Answer to Question #272 Submitted to "Ask the Experts"

Q

What is the relationship between 6 pCi/L wet to 6 pCi/g (dry) radium? How much water containing 6 pCi/L of radiation would it take to make 6 pCi/g (dry) radium? Would it be half that amount to get 3 pCi/g (dry) radium? Thanks.

A

The concentration of radium (pCi/g) in the residue from evaporation of a water sample would be the concentration of radium in the liquid sample (pCi/L) divided by the nonvolatile solids content of the liquid sample (g/L). Thus: Concentration in solid, pCi/g = (Concentration in liquid, pCi/L x Volume, L)/(Volume, L x solids content, g/L) = Concentration in liquid, pCi/L/solids content, g/L. Water with a radium concentration of 6 pCi/L would produce a residue with 6 pCi/g if the solids content was 1 g/L. (6 pCi/L)/(1g/L) = 6 pCi/g. Water with a radium concentration of 6 pCi/L would produce a residue with 3 pCi/g if the solids content was 2 g/L. (6 pCi/L)/(2g/L) = 3 pCi/g. In general, starting with a concentration in the liquid of 6 pCi/L, if the solids content is greater than 1 g/L, the concentration in the dry sample will be less than 6 pCi/g; if the solids content is less than 1 g/L, the concentration in the dry sample will be greater than 6 pCi/g. It doesn't matter what volume of water you use. If you desire a lower concentration of radium (pCi/g) in the solid residue than would result from the existing solids content of the water, this can be achieved by adding the appropriate amount of inert (nonradioactive) matter to the sample. C. E. Roessler, Ph.D., CHP