Fluorine-18 undergoes positron emission with a half-life of 1.10 × 102 If a patient is given a 248 mg dose for a PET scan, how long will it take for the amount of fluorine-18 to drop to 83 mg? (Assume that none of the fluorine is excreted from the body.)

Question

BlackTom AI · Accepted Answer

Radioactive decay follows N = N0 (1/2)^{t/t1/2}, where t1/2 is the half-life.

1) Relevant concepts
- Exponential decay of a radioisotope: N(t) = N0 (1/2)^{t/t1/2} or equivalently N = N0 e^{-λt} with λ = ln 2 / t1/2.
- Solve for time t given initial amount N0, remaining amount N, and half-life.

2) Step-by-step reasoning
Start with N = N0 (1/2)^{t/t1/2}. Rearranging to solve for t:
(1/2)^{t/t1/2} = N / N0
Take natural log: (t/t1/2) ln(1/2) = ln(N/N0)
So t = t1/2 × ln(N/N0) / ln(1/2).

3) Calculations (all intermediate steps)
- Given: t1/2 = 1.10 × 10^2 min = 110 min, N0 = 248 mg, N = 83 mg.
- N/N0 = 83 / 248 = 0.334677419.
- ln(N/N0) = ln(0.334677419) = −1.0959 (rounded).
- ln(1/2) = −0.693147.
- Ratio = ln(N/N0) / ln(1/2) = (−1.0959) / (−0.693147) = 1.5814.
- t = 110 min × 1.5814 = 173.95 min ≈ 1.74 × 10^2 minutes.

Alternative check using λ:
λ = ln 2 / 110 = 0.693147 / 110 = 0.0063013 min^−1.
t = ln(N0/N)/λ = ln(248/83)/0.0063013 = 1.0959 / 0.0063013 = 173.95 min.

4) Why this answer is correct
The computed time satisfies the exponential decay law: after ≈174 minutes (about 1.58 half-lives) the amount falls from 248 mg to 83 mg. The algebra and numerical evaluation directly follow from the decay formula.

5) Why other options are incorrect
- 99 min (too short): would leave ~133 mg, not 83 mg.
- 1.32 × 10^2 min (132 min): corresponds to ~0.435 fraction → ~108 mg.
- 3.00 × 10^2 min (300 min): far too long; would leave ~37 mg.
- 2.11 × 10^2 min (211 min): leaves ~66 mg, not 83 mg.

6) Conclusion
The correct time is 1.74 × 10^2 minutes (≈174 minutes), which matches the provided answer.

Fluorine-18 undergoes positron emission with a half-life of 1.10 × 102 If a patient is given a 248 mg dose for a PET scan, how long will it take for the amount of fluorine-18 to drop to 83 mg? (Assume that none of the fluorine is excreted from the body.)Single choice

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