Physics • Year 12 • Module 8 • Lesson 15
Radioactive Decay
Apply the exponential decay law, half-life calculations, and activity analysis to real data, graphs, and quantitative scenarios.
1. Interpret radioactive decay data
The table below shows the measured count rate (proportional to activity) of a radioactive sample over time. 9 marks
| Time (min) | Count rate (counts/s) | Fraction of initial count rate | ln(count rate) |
|---|---|---|---|
| 0 | 800 | 1 | |
| 10 | 566 | ||
| 20 | 400 | ||
| 30 | 283 | ||
| 40 | 200 | ||
| 50 | 141 | ||
| 60 | 100 |
1.1 Complete the “Fraction of initial count rate” and “ln(count rate)” columns. Show your working for the t = 20 min row. 3 marks
1.2 Use the data to determine the half-life of the sample. Show your reasoning. 2 marks
1.3 Calculate the decay constant λ of this isotope (in min−1 and s−1). 2 marks
1.4 If the initial number of nuclei N0 = 2.0 × 106, use A = λN0 to calculate the theoretical initial activity in Bq. Compare this to the count rate and comment on any discrepancy. 2 marks
2. Interpret a decay graph
The graph below shows the number of radioactive nuclei N remaining in two different samples (X and Y) over time. 7 marks
Figure 2.1. Nuclei remaining for samples X and Y. Illustrative data.
2.1 Read the half-life of sample X and sample Y from the graph. Show how you determined each value. 2 marks
2.2 At t = 20 h, compare the number of nuclei remaining in X and Y. Explain the difference in terms of number of half-lives elapsed. 2 marks
2.3 Calculate the initial activity of sample X, given N0 = 1000 nuclei. Use the half-life you read from the graph. Give your answer in Bq. 2 marks
2.4 Sample Y has a shorter half-life than sample X. If both samples contain the same number of nuclei at some instant, which has the higher activity? Justify. 1 mark
3. Compare alpha, beta-minus, and gamma radiation
Complete the table below. 9 marks (1 per cell)
| Property | Alpha (α) | Beta-minus (β−) | Gamma (γ) |
|---|---|---|---|
| Particle / radiation emitted | |||
| Change in Z | |||
| Change in A | |||
| Ionising ability | |||
| Penetrating power | |||
| Stopped by | |||
| Additional particle also emitted? | |||
| Typical situation | |||
| Charge of emitted particle |
4. Predict and justify — nuclear medicine scenario
Technetium-99m (99mTc) emits gamma radiation and has a half-life of 6.01 hours. A patient receives a dose containing 4.0 × 108 nuclei of 99mTc. 5 marks
4.1 How many half-lives pass in 24.04 hours? Calculate the number of 99mTc nuclei remaining after this time. 2 marks
4.2 Calculate the initial activity of the dose in becquerels. 2 marks
4.3 State one reason why gamma-emitting isotopes are preferred over alpha- or beta-emitting isotopes for medical imaging. 1 mark
Q1.1 — Complete columns (3 marks)
Fractions: 10 min: 0.708; 20 min: 0.500; 30 min: 0.354; 40 min: 0.250; 50 min: 0.176; 60 min: 0.125. [1 mark]
ln(count rate): t=0: 6.685; t=10: 6.338; t=20: 5.991; t=30: 5.645; t=40: 5.298; t=50: 4.948; t=60: 4.605. [1 mark]
Working for t=20: fraction = 400/800 = 0.50; ln(400) = 5.991. [1 mark]
Q1.2 — Half-life (2 marks)
Count rate halves from 800 to 400 at t = 20 min [1]. Therefore t1/2 = 20 min [1]. (Confirmed: 400 to 200 from t = 20 to t = 40 min = another 20 min.)
Q1.3 — Decay constant (2 marks)
λ = ln(2)/20 = 0.03466 min−1 [1]. In SI: λ = 0.03466/60 = 5.78 × 10−4 s−1 [1].
Q1.4 — Theoretical activity vs count rate (2 marks)
A = λN0 = 5.78 × 10−4 × 2.0 × 106 = 1156 Bq [1]. The table shows 800 counts/s, less than 1156 Bq. The discrepancy arises because the detector does not capture every decay — it has a detection efficiency of about 69% in this case. Count rate measures detected events; activity measures all decays [1].
Q2.1 — Half-lives from graph (2 marks)
X: N halves from 1000 to 500 at t = 10 h → t1/2(X) = 10 h [1]. Y: N halves from 800 to 400 at t = 5 h → t1/2(Y) = 5 h [1].
Q2.2 — Compare N at t = 20 h (2 marks)
X: 20/10 = 2 half-lives; N = 1000 × (1/2)2 = 250 [1]. Y: 20/5 = 4 half-lives; N = 800 × (1/2)4 = 50 [1]. Y has fewer nuclei because more half-lives have elapsed.
Q2.3 — Initial activity of X (2 marks)
λX = ln(2)/(10 × 3600) = 1.925 × 10−5 s−1 [1]. A = 1.925 × 10−5 × 1000 = 1.93 × 10−2 Bq [1]. (Small because N is tiny; in practice N ≫ 103.)
Q2.4 — Y has higher activity for same N (1 mark)
Y has higher activity. A = λN = (ln2/t1/2)N. Y has a shorter t1/2 (larger λ), so for equal N, Y has a higher decay rate [1].
Q3 — Comparison table
Particle emitted: α: He-4 nucleus ($^4_2$He); β−: electron (e$^-$); γ: high-energy photon. Change in Z: α: −2; β−: +1; γ: 0. Change in A: α: −4; β−: 0; γ: 0. Ionising ability: α: highest; β−: moderate; γ: lowest. Penetrating power: α: lowest; β−: moderate; γ: highest. Stopped by: α: paper/skin; β−: few mm aluminium; γ: several cm lead. Additional particle emitted? α: no; β−: yes — antineutrino ($\bar{\nu}_e$); γ: no. Typical situation: α: very heavy nucleus (Z > 82); β−: too many neutrons; γ: excited nuclear state. Charge of emitted particle: α: +2; β−: −1; γ: 0 (neutral).
Q4.1 — Nuclei after 24.04 h (2 marks)
n = 24.04/6.01 = 4.00 half-lives [1]. N = 4.0 × 108 × (1/2)4 = 4.0 × 108/16 = 2.5 × 107 nuclei [1].
Q4.2 — Initial activity (2 marks)
λ = ln(2)/(6.01 × 3600) = 3.20 × 10−5 s−1 [1]. A = λN0 = 3.20 × 10−5 × 4.0 × 108 = 1.28 × 104 Bq [1].
Q4.3 — Gamma preferred for imaging (1 mark)
Gamma radiation is highly penetrating and can exit the body to be detected externally; it deposits far less energy per unit path length in tissue than alpha or beta radiation, minimising patient dose while still allowing imaging [1]. (Also accept: short half-life minimises radiation dose.)