Radioactive Decay Model with a Measurement Discrepancy
A sample of a radioisotope has activity A(t) = A0 e^(-kt), with A in becquerels and t in days. Two measurements are recorded: A(0) = 800 Bq and A(10) = 500 Bq. A laboratory technician reports the half-life as 10 days, reasoning that activity fell from 800 to 500, 'a bit more than half'. A second sample of the same isotope, initially 2000 Bq, is measured at 3 Bq and declared 'effectively decayed'. The supervisor wants the exact half-life, the time for the 2000 Bq sample to reach 3 Bq, and an evaluation of the technician's reasoning.
1. Find the exact decay constant k and the half-life, and evaluate the technician's half-life claim.
Read the worked answer
From A(0) = 800: A0 = 800. From A(10) = 500: 800 e^(-10k) = 500, so e^(-10k) = 5/8, giving -10k = ln(5/8), k = -ln(5/8)/10 = ln(8/5)/10 ≈ 0.4700/10 ≈ 0.04700 per day.
Half-life: e^(-kt) = 1/2 → t = ln2/k = 10 ln2 / ln(8/5) ≈ 6.931/0.4700 ≈ 14.75 days.
The technician's claim of 10 days is wrong. At 10 days the activity is 500/800 = 0.625 of the original, not 0.5. The half-life is about 14.7 days. The error is treating 'more than half remaining' as 'half-life elapsed'; the half-life is the time to reach exactly half, which is longer than 10 days.
2. Find the time for a 2000 Bq sample of the same isotope to decay to 3 Bq, and comment on the 'effectively decayed' label.
Read the worked answer
Using the same k = ln(8/5)/10: 2000 e^(-kt) = 3, so e^(-kt) = 3/2000, giving t = ln(2000/3)/k = 10 ln(2000/3)/ln(8/5).
Numerically ln(2000/3) = ln(666.67) ≈ 6.502, and ln(8/5) ≈ 0.4700, so t ≈ 10 × 6.502/0.4700 ≈ 138.3 days.
Equivalently, the number of half-lives is log2(2000/3) ≈ 9.38, and 9.38 × 14.75 ≈ 138 days. The label 'effectively decayed' is a judgement, not a mathematical result: 3 Bq is small relative to 2000 Bq but the model predicts activity never reaches zero, only approaches it. A precise statement would be 'activity falls below 3 Bq after about 138 days'.
3. A proposed alternative model uses A(t) = 800 - ct for the first 10 days, matched to the same data. Compare its predictions at t = 20 with the exponential model and explain which is more plausible for radioactive decay.
Read the worked answer
Linear model: A(0) = 800, A(10) = 500, so c = 30 Bq/day and A(t) = 800 - 30t. At t = 20 this gives A = 200 Bq, and it predicts zero activity at t = 800/30 ≈ 26.7 days.
Exponential model: A(20) = 800 e^(-20k) = 800 (5/8)^2 = 800 × 25/64 = 312.5 Bq.
The two models agree at t = 0 and t = 10 but diverge sharply afterwards (200 vs 312.5 at t = 20). Radioactive decay is a constant-proportion process: the rate of decay is proportional to the amount present, which produces an exponential, not a linear, law. The linear model also predicts a finite time at which activity becomes exactly zero and then negative, which is physically impossible. The exponential model is therefore the appropriate choice, and the linear fit is only a local approximation valid near the two data points.