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DSE Physics | Radioactivity and Half-Life Lesson-Prep Guide (Tutor Edition) | 800 Bq to 100 Bq in Three Half-Lives and a Decay Constant of 0.139 per Day
Students know a half-life halves the amount, then subtract step by step when asked about fifteen days. Powers of two turn the whole topic into one line.
Objectives — Students can: (1) use the remaining fraction (1/2)^n to get activity or mass from elapsed time and half-life; (2) derive the decay constant from the half-life with lambda = ln2 / t-half and explain how activity relates to the number of atoms; (3) date a sample from its carbon-14 fraction and state the limits of the method.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Indices | What 2 to the power 3 is | 8 |
| Logarithms | Roughly what ln2 is | 0.693 |
| Units | What the becquerel measures | Decays per second |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-8 min | Three radiations | Alpha, beta, gamma penetration | Paper stops alpha |
| 8-20 min | Definition | Time to halve | Independent of starting amount |
| 20-32 min | Remaining amount | (1/2)^n with n = t over half-life | 800 to 100 Bq |
| 32-42 min | Decay constant | lambda = ln2 / t-half | 0.1386 per day |
| 42-52 min | Carbon dating | Two half-lives give 11,460 years | 25% remaining |
| 52-60 min | Exit check | Three questions marked live | All three right |
03 Teaching script (read it out)
Scene 1: The half-life ignores the starting amount
A sample starts at 800 Bq with a half-life of five days: 400 Bq after five days, 200 Bq after ten, and 100 Bq after fifteen, one eighth of the original.
The half-life does not change with the starting amount, temperature or pressure. A thousand atoms and a billion atoms take the same time to halve, because decay is a random process, and that constancy is exactly what makes dating possible.
Reduce the work to (1/2)^n, where n is elapsed time divided by the half-life. Fifteen over five is three, so the answer is one over 2 cubed, an eighth, with no table needed.
The half-life does not change with the starting amount, temperature or pressure. A thousand atoms and a billion atoms take the same time to halve, because decay is a random process, and that constancy is exactly what makes dating possible.
Reduce the work to (1/2)^n, where n is elapsed time divided by the half-life. Fifteen over five is three, so the answer is one over 2 cubed, an eighth, with no table needed.
Students must say: how many half-lives fifteen days holds and what fraction remains (three, one eighth).
Scene 2: The decay constant links half-life to activity
lambda = ln2 / t-half = 0.693 / 5 = 0.1386 per day, the probability that a given atom decays in a day.
Activity is A = lambda N, where N is the number of undecayed atoms. Since lambda is constant, activity falls in step with the number of atoms and both share the same half-life.
Units must match: with the half-life in days, lambda is per day. Convert the half-life to seconds (five days is 432,000 s) and lambda becomes about 1.6 x 10^-6 per second, a different number describing the same physics. Mismatched units are the commonest loss here.
Activity is A = lambda N, where N is the number of undecayed atoms. Since lambda is constant, activity falls in step with the number of atoms and both share the same half-life.
Units must match: with the half-life in days, lambda is per day. Convert the half-life to seconds (five days is 432,000 s) and lambda becomes about 1.6 x 10^-6 per second, a different number describing the same physics. Mismatched units are the commonest loss here.
Students must say: why a longer half-life means a smaller lambda (slower decay).
Scene 3: Carbon-14 only works on samples that exchanged carbon
Carbon-14 has a half-life of about 5,730 years. A sample retaining 25% has seen two half-lives, so 5,730 x 2 = 11,460 years; 12.5% means three half-lives, 17,190 years.
The method reaches roughly fifty thousand years. Beyond about ten half-lives the remainder is too small to separate from background radiation and contamination.
It also only applies to material that once exchanged carbon with the atmosphere, such as bone, wood or shell. Rock and metal never took carbon in, so carbon dating is meaningless for them, and this point appears almost every year.
The method reaches roughly fifty thousand years. Beyond about ten half-lives the remainder is too small to separate from background radiation and contamination.
It also only applies to material that once exchanged carbon with the atmosphere, such as bone, wood or shell. Rock and metal never took carbon in, so carbon dating is meaningless for them, and this point appears almost every year.
Students must say: why carbon dating cannot date a stone (it never exchanged carbon).
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Calculation | Activity after several half-lives | (1/2)^n, find n first | 800 to 100 Bq |
| Calculation | Decay constant | lambda = ln2 / t-half, units follow time | 0.1386 per day |
| Calculation | Age from a remaining fraction | Find n, multiply by the half-life | 25% gives 11,460 years |
| Explanation | Limits of the dating method | Carbon exchange, ten half-lives | Not for stone |
Sanity check (teach this): Find n before the power: fifteen over five is three, so the answer is one eighth, not one third.
The half-life is constant regardless of amount, temperature or pressure.
lambda falls as the half-life rises; the two are inversely related.
Carbon dating stops near ten half-lives, about fifty thousand years, and needs samples that exchanged carbon.
The half-life is constant regardless of amount, temperature or pressure.
lambda falls as the half-life rises; the two are inversely related.
Carbon dating stops near ten half-lives, about fifty thousand years, and needs samples that exchanged carbon.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| 800 Bq with a five-day half-life after fifteen days | 100 Bq | 800 divided by 2 cubed |
| Decay constant for a five-day half-life | 0.1386 per day | 0.693 divided by 5 |
| Age of a sample retaining 25% of its carbon-14 | 11,460 years | Two half-lives |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Subtracting linearly | Fifteen days answered as 800 minus 300 | Each five days divides by two; it is not a fixed subtraction. |
| Using a third instead of an eighth | Leaving 800 over 3 | Three half-lives means 2 cubed, so one eighth remains. |
| Mixed units in lambda | Days for the half-life, seconds for lambda | Match the units or lambda is out by tens of thousands. |
| Carbon dating a rock | Dating granite with carbon-14 | Rock never exchanged carbon, so carbon dating does not apply. |
07 Homework (paper, section and question numbers)
- Secondary physics chapter index — Radioactivity and nuclear practice
- All lesson-prep guides — Compare other subjects
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| Activity after fifteen days | 100 Bq | All three right: move to nuclear energy |
| Decay constant | 0.1386 per day | One wrong: reteach lambda |
| Age at 25% carbon-14 | 11,460 years | Two wrong: return to act three |
09 Tutor FAQ
Does heating shorten a half-life?
No. Nuclear decay is unaffected by temperature, pressure or chemical state.
How do activity and atom count relate?
A = lambda N, and with lambda fixed the activity falls as the undecayed atoms fall.
Why does carbon dating stop at fifty thousand years?
After ten half-lives the remainder sits below background, so the measurement is no longer reliable.
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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.
