Answer to Question #14101 Submitted to "Ask the Experts"

Category: Radiation Fundamentals — Radiation Shielding

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Q

Neutron shielding: What formula would I use to determine the thickness of borated polyethylene to shield a americium-241-beryllium (241Am-Be) Hydroprobe and case? I am designing a box to store the unit. The highest neutron dose rate is about 1.03 × 10-6 C kg-1 h-1 (4 mR h-1) on the outside of the case. 

A

While I cannot provide a necessarily exact answer to your question, I believe I can offer sufficient information for your purposes. You can make a reasonable estimate of the required thickness of borated polyethylene by using a simple exponential attenuation expression of the form

H = Hoe-Σx,

where H is the desired equivalent dose rate at the point of interest, Ho is the unattenuated dose rate at the point of interest, Σ is the linear attenuation coefficient for the borated polyethylene, and x is the thickness of the polyethylene. The inverse square law may be used to change unattenuated fluence rates in accounting for changes in source-to-dose point distances.

There have been some measurements made to assess the effectiveness of borated polyethylene in shielding neutrons. One study that I shall refer to is described in an available paper by Subramanian et al. 2018.The authors present data for the attenuation of epithermal and fast neutrons made available at reactor facilities. The neutrons produced by Am-Be sources are fast neutrons, having an average energy of about 4.5 MeV. The reactor fast neutrons may be a bit lower in average energy, but I expect results will be applicable. I used the authors’ fast neutron attenuation data to estimate the linear attenuation coefficients for borated polyethylene and unborated polyethylene. The value for polyethylene containing anywhere from 5% to 30% boron by weight (results did not change significantly in this boron range), was about 0.138 cm-1. The value for plain polyethylene was 0.115 cm-1.

You do not say how large the Am-Be source is, but I assume it is about 1.85 × 109 Bq (50 mCi), typical for this type of moisture gauge. To demonstrate the use of the above simple equation, I shall do two calculations, the first assuming that the source was removed from its case for storage in the shield enclosure, and the second assuming the source is kept in its case, as you indicate it would be, and the case placed in the shielded enclosure. For the first case, I shall assume that the neutron emission rate from the unattenuated source is 105 n s-1, in the ballpark for a 1.85 × 109 Bq source. I shall also assume that the dose point outside of the container is 15.24 cm from the source (you do not specify this, and you can adjust the calculation accordingly). Assuming isotropic neutron emission, the unattenuated neutron fluence rate at this distance would be given by

Φ = 105 s-1 (4π(15.24 cm)2)-1 = 34.3 cm-2 s-1.

The assumed source would yield an unattenuated equivalent dose rate of about 0.05 mSv h-1 at a distance of 15.24 cm from the source in air (using current US NRC 10CFR20 Table 1004(b)2 neutron fluence rate to dose equivalent rate conversion factors and considering the particular source, a fluence rate of about 6.8 cm-2 s-1 would yield an equivalent dose rate of about 0.01 mSv h-1). You do not specify what dose rate you require outside of the shielded container. For example purposes, I shall specify an equivalent dose rate of 0.01 mSv-1. For a borated polyethylene shield thickness of T, we then can write

0.01 mSv h-1 = 0.05 mSv h-1 e-0.138T

Solving for T, we obtain T= 11.7 cm. Adding a couple of additional centimeters of the borated polyethylene would be reasonable to account for possible scattered neutrons that might escape from the container.

In the second calculation, we need to know at what distance from the source the value that you cite of 0.04 mSv h-1 applies. For convenience, and considering likely case dimensions, I shall assume that this value applies at 17 cm from the source center. To accommodate the borated polyethylene, we would have to add additional distance to reach the dose point. If we add 5 cm to get a distance of 22 cm, the inverse square law would reduce the expected dose rate of 0.04 mSv h-1 at 17 cm to (17/22)2(0.04 mSv h-1) = 0.024 mSv h-1, If we fill the 5 cm added distance with borated polyethylene, the expected dose rate would be

Hshield = 0.024 mSv h-1 e-0.138(5) = 0.012 mSv h-1.

You can adjust assumed values according to your actual needs. The calculations are meant to be demonstrative.

I hope this is helpful to you.

George Chabot, CHP, PhD

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