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Radioactive decay is a random process, but when we have a large number of nuclei the statistical behaviour becomes highly predictable. This lesson covers the exponential decay law, the decay constant, activity, half-life, and their applications including carbon dating. This material is central to AQA section 3.8.1 and is examined in Paper 2.
Spec mapping (AQA 7408): This lesson develops the exponential decay law N = N₀ e^(−λt) and its activity form A = A₀ e^(−λt), introduces the decay constant λ as the probability of decay per nucleus per unit time, relates it to the half-life via t½ = ln 2 / λ, and applies the formalism to half-life experiments, carbon-14 dating, and background-corrected count-rate measurements. It maps to AQA 7408 section 3.8.1.3 (the radioactive decay equation, half-life, applications). (Refer to the official AQA specification document for exact wording.)
Synoptic links: (1) Exponential-decay maths is identical in form to RC-circuit discharge (section 3.7.4) and to thermal-equilibrium cooling — students who have met dQ/dt = −Q/τ in capacitors should recognise the kinship. (2) The decay constant connects to the more advanced activity-as-rate view in required practical 7 (RP7), where the inverse-square decline of gamma intensity with distance is studied separately from the time-dependent decay. (3) The natural-logarithm linearisation ln N = ln N₀ − λt is a standard data-analysis technique reused across mechanics (drag), thermal physics (cooling), and electronics (RC discharge); the gradient-of-ln-y-against-x method is a transferable skill.
If we start with N₀ undecayed nuclei at time t = 0, the number remaining undecayed at time t is:
N = N₀ e^(−λt)
where λ (lambda) is the decay constant, measured in s⁻¹ (or sometimes min⁻¹ or yr⁻¹). The decay constant represents the probability that any given nucleus will decay per unit time.
This equation can also be written:
N/N₀ = e^(−λt)
The graph of N against t is a smooth exponential curve that starts at N₀ and asymptotically approaches zero.
The activity (A) of a radioactive source is the number of decays per unit time. It is measured in becquerels (Bq), where 1 Bq = 1 decay per second.
Activity is directly proportional to the number of undecayed nuclei:
A = λN
Since N decreases exponentially, so does the activity:
A = A₀ e^(−λt)
where A₀ = λN₀ is the initial activity.
Similarly, the count rate (C) measured by a detector (which is proportional to activity) follows the same exponential law:
C = C₀ e^(−λt)
Key Point: Activity, the number of undecayed nuclei, and the count rate all follow the same exponential decay pattern. Any of them can be used to determine the half-life or decay constant.
The half-life (t½) is the time taken for:
Starting from N = N₀ e^(−λt), at time t = t½ we have N = N₀/2:
N₀/2 = N₀ e^(−λt½)
1/2 = e^(−λt½)
Taking natural logarithms: ln(1/2) = −λt½
−ln 2 = −λt½
t½ = ln 2 / λ ≈ 0.693 / λ
This is one of the most important equations in nuclear physics. It relates the measurable quantity (half-life) to the fundamental constant of the decay (decay constant).
Cobalt-60 has a half-life of 5.27 years. Find the decay constant in s⁻¹.
First convert half-life to seconds: t½ = 5.27 × 365.25 × 24 × 3600 = 1.663 × 10⁸ s
λ = ln 2 / t½ = 0.6931 / 1.663 × 10⁸
λ = 4.17 × 10⁻⁹ s⁻¹
A sample initially contains 5.00 × 10²⁰ atoms of cobalt-60. What is the initial activity?
A₀ = λN₀ = 4.17 × 10⁻⁹ × 5.00 × 10²⁰
A₀ = 2.08 × 10¹² Bq (or 2.08 TBq)
How many cobalt-60 atoms remain after 15.0 years?
Convert time: t = 15.0 × 365.25 × 24 × 3600 = 4.734 × 10⁸ s
N = N₀ e^(−λt) = 5.00 × 10²⁰ × e^(−4.17 × 10⁻⁹ × 4.734 × 10⁸)
N = 5.00 × 10²⁰ × e^(−1.974)
N = 5.00 × 10²⁰ × 0.1390
N = 6.95 × 10¹⁹ atoms
Alternatively, using half-lives: 15.0 years / 5.27 years ≈ 2.847 half-lives. N = 5.00 × 10²⁰ × (1/2)^2.847 = 5.00 × 10²⁰ × 0.1390 ≈ 6.95 × 10¹⁹ ✓
A detector records a count rate of 480 counts per minute from a radioactive source. After 6.0 hours, the count rate has fallen to 60 counts per minute. Find the half-life.
C = C₀ e^(−λt)
60 = 480 e^(−λ × 6.0)
60/480 = e^(−6.0λ)
0.125 = e^(−6.0λ)
ln(0.125) = −6.0λ
−2.079 = −6.0λ
λ = 0.347 h⁻¹
t½ = ln 2 / λ = 0.693 / 0.347
t½ = 2.0 hours
Check: 6.0 hours = 3 half-lives. 480 → 240 → 120 → 60 ✓
This gives the characteristic exponential decay curve. The half-life can be read directly by finding the time at which the value has halved.
Taking logarithms of N = N₀ e^(−λt):
ln N = ln N₀ − λt
This is a straight line with gradient −λ and y-intercept ln N₀. This method is useful because:
Exam Tip: In practical exam questions, you may be given data and asked to plot ln(count rate) against time to obtain a straight line. The gradient of this line gives −λ, from which you can calculate the half-life. Always subtract background count rate before taking logarithms.
Carbon-14 is produced in the upper atmosphere when cosmic-ray neutrons collide with nitrogen-14:
¹⁴₇N + ¹₀n → ¹⁴₆C + ¹₁p
Carbon-14 is incorporated into CO₂, which is absorbed by living organisms through photosynthesis and the food chain. While an organism is alive, the ratio of ¹⁴C to ¹²C in its body remains roughly constant (in equilibrium with the atmosphere). When the organism dies, it stops absorbing carbon, and the ¹⁴C decays with a half-life of 5730 years:
¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ⁰₀ν̄ₑ
By measuring the ratio of ¹⁴C to ¹²C in a sample (or equivalently, measuring the activity per gram of carbon), the time since death can be determined.
A sample of ancient wood has an activity of 0.160 Bq per gram of carbon. Living wood has an activity of 0.255 Bq per gram of carbon. The half-life of carbon-14 is 5730 years. Estimate the age of the sample.
A = A₀ e^(−λt)
λ = ln 2 / 5730 = 1.210 × 10⁻⁴ yr⁻¹
0.160 = 0.255 e^(−1.210 × 10⁻⁴ × t)
0.160/0.255 = e^(−1.210 × 10⁻⁴ × t)
0.6275 = e^(−1.210 × 10⁻⁴ × t)
ln(0.6275) = −1.210 × 10⁻⁴ × t
−0.4659 = −1.210 × 10⁻⁴ × t
t = 0.4659 / 1.210 × 10⁻⁴
t ≈ 3850 years
Common Misconception: Carbon dating is only reliable for samples up to about 50 000–60 000 years old (roughly 10 half-lives). Beyond this, the remaining ¹⁴C activity is too low to measure accurately. For older samples, other radiometric dating methods (e.g., potassium-argon dating) are used.
In experiments, a detector will always pick up some background radiation from natural sources (cosmic rays, rocks, radon gas, etc.). This must be subtracted before analysing the decay data:
Corrected count rate = Measured count rate − Background count rate
The background count rate should be measured separately (with no source present) and subtracted from all readings before applying the exponential decay law.
It is worth being explicit about the distinction between activity (decays per second emitted by the source, an intrinsic property) and count rate (events per second registered by a detector, which depends on the detector's geometry, efficiency, and distance from the source).
For an isotropic point source of activity A, a detector at distance r with active area a registers (after background subtraction):
C = A × (a / 4π r²) × η
where η is the intrinsic detection efficiency (typically 0.5 % to a few % for a GM tube detecting MeV γ photons; near 100 % for a NaI(Tl) scintillator at low energy).
Worked numbers: a Co-60 source of activity 200 kBq at 20 cm from a GM tube with active area 5 cm² and γ efficiency 1 % gives:
C = 200 × 10³ × (5 × 10⁻⁴) / (4π × 0.20²) × 0.01 ≈ 200 × 10³ × 5 × 10⁻⁴ / 0.503 × 0.01 ≈ 2 counts per second ≈ 120 /min — a moderate count rate.
Examiners often ask students to convert between activity (becquerels) and detector count rate, and to recognise that the count rate is always smaller than the activity (often by a factor of 10² to 10⁴ for school-lab geometries) because the detector only captures a small solid-angle fraction at modest efficiency.
Technetium-99m, used in roughly 30 million medical imaging scans per year worldwide, has a half-life of 6.0 hours. A hospital imaging department receives a fresh dose with an initial activity of 800 MBq.
(a) Number of Tc-99m nuclei in the dose at t = 0.
λ = ln 2 / t½ = 0.693 / (6.0 × 3600 s) = 3.21 × 10⁻⁵ s⁻¹.
N₀ = A₀ / λ = 800 × 10⁶ / 3.21 × 10⁻⁵ = 2.49 × 10¹³ nuclei.
(b) Activity 18 hours after delivery (3 half-lives).
A(18 h) = 800 × (1/2)³ = 100 MBq.
(c) Mass of Tc-99m in the dose at t = 0.
Mass per nucleus = 99 × 1.66 × 10⁻²⁷ kg = 1.64 × 10⁻²⁵ kg.
Total mass m = N₀ × mass per nucleus = 2.49 × 10¹³ × 1.64 × 10⁻²⁵ = 4.08 × 10⁻¹² kg ≈ 4 picograms.
Note how tiny the physical mass is — picograms of radioactive material produce hundreds of MBq of activity because the decay rate per nucleus is very high. This is why nuclear medicine is so dose-efficient: enormous diagnostic information per microgram of administered material.
A GM tube records 86 counts per minute with a radioactive source present. The background count rate is 6 counts per minute. After 20 minutes, the corrected count rate has fallen to 20 counts per minute. Find the half-life.
Corrected initial count rate: C₀ = 86 − 6 = 80 counts min⁻¹
Corrected final count rate: C = 20 counts min⁻¹
C = C₀ e^(−λt)
20 = 80 e^(−20λ)
0.25 = e^(−20λ)
ln(0.25) = −20λ
−1.386 = −20λ
λ = 0.0693 min⁻¹
t½ = 0.693 / 0.0693 = 10.0 minutes
Check: 20 minutes = 2 half-lives. 80 → 40 → 20 ✓
Specimen question modelled on the AQA 7408 paper format (9 marks):
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