by Michael V. McGloin
In field radiography, shielding is a very important part of what a radiographer must be concerned with. Shielding is considered to be any material that reduces the primary radiation intensity to less than what it would be without it—for example, a collimator, truck, wall, trash can, or any other material for that matter providing that there is enough of it to reduce the radiation intensity. Shielding is often expressed in “half-value layers” (HVLs), which is defined as the thickness of shielding material required to reduce the radiation intensity to one-half of its original value. In practice, radiographers should know how effective some existing shielding is before setting up an exposure.
Every radiographic source and shielding material combination has a unique HVL constant. For the purpose of this blog post, iridium-192 for the radiographic source and concrete for the shielding will be used. With this combination, the HVL of concrete is 1.7 in. (43 mm).
If an even number of HVLs were present—for example two HVLs, 3.4 in. (86 mm)—then the original intensity would simply be divided by the number 2 twice, once for each HVL. However, if there were 12.5 in. (317.5 mm) of concrete, the remaining intensity would have to be calculated. Here is one way to calculate that intensity.
In order to determine how many HVLs are present, the concrete thickness of 12.5 in. (317.5 mm) would be divided by the HVL for concrete (1.7 in. [43 mm]). Therefore, 12.5 in. (317.5 mm) of concrete would contain 7.35 HVLs. With an even 7 HVLs, the initial intensity would be divided by the number 2, seven total times. However, the 0.35 HVLs also needs to be applied.
A mathematical technique to account for multiple, or fractional, divisions by 2 is to use an exponent, such as 2 to the third power (23). This is the same as saying 2 x 2 x 2. On most calculators, there is either a ^ key or XY key. In this scenario, the exponent would be 7.35 or 27.35.
If an initial intensity was calculated to be 145 mR/h on one side of the concrete, then in order to calculate the intensity after passing through the concrete, 145 mR/h would be divided by 27.35. When entering this formula into the calculator (145 ÷ 2^7.35 =) the answer would be 0.89 mR/h.
This Tech Tip has been excerpted from the new book, ASNT Study Guide: Industrial Radiography Radiation Safety. This second edition is designed to help readers prepare for the ASNT Industrial Radiography Radiation Safety Personnel (IRRSP) exam. This edition is a revision of the 2009 first edition and was revised for ASNT by Michael V. McGloin. The content has been updated to reflect current technology and industry regulations and practices. The book includes new and revised chapters, and each chapter includes review questions. To purchase a copy of the book, go to https://ebooks.asnt.org/1pekeqm/.
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Michael V. McGloin, NDT Enterprises, LLC, www.NDTEnterprises.com.
Figure b provided by MarShield Radiation Protection Products.