To predict what is left after a stated time, divide the time by the half-life, then halve the starting amount that many times. The amount can be mass, number of nuclei or corrected count rate. This skill comes straight after reading a half-life from a graph.
It builds on determining half-life from a decay graph and is part of half-life and background.
Why does halving repeat?
In each half-life, half of the undecayed nuclei decay. The next half-life starts with fewer nuclei, and half of those decay. So the amounts form the pattern start, ½, ¼, ⅛, 1/16, and so on.
A table keeps this tidy. Write the number of half-lives in the first row and the amount in the second.
How do I work it out, step by step?
- Make the units match. Convert the time or the half-life so both use the same unit.
- Find the number of half-lives = time ÷ half-life.
- Halve the starting amount once for each half-life, writing each value.
- State the answer with its unit, and say whether it is the amount left or the amount that decayed.
If the time is not a whole number of half-lives, this course-level method is not enough on its own. A graph or a supplied relationship is then needed, so follow the question’s instruction.
Worked example
(Invented data.) A sample contains 64 g of a radioactive isotope with a half-life of 3 days. How much remains after 12 days?
Step 1, units: both are in days.
Step 2, number of half-lives: 12 ÷ 3 = 4.
Step 3, halve four times:
| Half-lives elapsed | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Time (days) | 0 | 3 | 6 | 9 | 12 |
| Mass left (g) | 64 | 32 | 16 | 8 | 4 |
Step 4, answer: 4 g of the isotope remains. That is 64 ÷ 16 = 4, so one-sixteenth of the original. The mass that has decayed is 64 − 4 = 60 g.
Check the fraction: (½)⁴ = 1/16, and 64 × 1/16 = 4. Both methods agree.
The mistakes to watch for
Two slips appear again and again.
Mistaken answer A: after 6 days (two half-lives) none is left.
The student treated “half-life 3 days” as “half gone in 3 days, the rest gone in the next 3.”
After 6 days, 16 g remains, not 0 g. Each half-life removes half of what is present.
Mistaken answer B: 64 ÷ 12 = 5.3 g
The student divided by the number of days rather than halving once for each half-life.
The correction is to count half-lives first (12 ÷ 3 = 4), then halve four times. The bounds and rounding explainer is useful if you want to practise how many figures an answer should carry.
Check yourself
1. A corrected count rate is 640 counts/min. The half-life is 20 min. What is the corrected rate after 1 hour?
Show answer
1 hour = 60 min, and 60 ÷ 20 = 3 half-lives. 640 → 320 → 160 → 80. The corrected rate is 80 counts/min.
2. A 5.0 g sample decays to 0.625 g in 15 years. What is the half-life?
Show answer
5.0 → 2.5 → 1.25 → 0.625 is 3 halvings. Half-life = 15 ÷ 3 = 5 years.
3. A half-life is 8 days. What fraction of the nuclei remains after 24 days, and what fraction has decayed?
Show answer
24 ÷ 8 = 3 half-lives, so the fraction remaining is (½)³ = 1/8. The fraction decayed is 1 − 1/8 = 7/8.
Where this leads next
Halving describes groups of nuclei well, but it says nothing about one nucleus. That is the point of explaining why one atom has no predictable decay time. If multi-step calculations still feel shaky, our teachers can rebuild them in online one-to-one Physics tuition.