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The following question was answered by an expert in the appropriate field:

Q

I will have 15 Troxler gauges; every gauge has two sealed sources, 137Cs and 241Am:Be. I want to make one storage room for those sources in Saudi Arabia. Could you please advise me on the design for a storage room with calculations. I want <0.83 nSv s-1 at the controlled area (a fence covers the area).

A

You do not specify the source strengths or the current dose rates for the devices so I cannot provide an exact solution, but I will attempt to demonstrate the types of calculations that might be helpful.

If you are using moisture/density gauges similar to the popular Troxler Model 3430, the typical source strengths are about 300 MBq of 137Cs and 1.5 x 103 MBq of 241Am (mixed with beryllium to provide neutrons). The sources for each device are contained in a partly shielded configuration, and the manufacturer specifications show an anticipated combined neutron plus gamma dose rate of 53 nSv s-1 at 5 cm from the device surface. Since the minimum dimension for the Model 3430 is given as 22.9 cm, if we assume the minimum distance from the source to the surface of the device is one half this distance, 11.5 cm, we can assume the distance from the source to the point where the dose rate of 53 nSv s-1 prevails is 11.5 cm + 5 cm = 16.5 cm. For 15 separate gauges all stored close together, the maximum expected dose rate at 16.5 cm (measured from the source closest to the dose point) would expectedly not exceed (15)( 53 nSv s-1) =795 nSv s-1. If we considered reduction in field intensity with distance alone, and we assumed that the dose rate would decrease inversely with the square of the distance from the source, we could estimate the distance, D, at which the dose rate would be 0.83 nSv s-1 as follows:

0.83 nSv s-1 = (795 nSv s-1)(16.5 cm/D)2,

and solving for D we obtain D = 510 cm, which implies that at about five meters from any source the dose rate would be acceptable. This is an approximation since (1) the distance of 16.5 cm was estimated, and the actual distance would depend on the exact position of the source within the gauge device, (2) the dose rate would likely be somewhat reduced because of shielding of radiation from one device by other devices present (an effect that would decrease the dose rate), and (3) because there are 15 devices that might be arranged in a variety of possible geometries within the room, the dose rate may not decrease with the inverse square of the distance but may decrease more slowly because of the distributed nature of the multiple sources; if the distance of five meters is measured from the source closest to the dose point, the actual dose rate will be less than that used (i.e., 0.83 nSv s-1) because the sources in the other 14 devices will be farther from the dose point.

Another approach is to begin by considering the sources in their unshielded configurations. If we look first at the 241Am:Be source and consider only the neutron radiation as significant, we can estimate the expected unshielded neutron dose rate. The gamma dose rate from the 60 keV 241Am gamma rays is insignificant because of the appreciable attenuation in the source encapsulation and their low dose consequences compared to the 137Cs.

The neutron yield from a typical 241Am:Be source is about 4.7 x 101 neutrons per second. For the 1.5 x 103 MBq source in a Model 3430, the expected neutron emission rate would then be about 7.1 x 104 s-1. If we assume isotropic neutron emission, the expected unattenuated neutron fluence rate at a distance of interest may be determined. For the 16.5 cm distance that we calculated above, the unshielded neutron fluence rate, f, would then be about

F = 7.1 x 104 s-1/4p(16.5 cm)2 = 2.1 x 101 cm-2 s-1.

For the fast neutrons produced by this source, the equivalent dose rate per unit fluence rate is about 0.83 nSv s-1/cm-2-s-1 (assuming a radiation weighting factor of 20 for the neutrons). Using this conversion factor, we determine that the unshielded fluence rate of 2.1 x 101 cm-2 s-1 would yield an equivalent dose rate of 6.3 x 101 µSv h-1.

The unattenuated gamma dose rate from the 300 MBq 137Cs source at the same distance may be estimated, using the gamma ray equivalent dose rate constant for 137Cs of 2.83 x 10-5µSv s-1/MBq at one meter:

H = (300 MBq)(2.83 x 10-5 µSv s-1/MBq)/0.165 m)2= 0.31 µSv s-1.

Clearly the 137Cs unshielded dose rate is much greater than the calculated neutron dose rate at the same distance. The estimated total unattenuated dose rate is then 0.33 µSv s-1. This compares to the manufacturer’s quoted dose rate of 0.053 µSv s-1, about 6.2 times lower than our calculated value. The difference is presumably due to attenuating material that is built into the device (the manufacturer cites the use of tungsten, lead, and cadmium) and possible differences in our calculated 16.5 cm distance from the actual distance from the source to the dose point.

The manufacturer does not specify what the thicknesses of specific shielding materials are, so it is not possible to calculate the extents of gamma and neutron dose attenuation. The tungsten, lead, and cadmium are not very effective attenuators for the fast neutrons from the 241Am:Be source; the gauges would require significant hydrogenous material to produce effective neutron attenuation. The gamma rays would be well attenuated by the tungsten, lead, and cadmium, but the extent depends on the specific thicknesses.

In any case, we may do reasonable calculations to determine what room dimensions and/or shielding might be appropriate. For example, if no additional attenuation were used, the maximum unattenuated neutron dose rate at five meters from the combined fifteen sources would be Hn(500):

Hn(500) = [(15)(7.1 x 104 s-1)/4π(500 cm)2](0.83 nSv s-1/ cm-2 s-1) = 0.28 nSv s-1.

Since we do not know how much fast neutron attenuation occurred in the gauge device, we might accept this value as true. If this is the case, we could add some effective neutron-attenuating material around the sources or in the walls of the room (e.g., polyethylene for localized shielding or concrete in the walls if the specified area had walls). We could also use lead or other reasonably dense material to provide shielding against the gamma radiation.

If we assume that the factor of 6.2 that we estimated above is an effective attenuation factor for the gamma dose rate from the 137Cs, we could estimate the expected maximum dose rate at five meters from the combined 15 sources as Hg(500):

Hg(500) = (15)(300 MBq)(2.83 x 10-5 µSv s-1/MBq)/ ((5 m)2(6.2)) = 8.22 x 10-4 µSv s-1.

If only increased distance from the sources is to be used for dose-rate reduction, it is easy to show, using the same rationale as above that the required distance from the sources would be about 5.7 m to obtain a total dose rate of 0.83 nSv s-1.

If it is desired to reduce the storage-area dimensions you may have to install some shielding material to maintain an acceptable dose rate. For example, if you require that the source to dose point distance not exceed 2.0 meters, the neutron dose rate, adjusted from 5 m to 2 m, would be (2.8 x 10-4 µSv s-1)(5 m/2 m)2 = 1.75 x 10-3 µSv s1. The expected gamma dose rate at the same 2 m distance would be (8.2 x 10-4 µSv s-1)(5 m/2 m)2 = 5.1 x 10-3 µSv s-1.

For estimative purposes, if polyethylene is used as a shielding material, we may assume an effective neutron attenuation coefficient, S, of about 0.1 cm-1. The value of µ, the linear attenuation coefficient of the shielding material for gamma radiation, can be obtained from various sources. A good one is NIST in the United States, which lists mass attenuation coefficients for elements and compounds or mixtures (use Table 3 for elemental data). The mass attenuation coefficient may be multiplied by the mass density of the material to obtain the linear attenuation coefficient. For example, if lead (mass density = 11.35 g cm-3) were to be used for shielding, the mass attenuation coefficient at 0.662 MeV (137Cs photon energy) is about 0.114 cm2 g-1, which yields a linear attenuation coefficient of 1.29 cm-1. If such attenuation materials were provided locally around the sources, such that the distance to the walls or fence from the added material was considerably greater than the lateral dimensions of the added shielding, the thickness, T, of required shielding could be estimated from a simple exponential relationship:

0.83 nSv s-1 = (1.7 x 10-3 µSv s-1)(e-STn) + (5.1 x 10-3 µSv s-1) e-µTg,

where Tn and Tg refer to the respective thicknesses of the neutron shield material (polyethylene) and the gamma shield material (lead). Since Tn and Tg are independent of each other, we can vary the respective thicknesses as fits our desires. For example, we might decide to choose thicknesses that yield equal dose rates of 4.7 x 104 µSv s-1 from neutrons and gamma rays. In that case we would have

4.17 x 10-4 µSv s-1 = 1.7 x 10-3 µSv s-1e-0.1Tn, and 4.17 x 10-4 µSv s-1 = 5.1 x 10-3 µSv s-1 e-1.29Tg.

Solving for Tn, we obtain Tn = 14.3 cm of polyethylene. Similarly solving for Tg we obtain Tg = 1.94 cm. These thicknesses take no account of scattered radiation that might reach the dose point, and some additional thicknesses could be required, depending on added shielding dimensions and source shield geometry. In general, it is desirable to have the polyethylene closer to the source with the lead on the outside so that the lead can reduce the capture gamma radiation that can result when neutrons are slowed and captured in the polyethylene.

In summary, you have the options of increasing storage-area dimensions or using added shielding, or both, to yield an acceptable dose rate outside the storage area. The above calculations are not intended as exact, but are provided to give you some guidance as you proceed. If you have access to neutron and gamma dose-measuring instruments, you can get firsthand dose information that will allow you to evaluate possible designs that you might pursue. Good luck in your efforts.

George Chabot, PhD, CHP

Answer posted on 22 July 2011. The information posted on this web page is intended as general reference information only. Specific facts and circumstances may affect the applicability of concepts, materials, and information described herein. The information provided is not a substitute for professional advice and should not be relied upon in the absence of such professional advice. To the best of our knowledge, answers are correct at the time they are posted. Be advised that over time, requirements could change, new data could be made available, and Internet links could change, affecting the correctness of the answers. Answers are the professional opinions of the expert responding to each question; they do not necessarily represent the position of the Health Physics Society.