Topic 5 of 7
Half-life from numbers and graphs
Half-life is the time taken for half the undecayed radioactive nuclei in a large sample of one nuclide to decay. Over that time, its activity falls to half.
Activity is decays per second. With unchanged detection conditions, the background-corrected source count rate follows the same proportion for the single-nuclide model considered here.
The symbol is t1/2. It is a time, with SI unit s. A problem may use minutes, hours or days consistently; compare the elapsed time and half-life in matching units.
Halve what remains each time
Repeated halving
800 Bq initially, with a half-life of 6 h
- After 6 h: 800/2 = 400 Bq.
- After 12 h: 400/2 = 200 Bq.
- After 18 h: 200/2 = 100 Bq.
Eighteen hours contains three half-lives. The remaining fraction is (1/2) x (1/2) x (1/2) = 1/8; the decayed fraction is 7/8. The same 6 h half-life is 6 x 3600 = 21600 s.
Half of the radioactive parent nuclei remain after one half-life on average. Half of the whole sample has not vanished: daughter material is still present. After the next half-life, half of the remaining parents decay.
Find half-life from corrected data
This separate supplied model follows one radioactive nuclide. Background has a constant mean of 20 counts/min; daughter contributions are neglected. The rates represent the labelled instants, with detection conditions unchanged.
| Time / min | Total / (counts/min) | Source / (counts/min) |
|---|---|---|
| 0 | 180 | 160 |
| 4 | 100 | 80 |
| 8 | 60 | 40 |
| 12 | 40 | 20 |
| 16 | 30 | 10 |
Remove the background before finding the half-life
Supplied smooth model: one radioactive nuclide, unchanged detection conditions, constant mean background of 20 counts/min and negligible daughter contribution. Real readings fluctuate around such a trend.
Total count rate includes the background
The dashed line is the 20 counts/min background. The total curve approaches that level, rather than zero.
Net source rate: subtract 20 counts/min
160 to 80 and 40 to 20 counts/min each take 4 min. The net rate approaches zero; it is still 10 counts/min at 16 min.
The axes use the same scales in both panels. Each marked dot matches a supplied table value; the curve represents the ideal average pattern between them.
- Correct the first pair: 180 - 20 = 160 and 100 - 20 = 80 counts/min.
- Read the time interval: 160 to 80 takes 4 - 0 = 4 min.
- Check another pair: the net rate falls from 40 at 8 min to 20 at 12 min, again taking 4 min.
A fall from 160 to 20 counts/min is three halvings: 160, 80, 40, 20. It therefore takes 3 x 4 = 12 min. It is not one half-life simply because the final reading is small.
Halving the uncorrected 180 to 90 counts/min would also halve the background contribution. The background does not follow the source's decay, so that would give the wrong interval.
Read a curve and interpret real measurements
Label elapsed time horizontally and the stated rate vertically. On a corrected curve, choose a source rate, find half that rate, and read the difference between their times. Use another well-separated pair as a check.
The source loses the same fraction over equal half-life intervals, not the same amount. The average curve bends as the rate falls; it does not become zero after two half-lives. Individual measurements scatter around this trend because decay and detection are random.
For a logged count series, retain the timestamps and counting durations. Calculate rates, subtract a comparable background estimate and keep source-detector geometry fixed. A short counting window can give an approximate rate at its labelled time, but an excessively long window blurs a rapid decay. Do not mistake a changed detector position or setting for a new half-life.