Answer to Question #4418 Submitted to "Ask the Experts"

I would like to know if there is a calibration chart between attenuation of radiation shielding thickness in lead and radiation energy. Also, how will this chart change if I used as attenuation material tungsten heavy metals and tungsten rubber.

There are a variety of sources scattered throughout the literature that provide information regarding photon attenuation in various materials. Most such information that provides transmission as a function of material thickness tends to be for specific energies or specific radionuclides and for particular irradiation geometries. One example of such a source is the Radiological Health Handbook (the edition I am referencing here is the 1970 edition published by the U.S. Department of Health, Education, and Welfare Public Health Service, Bureau of Radiological Health, Rockville, MD 20852). The transmission curves shown in the cited reference, pp 148-149, are for selected radionuclides, ²²⁶Ra, ⁶⁰Co, ¹³⁷Cs, ¹⁹⁸Au, ¹⁹²Ir, ¹⁸²Ta, and ²⁴Na in lead and for fewer radionuclides in iron and in concrete.

I am not aware of extensive compilations in graphical or tabular form that show transmission as a function of thickness for a wide range of energies for sources/radiation fields of various geometries. There are, however, extensive compilations of mass attenuation coefficients for many materials and a wide range of photon energies. National Institute of Standards and Technology (NIST) makes such compilations available on its Website. If you go to this site and select Table 3 from the Table of Contents, you will be directed to a table of elements from which you can individually select those of interest and find the mass attenuation coefficients as a function of energy. The mass attenuation coefficients, usually symbolized (μ/ρ, can be used in a simple evaluation to compare the transmission of the primary photons of specific energies in the materials of interest—i.e., T = e^{-(μ/ρ)(ρ)(x)}, where ρ is the material density and x is the linear thickness of the material. For example, the NIST data show the μ/ρ value for 1 MeV gamma rays in tungsten is 0.06618 cm²g^-1, and the density of tungsten is 19.3 g cm^-3. The transmission of 1 MeV gamma rays through 1 cm of tungsten would be given by

T = e^{-(0.06618 cm2/g)(19.3 g/cm3)(1 cm)} = 0.279.

For your information, I have performed similar calculations for lead and tungsten, using the NIST mass attenuation coefficients, for gamma ray energies of 0.1, 0.5, 1, 2, and 5 MeV and for thicknesses of 0.1 cm and 1.0 cm of each of the materials, individually; results are as follows:

Keep in mind that these results are for the primary photons and would apply to a case where a parallel beam of photons of the specified energy was incident normally on a fixed thickness of material; they do not include any possible effect of radiation buildup in the shielding material, associated with secondary photons from Compton scatter and other processes such as pair production that produces annihilation photons. The degree of buildup depends on the photon energy and the source geometry as well as the shield composition, shape and thickness and the distance of the point of interest from the shield.

In order to evaluate the effectiveness of the tungsten-rubber material you mention, you must know the mass fractions of the major constituents. The effective value of the mass attenuation coefficient may then be estimated from a mass-fraction weighted summation as

μ/ρ = Σ wi (μ/ρ)i,

where wi is the weight fraction of the i^th constituent in the material, and (μ/ρ)i is the mass attenuation coefficient of the same constituent. I hope this is useful to you.

George Chabot, PhD, CHP