Answer to Question #14414 Submitted to "Ask the Experts"

Category: Radiation Fundamentals

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

Q

I have some questions regarding the level of protection offered from commonly used shielding materials for shelter from radiological disasters.

First, based on my initial research, it seems the thickness of any type of shielding required to attenuate gamma radiation by a given percentage depends on the energy of the gamma rays. What are the primary isotopes and their respective gamma energies contributing to the radiation risk following the detonation of a nuclear weapon?

Second, what is the general recommendation for the attenuation/protection factor (PF) necessary for adequate protection from the effects of a nuclear weapon? The most oft-cited figure seems to be a PF of 1,000 (or an attenuation to 1/1000th of ambient radiation levels outside of the sheltered area). However, attempting to find anything close to a definitive answer to this, along with my next question, is proving futile. Part of this may depend on what the intensity of the ambient radiation would be soon after a nuclear explosion, which this too, I am unsure of and am curious about.

Third, I would like to know the level of attenuation some common shielding types offer. I believe this is otherwise known as "halving thickness." Wood, earth, concrete, steel, and lead seem to be the most commonly studied and commercially available shielding types. As to the thickness of a given shielding to offer adequate protection, I have seen this described as "layers" of halving thickness. For example, 10 layers of halving thickness equates to shielding that attenuates ambient radiation to 1/1024th of its intensity (2 to the power of 10). Some calculations I've seen online express a requirement of 60 cm of concrete to achieve necessary protection if we assume 10 "layers" of concrete are needed, but this seems excessive and impractical at first glance. Looking at home fallout shelter plans from organizations such as Federal Emergency Management Agency (FEMA), one such plan they offer involves stacking solid concrete blocks 23 cm high across a basement ceiling, supported by additional joists, to shield the occupants from radioactive fallout. Clearly, there is a disparity in what is recommended across different sources.

A

I will try to answer them in order.

First, a shield's ability to attenuate gamma radiation depends on three factors: energy of the gamma rays, the density of the shield's material, and the thickness of the shield. The unsatisfying answer to your first question is that the radionuclides of concern depend on how much time has passed since the nuclear detonation. Immediately after detonation, hundreds of radionuclides are created. The majority of these isotopes are very radioactive and very short-lived; they decay away within a few seconds or a few days depending on their specific half-life. Over time as the short-lived isotopes decay away, cesium-137 (137Cs) which has a gamma energy of 662 kiloelectron volts {keV), starts to become the main contributor of gamma radiation exposure.

Regarding your second question, in 1962 the Office of Civil Defense published "Family Shelter Designs," which recommended a minimum protection factor of 100; however, more recent publications recommend a minimum protection factor of 1,000. Considerations such as likely bomb size, distance from the event, time since the event, and acceptable dose impacts may all affect decisions about the values of protection factors.

To answer your third question, in 2002 the US Army published Field Manual FM 3-05.70 which has half-value layers (HVL) for a variety of materials (Figure 23-1). I have listed a few below, converted to integer centimeters.

            Brick – 5 cm
            Concrete – 6 cm
            Dirt – 8 cm
            Wood – 22 cm.

In order to achieve a protection factor of 1,000, at least 10 HVLs must be used. The layers do not all have to be the same material. As you noted, 10 HVLS of concrete would be impractical. Instead, you could use 5 HVLS of concrete (30 cm) with 5 HVLs of dirt (40 cm) or any combination of materials.

I hope this answers your questions. Thanks for submitting your question to the Health Physics Society.

Anthony R. Davila, MS

Ask the Experts is posting answers using only SI (the International System of Units) in accordance with international practice. To convert these to traditional units we have prepared a conversion table. You can also view a diagram to help put the radiation information presented in this question and answer in perspective. Explanations of radiation terms can be found here.
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