)  Martin, A., Harbison, S. A., Beach, K., Cole, P., An Introduction to Radiation Protection, 6 th Edition, Hodder Arnold, London (2012)  Attix, F. H., Introduction to Radiological Physics and Radiation Dosimetry, Wiley and Sons, Chichester (1986)  Firestone, R.B., Baglin, C.M., Frank-Chu, S.Y., Eds., Table of Isotopes (8 th Edition, 1999 update), Wiley, New York (1999)