Is This Filter Sample Above the Reporting Level?
A health physics technician counts an air-filter sample for 10 minutes on a NaI(Tl) detector and obtains 400 gross counts. A 10-minute background count taken the same morning gives 250 counts. The detector efficiency for the gamma energy of interest, in the counting geometry used, is 0.12 counts per emitted photon. The filter was collected 72 hours before counting, and the nuclide is assumed to be I-131 with a half-life of 8.04 days. The procedure requires a result to be reported as 'above the reporting level' only if the decay-corrected activity at the end of sampling exceeds 5 Bq, and requires the decision limit to be derived from the measured background. A 20-minute background taken a week earlier had given 380 counts. The technician wants to release a positive result immediately because the gross count is well above background.
1. Calculate the net count rate, the decision limit, the activity at count time and at end of sampling, with its counting uncertainty. Is the sample above the reporting level?
Read the worked answer
Net rate = 400/10 - 250/10 = 40 - 25 = 15 cpm = 0.25 cps. Activity at count time = 0.25/0.12 = 2.08 Bq. Decay correction: 72 h = 3.0 d = 3/8.04 = 0.373 half-lives, so the factor is 2^0.373 = 1.295. Activity at end of sampling = 2.08 × 1.295 = 2.70 Bq.
Decision limit (Currie, 95%): Lc = 2.33√B = 2.33√250 = 36.8 counts = 3.68 cpm. The 15 cpm net exceeds this, so a detection is claimed. The detection limit is 2.71 + 4.65√250 = 76.2 counts = 7.62 cpm, about 1.06 Bq at count time.
Uncertainty: σ(net) = √(400+250)/10 = 2.55 cpm, which converts to 0.35 Bq at count time and 0.46 Bq decay-corrected. Result = 2.70 ± 0.46 Bq (1σ).
The value is about five standard deviations below 5 Bq, so it is not above the reporting level. Report the measured value with its uncertainty and the 'below reporting level' conclusion; do not present it as a positive release.
2. The technician proposes to count for 60 minutes instead. Show what that changes, and state the correct reporting statement.
Read the worked answer
Scaling the observed rates to 60 minutes: gross ≈ 2400 counts, background ≈ 1500 counts, net ≈ 900 counts. σ(net) = √(2400 + 1500) = 62.4 counts = 1.04 cpm, compared with 2.55 cpm for the 10-minute count - a factor of about 2.5 improvement for six times the counting time, because uncertainty falls as the square root of time. Activity at count time = 15/60/0.12 = 2.08 Bq (unchanged in expectation), with σ = 1.04/60/0.12 = 0.14 Bq, or 0.19 Bq decay-corrected.
Reporting statement: activity at end of sampling 2.7 ± 0.2 Bq, below the 5 Bq reporting level, nuclide assumed I-131. State the assumption explicitly, because the decay correction depends on it; if the true nuclide were shorter-lived the corrected value would be higher.
The longer count is still the right call: it turns a marginal, assumption-driven number into a defensible one and reduces the chance that a genuinely elevated sample is misclassified.
3. The earlier 20-minute background gave 380 counts. Is the background consistent, and what should be done before any sample result is released?
Read the worked answer
Convert to rates: today 250/10 = 25.0 cpm; last week 380/20 = 19.0 cpm. Difference = 6.0 cpm. σ of the difference = √(25/10 + 19/20) = √(2.5 + 0.95) = 1.86 cpm. The difference is 6.0/1.86 = 3.2σ, which is unlikely to be a chance fluctuation.
A drifting background matters because the background is subtracted: if today's 25 cpm is artificially high, the net rate and hence the activity are underestimated, which is a non-conservative error. Possible causes are contamination on the detector face or inside the shield, a source left near the detector, a change in shielding, or a faulty high-voltage or gain setting.
Before releasing any result: repeat the background with a long count, run a blank filter, inspect and wipe-test the shield interior and detector window, check the calibration and gain, and perform a chi-square test on a series of background counts. Until the background is verified as stable, the sample result should be treated as provisional and the discrepancy logged.