Answer to Question #7922 Submitted to "Ask the Experts"

Category: Environmental and Background Radiation — Building and Construction Material

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


I have used a liquid scintillation counter to measure the activity of a 1 cm3 piece of some sort of "volcanic" tile. The reading was about 200 cpm while the reference vial read 10 cpm. Using a G-M detector on the tile, the reading was 150 cpm and using a compensated G-M detector, the reading was about 40 µR h-1. I would like to know how safe it would be to use this type of tile to cover the floor in all the rooms of a house.


Given that there has been considerable interest among members of the public regarding the potential for radiation exposure associated with the use of natural materials such as granite in their homes, this is a relevant and good question. I do not know what technical background you have, but since you have access to and presumably know how to use the instruments you describe, I shall assume that you have some technical training so that you can reasonably follow the ensuing discussion.

While the information you provide is informative, it is not of sufficient detail to lead to an absolute recommendation as to the use of the tile. I shall, however, attempt to do a representative estimation of the possible significance of any excess radiation exposure that might ensue from use of the tile, and you may be able to pursue a similar estimation with corrected data as necessary. In order to do so I shall have to make some assumptions regarding your measurements; among the most significant are the following:

  1. The measurements you made were with the detector in contact or in very close proximity to the surface of the tile; when the detector is in contact with the tile surface, the effective distance of the detector volume depends on the detector dimensions. In the estimation that follows I assume an effective distance of the detector from the tile surface of 1 cm when the measurement was made.
  2. I assume the tile you measured had dimensions of 12 in x 12 in; for purposes of this estimation I have converted the 144 in2 area to an equivalent area disc with a radius of 17.2 cm.  
  3. The energy-compensated G-M detector you used for the exposure rate measurement had sufficient wall thickness to ensure that no beta radiation was contributing to the response (this is important because the presence of 238U leads to 234mPa, which emits relatively high-energy beta particles, some of which can penetrate somewhat more than 1 cm of Lucite plastic or about 2 mm of steel.) You can check whether beta radiation is a significant contributor to the detector reading by placing a sufficiently thick absorber, such as about 1 cm of Lucite, between the tile surface and the detector.
  4. For purposes of the calculation that follows, I assume that gamma radiation contributing to the reading is coming from the surface of the tile; this may be true or not depending on whether the tile matrix or glazing on the tile is primarily responsible for the readings obtained.
  5. I assume a background exposure rate of 10 µR h-1; this will be subtracted from your reading to yield a net reading of 30 µR h-1 associated with the tile.
  6. I assume no significant attenuation of gamma radiation between the tile source and the measurement point.

As noted above, we are treating the square tile as a disc. It is easy to show that the gamma exposure rate at a point centered above a disc source is proportional to the quantity ln[(R2 + H2)/H2], where R is the disc radius, H is the height of the measurement point above the disc, and ln refers to the natural logarithm. We are also assuming that the measured 40 µR h-1 is due only to gamma radiation, radius R is equal to 17.2 cm, and height H is equal to 1 cm. Calculation then yields:

ln[(R2 + H2)/H2] = ln[(17.22 + 12)/12] = 5.69.

We may now consider the exposure rate from a larger-area source as might be associated with the entire floor of one room; let us assume a room 12 feet by 12 feet, for which the radius of an equivalent-area disc would be 207 cm. Calculating the same quantity as above for this new radius would then yield:

ln[(R2 + H2)/H2] = ln[(2072 + 12)/12] = 10.7,

which implies that the exposure rate at 1 cm above the floor would be (10.7/5.69)30 + 10 µR h-1 = 66 µR h-1, which is roughly seven times what a typical background level might be. A distance of 1 cm above the floor may be somewhat reasonable to use if one is concerned about potential dose to an infant who spends a lot of time crawling or lying on the floor, but it probably does not relay a very realistic estimate of impact to an adult, the major part of whose body may spend little time in such proximity to the floor surface. A distance of perhaps 1 m above the floor might be more sensible for the adult. At a distance of 1 m above the floor, the similarly calculated exposure rate would be about 19 µR h-1, about twice the usual background rate.

While the excess exposure rates estimated here are not sufficiently large to produce any biological impact that could be ascribed to the radiation, if such numbers were realistic I would not personally use the tiles since they would be producing a dose impact that is roughly 2 to 10 times the expected background value, and this simply represents unnecessary exposure that can readily be averted or reduced by using a different tile.

Assuming that a significant part of the measured exposure rate is associated with gamma radiation from the decay progeny of 238U, there would also be some radon gas generated, and this could add somewhat to the inhalation intake of radon and its progeny. There is a calculation done in a document on the Health Physics Society website that estimates potential radon concentration from granite countertops (see Radiation from Granite Countertops). If we made similar assumptions here, taking the measured value of 40 µR h-1 as true (with 10 µR h-1 background), the exposure rate at 1 cm from the floor tile in a 12-foot by 12-foot room would translate to an airborne radon concentration of about 0.7 pCi L-1, which is about 18 percent of the 4 pCi L-1 value that the Environmental Protection Agency uses as an action level. This number is not high, but it would add to whatever radon levels are already present in the home and is an exposure that is avoidable.

Please keep in mind that the estimations done here are representative but not necessarily realistic, since they depend on a number of assumptions made in the absence of more precise information. I hope you are able to take this example evaluation and modify it as might be appropriate to more closely simulate your actual situation.

I hope this is helpful.

George Chabot, PhD, CHP

Answer posted on 19 January 2009. 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.