Answer to Question #11839 Submitted to "Ask the Experts"

Category: Instrumentation and Measurements — Surveys and Measurements (SM)

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

Q

The tritium (3H) activity from a bubbler can be estimated by adding the activity from vial 1 and vial 2, when each vial is counted in a liquid scintillation counter for five minutes (10 minutes in total). If the background counting time is only five minutes, can we use the following equation to estimate the lower limit of detection (LLD)?

LLD = 2.71 + 3.29 {Rb × Ts [1 + (Ts/Tb)]}½ ÷ (Ts × E × V)
Rb = background count rate in counts per minute (cpm)
Ts = sample counting time in minutes
Tb = background counting time in minutes
E = detector efficiency
V = air sample volume
A

There are a couple of points that we should note that make for some differences from the equation that you present. A minor definitional clarification relates to your use of the quantity LLD, which is actually expressed fundamentally as a count rate in cpm, in this case. By dividing the count rate by efficiency, E, the count rate is converted to activity in disintegrations per minute (dpm), which would represent what is commonly referred to as the minimum detectable activity (MDA); dividing by the air sample volume yields the minimum detectable air activity concentration (MDC). You do not specifically state it, but the use of 2.71 as the first term and 3.29 as the multiplier in the second term assumes equal values of 0.05 for the acceptable rate of false positive and false negative conclusions regarding the presence of activity; this assumption is associated with a k-value of 1.645. Further, it is assumed that the statistical distribution of counts is Poisson so that the standard deviation in a given count is equal to the square root of the count, and the standard deviation in the count rate is given by the square root of the quotient of the count rate by the counting time. The equation for LLD (a count rate) is often written:

LLD =
k2
Ts
+ 2Lc,

where Lc is the critical level, which is given by:

Lc = k (
Rb
Ts
+
Rb
Tb
)½,

and k is the number of standard deviations above zero that the critical level lies (in a Gaussian distribution with a mean net count rate of zero); Rb is the background count rate, Tb is the background count rate, and Ts is the gross sample counting time. (The above equation for LLD is the same as the expression you cited if the efficiency and volume terms in your equation are disregarded, and the 1/T term in the denominator of your equation is multiplied into the radical term as 1/T2.)

These equations are based on original work by Currie (1968). The equations are based on a gross count of a single sample and a single (constant) background. For the case that you describe, the total net count rate is determined by counting two samples and adding the results together. This process of counting two samples results in a propagated uncertainty that is different from the case in which a single sample is counted as discussed below. I have not dealt with this latter case in the past, and I believe it is worth considering.

For the case in which a single sample is counted for time T1 and a gross count rate of R1 is obtained, and the background rate is Rb and background counting time is Tb, the one-sigma uncertainty in the net count rate, R1 - Rb, is obtained from typical propagation of errors, as:

σn = (
R1
T1
+
Rb
Tb
)½,

For the current situation in which two samples are counted with respective gross count rates and counting times of R1 and R2 and T1 and T2 and a background rate and time of Rb and Tb, the total net count rate is given by R1 - Rb + R2 - Rb = R1 + R2 - 2Rb. This has the effect that the propagated one-sigma uncertainty in the net count rate is now given by:

σn = (
R1
T1
+
R2
T2
+
4Rb
Tb
)½,

The critical level is then calculated (from the usual assumption that the sample contains zero net activity) as:

Lc = k (
Rb
T1
+
Rb
T2
+
4Rb
Tb
)½,

If T1 = T2 = T, as in your instance, then

Lc = k (
2Rb
T
+
4Rb
Tb
)½,

and the consequent value of LLD would be given by:

LLD =
k2
Ts
+ 2k (
2Rb
T
+
4Rb
Tb
)½,

Dividing the LLD by the radionuclide counting efficiency and the air volume would yield the MDC for specified conditions. Based on the above results, the values generated using the modified expressions for Lc and LLD would be greater than those obtained for the case in which a single count of a sample is used rather than two counts that are added together.

George Chabot, CHP, PhD

Reference
Currie LA. Limits for qualitative detection and quantification determination. Analytical Chemistry 40(3):587–593; 1968.

Answer posted on 3 January 2017. 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.