Topic 5 of 9
Half-life and corrected curves
Half-life is the time for the expected number of undecayed parent nuclei, or the activity, to fall to half its initial value. For a detector reading, use the source contribution and check that the conditions make it proportional to the decaying population.
For a single unchanged radionuclide, the expected number of undecayed parent nuclei and the activity halve over the same interval. A proportional net source count rate does too when detection conditions remain fixed. Half-life is not the lifetime of every individual nucleus.
After one, two and three half-lives, the remaining fractions are 1/2, 1/4 and 1/8. A few half-lives leave a fraction, not zero. Match time units before comparing: 0.2 min is 12 s, while 2 min is 120 s.
Separate total rate from the source trend
In this supplied model, net source rate starts at 80.0 counts/s and halves every 120 s. Background remains 4.0 counts/s. These are instantaneous expected rates at the named times, not long-window measured averages.
| t / s | Net / counts s-1 | Total / counts s-1 |
|---|---|---|
| 0 | 80 | 84 |
| 120 | 40 | 44 |
| 240 | 20 | 24 |
| 360 | 10 | 14 |
| 480 | 5 | 9 |
Total expected rate approaches the background
The same fixed background is added at every time. Half of the initial raw total is 42 counts/s; that is not the correct halving target for the source. At 120 s the total is 44 counts/s.
Corrected expected rate halves in equal times
The two marked intervals each last 120 s even though the absolute decrease is smaller in the second interval. A proportional corrected rate shares the half-life only while source and detection conditions remain fixed.
Half of the initial raw total is 42 counts/s, but the total at one half-life is 44 counts/s: 40 from the source plus 4 background. Read half-life from the corrected quantity. The falling trend becomes less steep as fewer parents remain; it is not a constant rate of loss.
Worked transfer problem
Count the halvings
A corrected rate decreases from 72 to 9 counts/s in 15 min under fixed detection conditions:
Three half-lives = 15 min
Half-life = 5.0 min
One eighth of the initial rate represents three halvings, not eight. A small negative background-subtracted estimate near zero can arise from fluctuations; it is not negative physical activity.
Supplied fractional model
Use an exponential between whole half-lives
For the 120 s model, use the following supplied continuous relationship, with t in seconds and r0 = 80 counts/s:
= exp[-ln(2)t/120]
The exponent and the rate ratio are dimensionless. At 60 s, calculate EXP(-LN(2)*60/120) = 0.7071068. The expected net rate is 56.5685 counts/s and total rate is 60.5685 counts/s. Half a half-life does not remove half of the amount lost during a whole half-life.
At 180 s the fraction is 0.3535534 and net rate is 28.2843 counts/s. To find when only 10.0% remains, take logarithms of the supplied ratio:
t = -120 ln(0.10)/ln(2)
= 398.631 s ≈ 399 s
The same model gives a straight transformed relationship:
A logarithm turns the supplied fractional model into a straight line
This graph is the supplied model ln(r/r0) = -ln(2)t/120, with time in seconds. Its gradient is about -0.00578 per second. The logarithm acts on a dimensionless positive ratio; the zero ordinate means r = r0, not r = 0.
The logarithm requires a positive corrected rate. Keep seconds in this expression; substituting a time in minutes without converting changes the result.