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

These notes condense 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, and section 3.3.3 says the exponential decay equations are required at International A2 only. For full explanations and worked examples, read the study guide.

Use the OxfordAQA Physics hub and the printable checklist to track your revision. Test yourself with the practice set and the free 10-minute diagnostics. Earlier decay ideas (modes, half-life by whole numbers) are in the decay modes and half-life guide.

Key ideas

• Decay is spontaneous (not affected by temperature, pressure or chemical state) and random (you cannot predict which nucleus decays, or when).
• For a given nuclide, each nucleus has one fixed chance of decaying in each second, whatever its age.
• The decay constant λ is that probability per unit time. Unit: s⁻¹, h⁻¹, y⁻¹.
• Large samples follow smooth curves. Counts for a single measurement scatter by about √C for a count of C.

Formulas

Quantity	Equation	Notes
Decay rule	ΔN/Δt = −λN	Negative: N falls
Activity	A = −ΔN/Δt = λN	Unit becquerel, Bq = s⁻¹
Number of nuclei	N = N₀e^(−λt)	N₀ at t = 0
Activity law	A = A₀e^(−λt)	Also true for corrected count rate
Half-life	T½ = ln2/λ	ln2 = 0.693; independent of N₀
Whole half-lives	N = N₀(½)ⁿ	n = t/T½
Nuclei in a mass	N = (m/M) N_A	m in g, M in g mol⁻¹
Time to reach A	t = (1/λ) ln(A₀/A)	Natural log
Straight line	ln A = ln A₀ − λt	Gradient −λ, intercept ln A₀
Numerical step	N(new) = N(old)(1 − λΔt)	Small Δt only

N_A is 6.02 × 10²³ mol⁻¹ in the calculations on this page. In an exam question, use the values provided with the question.

Must-know distinctions

• λ and A. λ is a probability per unit time for one nucleus, the same for every sample of the nuclide. A = λN is the decays per second for a particular sample, and falls as N falls.
• Activity and count rate. The detector sees only part of the decays, and also records background. Corrected count rate is proportional to activity, so it follows the same law.
• T½ and λ. T½ is long when λ is small. Doubling λ halves T½.
• N–t graph and ln A–t graph. The first is a curve with gradient −λN. The second is a straight line with gradient −λ.
• ln and log. The exponential law needs ln. Base 10 logs give gradient −λ log₁₀e, so λ = −gradient × 2.303.

Methods in steps

Activity or nuclei after a time t

1. Convert T½ to λ = ln2/T½, keeping the time unit of the question, or convert to seconds if the answer must be in Bq.
2. Convert t to the same unit.
3. Work out λt, then e^(−λt).
4. Multiply by N₀ or A₀.
5. Sense check: is t about a whole number of half-lives? If so, N ≈ N₀(½)ⁿ.

Time to reach a given activity

1. Form the ratio A₀/A (bigger than 1).
2. Take ln of the ratio.
3. Divide by λ in the matching unit.

Nuclei from a mass

1. Convert mass to grams.
2. Divide by M to get moles.
3. Multiply by N_A.
4. Use A = λN if the activity is needed.

Half-life from a log graph

1. Subtract the background rate from each reading.
2. Take ln of each corrected value.
3. Plot ln against t and draw a best-fit line.
4. Use a large triangle to find the gradient (negative).
5. λ = −gradient, then T½ = ln2/λ.

Half-life from a decay curve

1. Subtract the background.
2. Sketch a smooth curve through the points.
3. Measure how long the value takes to halve, from two or more starting heights.
4. If these agree, the decay is exponential, and the half-life is their mean.

Small worked reminders

Fraction left. A nuclide has λ = 0.12 day⁻¹. After 10 days, N/N₀ = e^(−1.2) = 0.30. Its half-life is 0.693 ÷ 0.12 = 5.8 days.

Activity from nuclei. A sample has 2.0 × 10¹⁴ nuclei and λ = 4.0 × 10⁻⁸ s⁻¹. Then A = 4.0 × 10⁻⁸ × 2.0 × 10¹⁴ = 8.0 × 10⁶ Bq.

Number of nuclei in a mass. 5.0 µg of a nuclide with M = 150 g mol⁻¹ contains (5.0 × 10⁻⁶ ÷ 150) × 6.02 × 10²³ = 2.0 × 10¹⁶ nuclei.

Numerical model. N₀ = 800, λ = 0.050 s⁻¹, Δt = 2.0 s. Each step multiplies by 1 − 0.050 × 2.0 = 0.90, so N = 720 after one step and 648 after two. The exact value at 4.0 s is 800e^(−0.20) = 655, so the model is slightly low.

Time to fall to a fraction. With λ = 0.12 day⁻¹, the time for N to fall to 5% of N₀ is t = ln(20) ÷ 0.12 = 2.996 ÷ 0.12 = 25 days. That is about 4.3 half-lives, and (½)^4.3 is about 0.05, so the answer is sensible.

Log graph. A best-fit line on a graph of ln A against t passes through (0, 7.20) and (14 days, 5.10). The gradient is (5.10 − 7.20) ÷ 14 = −0.150 day⁻¹. So λ = 0.150 day⁻¹ and T½ = 0.693 ÷ 0.150 = 4.6 days. The starting activity is A₀ = e^7.20, which is about 1.3 × 10³ in whatever unit the readings used.

Which representation to use. Use a decay curve when you want a half-life from a quick read, because only the halving time is needed. Use a log graph when you want the best value of λ from many points, because a straight line averages out the scatter. Use the equation when you are given values and asked for one unknown.

Quick self-test

1. What does the decay constant λ represent?
2. A nuclide has a half-life of 4.0 h. Find λ in h⁻¹ and in s⁻¹.
3. A sample has λ = 0.25 h⁻¹. What fraction of the nuclei remains after 6.0 h?
4. A nuclide has λ = 3.0 × 10⁻⁴ s⁻¹. Find its half-life in seconds and in minutes.
5. A sample has 5.0 × 10¹² undecayed nuclei and λ = 2.0 × 10⁻⁶ s⁻¹. Find its activity.
6. A nuclide has a half-life of 8.0 h. How long does its activity take to fall to 10% of its starting value?
7. How many nuclei are in 2.0 mg of a nuclide with molar mass 32 g mol⁻¹?
8. A graph of ln A against t, with t in days, has gradient −0.080 day⁻¹. Find the half-life.
9. A source has an activity of 480 Bq. Its half-life is 6.0 h. Find its activity 21 h later.
10. A numerical model starts with N = 500, uses λ = 0.10 s⁻¹ and Δt = 2.0 s. Find N after two steps.
11. Why must you subtract the background before plotting ln of a count rate?
12. Why does a graph of ln A against t give a straight line?

Answers

1. The probability that a given nucleus decays per unit time. It is constant for a nuclide.
2. λ = 0.693 ÷ 4.0 = 0.17 h⁻¹; ÷ 3600 gives 4.8 × 10⁻⁵ s⁻¹.
3. e^(−0.25 × 6.0) = e^(−1.5) = 0.22.
4. T½ = 0.693 ÷ 3.0 × 10⁻⁴ = 2.3 × 10³ s, which is 39 minutes.
5. A = λN = 2.0 × 10⁻⁶ × 5.0 × 10¹² = 1.0 × 10⁷ Bq.
6. t = (T½/ln2) × ln(10) = 8.0 × 3.32 = 27 h (about 3.3 half-lives).
7. (2.0 × 10⁻³ ÷ 32) × 6.02 × 10²³ = 3.8 × 10¹⁹.
8. λ = 0.080 day⁻¹, so T½ = 0.693 ÷ 0.080 = 8.7 days.
9. 21 h is 3.5 half-lives: 480 × (½)^3.5 = 42 Bq.
10. Each step multiplies by 1 − 0.10 × 2.0 = 0.80, giving 400, then 320.
11. Measured rate = source rate + background. The ln of a sum is not the sum of the ln values, so with background left in the points curve and the gradient is not −λ.
12. Taking ln of A = A₀e^(−λt) gives ln A = ln A₀ − λt, which is of the form y = mx + c.

Where marks are usually lost

• Using the exponential law with λ in one time unit and t in another.
• Writing the exponent as e^(λt) with the minus sign missing, so the activity grows.
• Quoting a negative decay constant, or a positive gradient from a ln A graph.
• Giving a time for activity to fall from the ratio A/A₀ instead of A₀/A, which flips the sign of the log.
• Forgetting to convert micrograms or milligrams to grams before finding N.
• Reading a half-life from one starting point on a noisy curve.
• Plotting ln of measured, not corrected, count rate.
• Saying the decay rate of a sample is “constant” when it is the decay probability that is constant.
• Not showing the substitution before the answer when asked to show a result.

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).