A radioactive sample initially contains 8.0 x 1012 undecayed nuclei and has a half-life of 6.0 h. Calculate the number of undecayed nuclei after 15 h.
[4 marks]
Total for this question: 4
All equation deep-dives
Radioactive decay and half-life
Nuclear physics
AQA 7408
In the data booklet
ORIGINAL
N = N0 e−lambda t and T1/2 = ln(2) / lambda
Keep time and decay constant in matching units; hours with per hour is just as valid as seconds with per second.
Board and spec code confirmed against AQA 7408 · registry checked 2026-07-11How this checking works
Know the equation
| Symbol | Quantity | Unit |
|---|---|---|
| N | number of undecayed nuclei after time t | no unit |
| N0 | initial number of undecayed nuclei | no unit |
| lambda | decay constant | s−1 (or matching time−1) |
| t | time | s (or matching time unit) |
| T1/2 | half-life | s (or matching time unit) |
Rearrangements
- lambda = -ln(N / N0) / t
- t = -ln(N / N0) / lambda
- N0 = N elambda t
- lambda = ln(2) / T1/2
Apply it — mark your own working
Work each one out on paper first, then reveal the mark scheme and tick the marks you actually earned. That is exactly how you should mark past papers.
Q1
Do the calculation on paper first — then mark it.
Q2
The activity of a source falls from 12 MBq to 3.6 MBq in 8.0 h. Calculate its half-life.
[4 marks]
Total for this question: 4
Do the calculation on paper first — then mark it.
Where the marks get lost
- Mixing hours with a decay constant in s−1 without converting one of them.
- Using log base 10 instead of ln when undoing e.
- Assuming a non-integer number of half-lives must be rounded; use the exponential for the exact fraction remaining.
Exam tip: Activity A obeys the same exponential ratio as N because A = lambda N. In ratio calculations, MBq can stay as MBq because the common factor cancels.
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