Answer to Question #14205 Submitted to "Ask the Experts"

I have a tiny, high activity sample, weighing in micrograms (µgm) to be subjected to gamma spectrometry. I am encountering a detector dead time of about 50% when counting the sample using HPGe at a distance of 50 cm from the detector. Increasing the distance further might help but the efficiency calibration is prone to errors when I perform the calibration with sources, which are low activity. The same would be the case, if I introduce shielding between the detector and sample. Kindly advise me as to how to overcome the limitations.

Unfortunately, I cannot provide a decisive answer to your question; there are limited options available, but I will attempt to offer some guidance. I shall assume that obtaining higher activity calibration sources is not an option available to you.

The way the detector response varies with distance in air from source to the detector depends on a number of factors, including the geometries of the detector and of the source, the energies of the photons, and the level of accuracy required in the measurement. If the energies of the photons are relatively high, the attenuation of photons in the air may be negligible, or at least acceptable. For example, at 1 megaelectron volt (MeV), 1 meter of air will attenuate the primary photons by 0.76%, based on simple exponential attenuation.

If both the high and low activity sources are small in dimensions compared to both the distance from the detector and to the detector dimensions, and if the air attenuation is acceptably low, you may be able to take advantage of an expected inverse square dependence of the primary photon intensity with distance between source and detector. This discussion assumes that you are doing photopeak counting for the primary photons. The recommendations I am making will likely not apply if you are doing gross gamma counting, most notably because the influence of scattered radiation changes with distance between source and detector.

If you take a measurement with the low activity source at the maximum distance from the detector at which you can obtain statistically acceptable counts in the photopeak region of interest (ROI), and then take a few sequential measurements, decreasing the spacing by perhaps 10% each time and comparing results with expected values, you can judge whether the inverse square law holds—at least for the two greatest distances (because typical HPGE detectors are rather large, it is probably preferable to make the distance measurement from the center of the detector to the center of the source, especially when dealing with high energy photons). As distances between detector and source become shorter, you will likely see greater deviations from inverse square expectations. This is because at greater distances most photons are incident on the detector at approximately 90^o and traverse the same potential pathlength in the detector, but as the distance shortens, more photons are incident at various angles different from 90^o, resulting in variations in potential pathlengths.

As an example, to demonstrate inverse square dependence, if a 10-minute count yielded 1,000 + 32 net counts (one sigma uncertainty) in the photopeak ROI at 50 cm distance, and if the inverse square law held, the expected ROI net counts in 10 minutes at 45 cm would be 1,235 + 35 (i.e., (50/45)²(1,000)). You may have to count for longer times to ensure adequate counts. If the inverse square law continued to hold at 40 cm, a similar 10-minute count would yield an expected 1562 + 40 net counts. If the inverse square law appears to be applicable between the last two distances, it is likely safe to assume that it would also apply beyond this distance for both the standard and the high activity sample that you have. Thus, you could calculate the expected net ROI counts for the calibration standard at the distance to be used for the sample measurement. From the above, I would judge that you might have to get to perhaps 150 cm distance between sample and detector to reduce the dead time to an acceptable level. At 150 cm, the expected net ROI count rate would be nine times greater than that at 50 cm. Assuming the photon energies are high, the attenuation in air will not be great, but you could correct for it, if desired. If you had 1 MeV photons, the linear attenuation coefficient for air at 20^o C would be 7.63 x 10^-5 cm^-1 and the transmission factor for 1 MeV photons through 150 cm air would be exp((-7.63 x 10^-5)(150)) = 0.9886. Dividing the net ROI counts by this value would yield the corrected count. You can find values for the mass energy absorption coefficients for gamma radiation in air at this NIST link; multiply the mass attenuation coefficient by the mass density of air to get the linear attenuation coefficient. I used an air density at 20^o C of 0.0012 g cm^-3. I hope this is helpful to you.

George Chabot, PhD, CHP