Dead-Time Loss in a High-Count Rate Well Counter
A nuclear medicine department uses a well counter with a paralyzable detection system. The measured dead time is 2.5 µs. A technologist counts a sample of Tc-99m that was expected to contain 1.8 MBq based on the preparation record. The observed count rate is 92,000 cps. The background count rate in the same well is 60 cps. The technologist is concerned that the sample is decaying faster than expected and wants to recalibrate the counter. The department's protocol requires that count-rate losses remain under 5% for quantitative assays.
1. Calculate the true count rate and the percent loss due to dead time. Is the observed count rate consistent with the expected activity?
Read the worked answer
For a paralyzable system, the relationship between observed rate m and true rate n is m = n·e^(−nτ).
Here τ = 2.5 µs = 2.5×10⁻⁶ s, m = 92,000 cps.
Solve iteratively:
- First guess: n ≈ m/(1 − mτ) = 92,000/(1 − 0.23) = 119,480 cps.
- Refine: n·e^(−nτ) = 119,480·e^(−0.2987) ≈ 119,480 × 0.7418 ≈ 88,630 cps (too low).
- Increase n to 124,000: 124,000·e^(−0.31) ≈ 124,000 × 0.7334 ≈ 90,940 (still low).
- n = 127,000: 127,000·e^(−0.3175) ≈ 127,000 × 0.7280 ≈ 92,460 (slightly high).
- n ≈ 126,300 cps gives m ≈ 92,000 cps.
Percent loss = (n − m)/n × 100 = (126,300 − 92,000)/126,300 × 100 ≈ 27.2%.
Expected activity is 1.8 MBq. If counting efficiency were near 100% (well counters typically have high geometric efficiency but not 100% photon yield per decay), the expected count rate would be roughly 1.8×10⁶ cps × efficiency. Even at a generous 10% efficiency, that is 180,000 cps, far above the observed rate. The observed rate is not consistent with a decay problem; it is dominated by dead-time loss. The counter must not be recalibrated. The sample should be diluted or counted at a lower activity so that the loss is under 5% (mτ < ~0.05, i.e., m < ~20,000 cps for τ = 2.5 µs).
2. What is the maximum observed count rate at which the 5% loss criterion is met, and what action should the technologist take?
Read the worked answer
The 5% loss criterion requires (n − m)/n ≤ 0.05, i.e., m/n ≥ 0.95.
For a paralyzable system, m/n = e^(−nτ). Setting e^(−nτ) = 0.95 gives nτ = −ln(0.95) = 0.0513, so n = 0.0513/2.5×10⁻⁶ ≈ 20,520 cps. Then m = 0.95 × 20,520 ≈ 19,490 cps.
So the observed rate must stay below roughly 19,500 cps.
Action: dilute the sample (or count a smaller aliquot) so the observed rate is below ~19,500 cps. Re-count and apply the dead-time correction if needed. Do not recalibrate the counter based on this measurement. Document the dilution and correction in the worksheet.