Answer to Question #7150 Submitted to "Ask the Experts"

I had been calibrating alpha meters with a ²³⁹Pu standard source. That was replaced with a ²³⁸Pu standard source. Now the meters read high for equivalent values. How do I convert a ²³⁹Pu meter reading to the ²³⁸Pu value equivalence?

While the two plutonium isotopes decay essentially 100% of the time by alpha emission, the alpha energies are slightly different, the average alpha energy from the ²³⁸Pu being 5.49 MeV and that from ²³⁹Pu being 5.15 MeV. Assuming that the two sources are pretty much identical in physical characteristics other than the specific isotope, we would expect that the higher-energy ²³⁸Pu source would produce somewhat greater alpha count rates than would the ²³⁹Pu source when the sources are individually measured with the same detector under the same geometry conditions. Because most alpha detectors have a thin window or a layer of attenuating material over the active volume of the detector, there will be alpha particle attenuation in these components, and when a source is placed close to the detector window, alpha particles traveling at other than normal incidence to the window will traverse an increased pathlength through the window, and some alpha particles may be stopped in the window or lose sufficient energy that they are not able to produce a pulse greater than the threshold for detection. Such effects are greater for lower energy sources than for higher energy sources. Additionally, even when the two sources of interest are prepared in as similar a fashion as possible, there may be measurable differences in the degradation of alpha particle energies that result from self-absorption in the source and any possible source covering.

To this point we have probably not said anything new to you, but have simply stated some of the factors that produce the effect you observed. The question does arise as to why you have decided to replace the ²³⁹Pu source with a ²³⁸Pu source. What alpha emitter(s) are you actually interested in measuring in the field? You should select the calibration source that is closest in energy to what you are concerned with in the field. If you are concerned with multiple alpha emitters in the field, and you do not know which of these you are measuring at any given time or location, you should err on the conservative side and use an alpha emitter for calibration that has an alpha energy close to that of the lowest-energy alpha emitter you expect in the field.

While there are theoretical methods to predict what the alpha counting efficiency will be for a given source and detector in a given configuration and to relate efficiencies for the two sources you refer to, such calculations require detailed knowledge of the source and detector characteristics, and I don't believe they are necessary for your case. If you have data for both sources evaluated with a given detector, you can use the data you have obtained for the two sources for that detector and simply use the ratio of ²³⁹Pu to ²³⁸Pu efficiencies as a multiplier to correct future ²³⁸Pu counts to get equivalent ²³⁹Pu counts for the same instrument. The reciprocal ratio would apply if you wish to convert ²³⁹Pu counts to equivalent ²³⁸Pu counts. This approach will not necessarily work if you are trying to assess the counting efficiency for ²³⁹Pu using the ²³⁸Pu source for a new instrument that has not earlier been evaluated using ²³⁹Pu. As noted above, if you are interested in a specific radionuclide in the field, you should use the alpha emitter for calibration that has an energy closest to the nuclide of interest or, conservatively, use a radionuclide that has an alpha energy close to the lowest-energy alpha emitter of field concern.

If you feel that you do want to pursue theoretical methods to compare efficiencies, you might want to review a paper by Skrable et al. ("Theoretical Response of a ZnS (Ag) Scintillation Detector to Alpha-emitting Sources and Suggested Applications" in Health Physics 60(3):381–392; March 1991. Skrable KW, Phoenix KA, Chabot GE, French CS, Jo M, Falo GA). It provides some insight into such calculations as applied in the case of a ZnS(Ag) detector.

Good luck.

George Chabot, PhD, CHP