Before finding half-life, express the background and detector readings in the same units, such as counts per minute. Subtract the background rate from each total rate, then see how long the corrected rate takes to halve.
The 2027 SEC G3 Physics K323 syllabus covers background radiation and simple half-life calculations. This example uses a two-minute background reading and shorter detector readings, so it also shows why the counting time matters before you subtract.
- Separate elapsed time from the time used to collect each reading.
- Convert both background and total counts into counts per minute.
- Subtract the background rate before halving.
- Read the elapsed time between the two matching corrected rates.
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Keep the two kinds of time separate
Elapsed time tells you where you are in the decay record. Counting duration tells you how long the detector ran to produce a particular count. A table can say that a reading was taken at the six-minute mark using a detector window of 30 seconds.
Suppose the detector records 48 background counts in 2 minutes. That gives a background rate of 24 counts per minute, or an expected 12 background counts in 30 seconds. You can correct in either unit if both terms in the subtraction refer to the same duration.
When a question states that readings are already corrected for background, use those readings directly. If the heading says total or observed count rate, check whether it includes background.
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Worked example | Read the corrected half-life
A practice table shows detector counts taken over 30 seconds at each labelled elapsed time. Use the background reading of 48 counts in 2 minutes. Assume steady background and unchanged detector geometry. The numbers are idealised for practice.
| Elapsed time (min) | Counts in 30 s | Total rate (counts/min) | Corrected rate (counts/min) |
|---|---|---|---|
| 0 | 92 | 184 | 160 |
| 6 | 52 | 104 | 80 |
| 12 | 32 | 64 | 40 |
At the start, double 92 to get 184 counts per minute, then subtract 24 to get 160 counts per minute from the source. At 6 minutes, the corresponding calculation is 104 − 24 = 80 counts per minute. At 12 minutes it is 64 − 24 = 40 counts per minute.
The corrected rate halves from 160 to 80 in 6 minutes and from 80 to 40 over the next 6 minutes. The half-life is 6 minutes. On a graph of the total rate, those same points sit at 184, 104 and 64 counts per minute.
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Practise with a target on the uncorrected graph
Imagine a total-rate graph starts at 224 counts per minute with a background of 24. What total rate represents one half-life? The starting source rate is 200, its half is 100, and adding 24 back gives a target total rate of 124 counts per minute.
Halving the starting total directly would give 112 counts per minute. Its corrected rate is 88, which is below half the starting source rate. Using that target on a smooth decreasing curve would give a time that is too long.
If a noisy graph asks for an estimate, read from the supplied smooth decay curve or follow the question’s instruction. Use the same rate scale for both points, and state the interval between their times.
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CHAPTER 4 OF 4 · Back to contents
Frequently asked questions
Could I subtract 12 counts from each 30-second count instead?
Yes. That gives 80, 40 and 20 counts per 30 seconds. The rates are expressed per the same 30-second interval, so they still show a half-life of 6 minutes.
Does the total rate tend towards zero?
With a steady background and a source contribution that becomes very small, the total rate tends towards the background rate. The corrected source rate tends towards zero.
How can I check the units in my working?
Write counts per minute after every rate and minutes after each elapsed time. Read the subtraction out loud. Both terms should describe counts per the same time unit.
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