2.2 Decay Calculations and Counting Statistics
Key Takeaways
- Activity decays exponentially: A = A₀ e^(−λt) with λ = 0.693 / T½ (or ln 2 / T½)
- Tc-99m T½ ≈ 6.02 h: after one half-life activity is 50%; after 3 half-lives (~18 h) about 12.5% remains
- F-18 T½ ≈ 110 min: a 10 mCi dose at calibration has ~5 mCi after 110 min and ~2.5 mCi after 220 min
- Counting follows Poisson statistics; percent error ≈ 100 / √N, so quadrupling counts halves the percent error
- Longer count time or higher activity improves precision until dead-time or background limits the gain
Decay Calculations and Counting Statistics
Quick Answer: Use A = A₀ e^(−λt) with λ = 0.693/T½. For Tc-99m (T½ ≈ 6.02 h) activity halves every ~6 h; for F-18 (T½ ≈ 110 min) it halves every ~1.8 h. Counting error is approximately 100/√N percent — more counts (higher activity or longer time) improve precision.
The Decay Law
Radioactive decay is a first-order process. If A₀ is activity at a reference (calibration) time t = 0, activity at later time t is:
A = A₀ e^(−λt)
where the decay constant λ is related to half-life by:
λ = ln(2) / T½ ≈ 0.693 / T½
Units of λ are inverse time (h⁻¹, min⁻¹). Keep t and T½ in the same units. You may also write remaining fraction as (1/2)^(t/T½) — often faster for mental half-life multiples:
- t = 1 × T½ → 50% remains
- t = 2 × T½ → 25% remains
- t = 3 × T½ → 12.5% remains
- t = 0.5 × T½ → √0.5 ≈ 70.7% remains
Pre-calibration (future calibration time): if the dose is calibrated for a later time, activity now is higher: A_now = A_cal / e^(−λt) = A_cal × e^(+λt), where t is time until calibration.
Worked Example 1 — Tc-99m Dose Decay
A Tc-99m MDP kit is calibrated as 25.0 mCi at 08:00. The patient is injected at 11:00. T½ = 6.02 h.
- Elapsed time t = 3.0 h
- λ = 0.693 / 6.02 h ≈ 0.1151 h⁻¹
- λt = 0.1151 × 3.0 ≈ 0.345
- e^(−λt) ≈ e^(−0.345) ≈ 0.708
- A = 25.0 × 0.708 ≈ 17.7 mCi at 11:00
Sanity check: 3 h is half of one half-life, so remaining fraction should be near √0.5 ≈ 0.707 — matches.
Another Tc-99m check: at 08:00 next day (24 h later), t/T½ ≈ 24/6.02 ≈ 4.0 half-lives → remaining ≈ (1/2)⁴ = 6.25% → 25 × 0.0625 ≈ 1.6 mCi. Overnight residual in a syringe shield is non-zero; treat waste and surveys accordingly.
Worked Example 2 — Multi-Step F-18 Decay
An FDG unit dose is 10.0 mCi at 10:00 (calibration). T½ ≈ 110 min.
Part A — Activity at 11:50 (110 min later):
t = 110 min = 1 half-life → A = 10.0 × 0.5 = 5.0 mCi.
Part B — Activity at 12:45 (165 min after calibration):
t = 165 min; t/T½ = 165/110 = 1.5
Remaining = (1/2)^1.5 = 0.5 × √0.5 ≈ 0.5 × 0.707 = 0.354
A ≈ 10.0 × 0.354 = 3.54 mCi.
Part C — How early must you assay if you need ≥8 mCi at injection and you will inject at 10:40?
Required at 10:40: 8 mCi; t from calibration to injection = 40 min if calibrated at 10:00.
λ = 0.693/110 min⁻¹ ≈ 0.0063 min⁻¹; e^(−λ·40) ≈ e^(−0.252) ≈ 0.777
A₀ needed at 10:00 = 8 / 0.777 ≈ 10.3 mCi. If the delivered dose is only 10.0 mCi at 10:00, at 10:40 you have ~7.8 mCi — slightly short; reschedule or request a higher calibrated amount.
Remaining Activity Tables (Quick Reference)
Tc-99m (T½ = 6.0 h used for rounded table)
| Time after calibration | Fraction remaining | 20 mCi becomes |
|---|---|---|
| 0 h | 1.00 | 20.0 mCi |
| 1 h | 0.89 | 17.8 mCi |
| 2 h | 0.79 | 15.9 mCi |
| 3 h | 0.71 | 14.1 mCi |
| 6 h | 0.50 | 10.0 mCi |
| 12 h | 0.25 | 5.0 mCi |
| 18 h | 0.125 | 2.5 mCi |
| 24 h | 0.0625 | 1.25 mCi |
F-18 (T½ = 110 min)
| Time after calibration | Fraction remaining | 10 mCi becomes |
|---|---|---|
| 0 min | 1.00 | 10.0 mCi |
| 55 min | 0.707 | 7.1 mCi |
| 110 min | 0.50 | 5.0 mCi |
| 165 min | 0.354 | 3.5 mCi |
| 220 min | 0.25 | 2.5 mCi |
| 330 min | 0.125 | 1.25 mCi |
Exam tip: when times are exact multiples of T½, prefer the half-life power method over calculator work — faster and less error-prone under timed conditions.
Counting Statistics (Poisson)
Nuclear decay and detector counts are random. For a large number of independent decays, counts follow a Poisson distribution. For mean count N, the standard deviation σ ≈ √N, and the percent error (coefficient of variation × 100) is:
% error ≈ 100 / √N
| Total counts N | √N | Approx. % error |
|---|---|---|
| 100 | 10 | 10% |
| 1,000 | 31.6 | 3.2% |
| 10,000 | 100 | 1% |
| 40,000 | 200 | 0.5% |
Effect of count time: If the true count rate is constant (neglecting decay during the count and dead time), counts N = rate × time. Doubling count time doubles N and multiplies % error by 1/√2 ≈ 0.71 (about a 29% relative reduction in percent error). Quadrupling counts halves percent error.
Clinical implications:
- Well-counter assays of wipe tests or blood samples: choose count time so expected net counts give acceptable % error (often aiming for a few percent or better for quantitative work).
- Imaging: longer acquisition or more administered activity increases counts and improves signal-to-noise, balanced against patient dose, motion, and throughput.
- Background: net counts = gross − background; low-level contamination surveys need adequate time or you risk false negatives from large % error.
- Decay during counting: for very short T½ (Rb-82) or long counts, apply decay correction to the count interval — beyond intro CNMT math but know the concept.
Putting It Together
A technologist who can (1) convert mCi ↔ MBq, (2) decay-correct Tc-99m and F-18 across clinic schedules, and (3) estimate whether a 1-minute wipe count with 100 net counts (~10% error) is good enough for a release decision is operating at Domain I competence. Practice both formula and half-life-table paths until either is automatic.
A Tc-99m dose is 20 mCi at 09:00. Using T½ = 6.0 h, what is the approximate activity at 15:00 the same day?
Using A = A₀ e^(−λt) with λ = 0.693/T½, which value of λ is correct for F-18 if T½ = 110 minutes?
A sample is counted long enough to collect 10,000 net counts. Approximately what is the percent error from Poisson counting statistics?