OxfordAQA International AS and A-level Physics: Exponential radioactive decay (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.9.2 (Exponential changes in radioactivity) of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4, for exams from May/June 2020 onwards. Section 3.9 is International A-level only. Where needed, take 1 year = 3.156 × 10⁷ s and N_A = 6.02 × 10²³ mol⁻¹.

Links: study guide, revision notes, the OxfordAQA Physics hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. (a) Say what cannot be predicted about one nucleus in a radioactive sample. [1] (b) Define the decay constant of a nuclide. [1] (c) Explain why the half-life of a nuclide is the same for a large sample as for a small one. [1]

2. A nuclide has a half-life of 3.8 days. A sample contains 2.4 × 10¹² undecayed nuclei. (a) Calculate the decay constant in s⁻¹. [2] (b) Calculate the activity of the sample. [2]

3. A sample initially contains 6.0 × 10¹⁰ undecayed nuclei of a nuclide with decay constant 0.018 h⁻¹. (a) Calculate the number of undecayed nuclei after 48 h. [2] (b) Calculate the gradient of a graph of N against t at t = 0, in nuclei per second. [2]

4. A source has an initial activity of 7.5 × 10⁵ Bq and a half-life of 2.4 h. Calculate how long it takes for the activity to fall to 1.2 × 10⁴ Bq. [4]

5. A small power source contains 0.50 mg of plutonium-238, which has a molar mass of 238 g mol⁻¹ and a half-life of 87.7 years. (a) Find how many plutonium-238 nuclei the source contains. [2] (b) Find the initial activity. [2] (c) Find the activity 25 years later. [2]

6. A sample of mass 2.0 µg of a pure nuclide, with molar mass 210 g mol⁻¹, has a measured activity of 3.3 × 10⁸ Bq. (a) Find how many nuclei the sample contains. [2] (b) Find the decay constant, and from it the half-life in days. [3]

7. A detector placed near a decaying source recorded these measured count rates. The background rate, measured with the source removed, was 30 counts per hour.

t / h	0	2	4	6	8	10	12
Measured rate / counts per h	930	738	587	468	375	301	243

(a) Explain why the background rate must be subtracted before taking logarithms. [1] (b) Find the corrected rate at t = 0, 6 and 12 h, and the natural logarithm of each. [3] (c) Use these three points to find the decay constant. [2] (d) Calculate the half-life. [1]

8. A numerical model of decay starts with N = 2000 nuclei and uses λ = 0.30 s⁻¹ with a time step Δt = 0.5 s. (a) Calculate the change in N, and the new value of N, in the first step. [2] (b) Find N after the second and third steps. [2] (c) The exact value of N at t = 1.5 s is given by N = N₀e^(−λt). Calculate it, and the percentage by which the model value is too low. [2] (d) State one way to make the model more accurate. [1]

9. A program simulates 800 nuclei. In each 1.0 s tick, every nucleus still present has a probability of 0.040 of decaying. (a) Calculate the expected number of nuclei left after 10 ticks. [2] (b) Treating each tick as 1.0 s, find the decay constant and the half-life of the simulated nuclide. [2] (c) One run leaves 519 nuclei after 10 ticks. Explain why this differs from your answer to (a), and why a real sample of 10¹⁵ nuclei shows far smaller relative differences. [2]

10. A luminous sign contains 0.30 µg of tritium. Tritium has a molar mass of 3.02 g mol⁻¹ and a half-life of 12.3 years. (a) Calculate the number of tritium nuclei. [2] (b) Find the decay constant in s⁻¹ and the initial activity. [3] (c) Calculate the activity after 5.0 years. [2] (d) Find how many years pass before the activity is 60% of its initial value. [2]

11. An industrial radiography source contains iridium-192, with a half-life of 73.8 days and a molar mass of 192 g mol⁻¹. When new, its activity is 3.7 × 10¹² Bq. (a) Calculate the decay constant in s⁻¹. [2] (b) Calculate the number of iridium-192 nuclei in the new source. [1] (c) Calculate the mass of iridium-192 in the source, in mg. [2] (d) The source must be replaced when its activity falls to 2.0 × 10¹¹ Bq. Calculate the time, in days, after which it must be replaced. [3]

Answers

1. (a) Whether it will decay in the next moment, so which nucleus goes next and when [1]. (b) The chance, per unit time, that one given nucleus decays [1]. (c) T½ = ln2/λ and λ is a property of the nuclide, so N₀ does not appear in it [1]. Examiner insight: A good answer on randomness says what cannot be predicted for one nucleus.

2. (a) T½ = 3.8 × 24 × 3600 = 3.28 × 10⁵ s [1]. λ = 0.693 ÷ 3.28 × 10⁵ = 2.1 × 10⁻⁶ s⁻¹ [1]. (b) A = λN [1] = 2.11 × 10⁻⁶ × 2.4 × 10¹² = 5.1 × 10⁶ Bq [1]. Examiner insight: Show the days-to-seconds conversion line.

3. (a) λt = 0.018 × 48 = 0.864 [1]. N = 6.0 × 10¹⁰ × e^(−0.864) = 2.5 × 10¹⁰ [1]. (b) Gradient = −λN₀ [1] = −(0.018 ÷ 3600) × 6.0 × 10¹⁰ = −3.0 × 10⁵ nuclei per second [1]. Examiner insight: In (b) the sign and the unit both matter.

4. λ = 0.693 ÷ 2.4 = 0.289 h⁻¹ [1]. A₀/A = 7.5 × 10⁵ ÷ 1.2 × 10⁴ = 62.5 [1]. ln(62.5) = 4.135 [1]. t = 4.135 ÷ 0.289 = 14 h [1]. Examiner insight: Inverting the ratio gives a negative time. Check the size against whole half-lives: 2⁶ = 64 is close to 62.5, so about 6 × 2.4 = 14.4 h.

5. (a) Moles = 0.50 × 10⁻³ ÷ 238 [1]. N = 2.10 × 10⁻⁶ × 6.02 × 10²³ = 1.3 × 10¹⁸ [1]. (b) λ = 0.693 ÷ (87.7 × 3.156 × 10⁷) = 2.50 × 10⁻¹⁰ s⁻¹ [1]. A = λN = 2.50 × 10⁻¹⁰ × 1.26 × 10¹⁸ = 3.2 × 10⁸ Bq [1]. (c) 25 ÷ 87.7 = 0.285 half-lives, so A/A₀ = 2^(−0.285) = 0.821 [1]. A = 3.17 × 10⁸ × 0.821 = 2.6 × 10⁸ Bq [1]. Examiner insight: Convert milligrams to grams before dividing by molar mass, and keep extra digits in λ until the final answer.

6. (a) Moles = 2.0 × 10⁻⁶ ÷ 210 [1]. N = 9.52 × 10⁻⁹ × 6.02 × 10²³ = 5.7 × 10¹⁵ [1]. (b) λ = A/N = 3.3 × 10⁸ ÷ 5.73 × 10¹⁵ = 5.8 × 10⁻⁸ s⁻¹ [1]. T½ = 0.693 ÷ 5.76 × 10⁻⁸ = 1.2 × 10⁷ s [1]. Dividing by 86 400 gives 1.4 × 10² days [1]. Examiner insight: Rearrange A = λN for λ first, then use the half-life equation.

7. (a) Measured rate = source rate + background. ln of a sum is not the sum of the ln values, so the graph would curve [1]. (b) Corrected rates are 900, 438 and 213 counts per hour [1]. The ln values are 6.802, 6.082 and 5.361 [2]. (c) Gradient = (5.361 − 6.802) ÷ 12 = −0.120 h⁻¹ [1], so λ = 0.120 h⁻¹ [1]. (d) T½ = 0.693 ÷ 0.120 = 5.8 h [1]. Examiner insight: Use points far apart on the line; the gradient is negative and λ is its magnitude.

8. (a) ΔN = −λNΔt = −0.30 × 2000 × 0.5 = −300 [1]. New N = 1700 [1]. (b) Step 2: ΔN = −0.30 × 1700 × 0.5 = −255, so N = 1445 [1]. Step 3: ΔN = −0.30 × 1445 × 0.5 = −216.75, so N = 1228 [1]. (c) Exact N = 2000e^(−0.45) = 1275 [1]. Model is too low by (1275 − 1228) ÷ 1275 = 3.7% [1]. (d) Use a smaller time step [1]. Examiner insight: The model uses the rate at the start of each step, which overestimates the loss.

9. (a) Each tick leaves 0.96 of the nuclei [1]. N = 800 × 0.96¹⁰ = 532 [1]. (b) e^(−λ) = 0.96, so λ = −ln0.96 = 0.041 s⁻¹ [1]. T½ = 0.693 ÷ 0.0408 = 17 s [1]. (c) Each nucleus decays by chance, so the count scatters about the expected value, here by about 13 either way [1]. With 10¹⁵ nuclei the scatter is about √N, a tiny fraction of N, so the curve is smooth [1]. Examiner insight: Link the scatter to the small number of nuclei, not to a faulty program.

10. (a) Moles = 0.30 × 10⁻⁶ ÷ 3.02 [1]. N = 9.93 × 10⁻⁸ × 6.02 × 10²³ = 6.0 × 10¹⁶ [1]. (b) T½ = 12.3 × 3.156 × 10⁷ = 3.88 × 10⁸ s [1]. λ = 0.693 ÷ 3.88 × 10⁸ = 1.8 × 10⁻⁹ s⁻¹ [1]. A = λN = 1.79 × 10⁻⁹ × 5.98 × 10¹⁶ = 1.1 × 10⁸ Bq [1]. (c) 5.0 ÷ 12.3 = 0.407 half-lives, so A/A₀ = 2^(−0.407) = 0.754 [1]. A = 1.07 × 10⁸ × 0.754 = 8.1 × 10⁷ Bq [1]. (d) λ in years = 0.693 ÷ 12.3 = 0.0564 y⁻¹, and ln(1/0.6) = 0.511 [1]. t = 0.511 ÷ 0.0564 = 9.1 years [1]. Examiner insight: In (d), using λ in y⁻¹ avoids converting to seconds and back.

11. (a) T½ = 73.8 × 86 400 = 6.38 × 10⁶ s [1]. λ = 0.693 ÷ 6.38 × 10⁶ = 1.09 × 10⁻⁷ s⁻¹ [1]. (b) N = A/λ = 3.7 × 10¹² ÷ 1.09 × 10⁻⁷ = 3.4 × 10¹⁹ [1]. (c) Moles = 3.40 × 10¹⁹ ÷ 6.02 × 10²³ = 5.65 × 10⁻⁵ [1]. Mass = 5.65 × 10⁻⁵ × 192 = 1.09 × 10⁻² g = 11 mg [1]. (d) A₀/A = 3.7 × 10¹² ÷ 2.0 × 10¹¹ = 18.5 [1]. λ = 0.693 ÷ 73.8 = 9.39 × 10⁻³ day⁻¹ and ln(18.5) = 2.918 [1]. t = 2.918 ÷ 9.39 × 10⁻³ = 3.1 × 10² days [1]. Examiner insight: Set the chain out line by line so each result is visible.

Where marks are usually lost

• Substituting a half-life in days, hours or years into an equation that needs seconds.
• Using log to base 10 where the exponential needs natural logs.
• Inverting the activity ratio, which makes the time negative.
• Not converting micrograms and milligrams to grams before using molar mass.
• Taking ln of measured rather than corrected count rates.
• Confusing the decay constant, a probability per unit time, with the activity of a sample.
• Rounding λ early.

Next steps

• Revision notes
• Study guide
• OxfordAQA Physics hub and printable checklist
• All free 10-minute diagnostics
• Book a free trial class: /trial/

Official syllabus

OxfordAQA International AS and A-level Physics (9630), specification version 4.4, for exams from May/June 2020 onwards, published by Oxford International AQA Examinations: section 3.9.2 Exponential changes in radioactivity (International A-level only).