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Although we cannot predict when any individual nucleus will decay, the mathematics of large numbers gives us powerful tools for describing radioactive decay quantitatively. The key concepts are activity, decay constant, and half-life — and they are all connected by elegant exponential relationships.
The activity of a radioactive source is the number of nuclei that decay per unit time. It is measured in becquerels (Bq), where 1 Bq = 1 decay per second.
Activity depends on two things: how many undecayed nuclei are present, and how quickly each one is likely to decay. This gives us the fundamental equation:
A = λN
where:
As nuclei decay, N decreases, so activity decreases over time. A source does not have a constant activity — it weakens as the radioactive atoms are used up.
| Unit | Equivalent |
|---|---|
| 1 Bq | 1 decay per second |
| 1 kBq | 10³ Bq |
| 1 MBq | 10⁶ Bq |
| 1 GBq | 10⁹ Bq |
| 1 Ci (curie, older unit) | 3.7 × 10¹⁰ Bq |
The decay constant λ (lambda) represents the probability that any given nucleus will decay per unit time. It is a fixed property of a particular isotope and does not change with external conditions.
The units of λ are s⁻¹ (or sometimes min⁻¹, hr⁻¹, yr⁻¹ depending on context).
Since each decay removes one nucleus from the undecayed population, and the rate of decay is proportional to the number remaining (A = λN), the result is exponential decay:
N = N₀ e^(−λt)
where:
Because A = λN, activity follows the same exponential pattern:
A = A₀ e^(−λt)
where A₀ = λN₀ is the initial activity.
These equations describe the same fundamental process: the number of undecayed nuclei (and therefore the activity) decreases exponentially with time.
Starting from the statement that the rate of decay is proportional to the number of undecayed nuclei:
dN/dt = −λN
This is a first-order differential equation. Separating variables:
dN/N = −λ dt
Integrating both sides:
ln(N) = −λt + constant
At t = 0, N = N₀, so constant = ln(N₀):
ln(N) − ln(N₀) = −λt
ln(N/N₀) = −λt
N = N₀ e^(−λt)
The half-life (t½) is the time taken for half the undecayed nuclei in a sample to decay. Equivalently, it is the time for the activity to fall to half its initial value.
Starting from N = N₀ e^(−λt), we set N = N₀/2 at t = t½:
N₀/2 = N₀ e^(−λt½)
1/2 = e^(−λt½)
ln(1/2) = −λt½
−ln 2 = −λt½
t½ = ln 2 / λ
Since ln 2 ≈ 0.693:
t½ = 0.693 / λ
This is one of the most important relationships in nuclear physics. It connects the half-life (which is easy to measure experimentally) to the decay constant (which appears in the mathematical equations).
Half-lives vary enormously across different isotopes:
| Isotope | Half-life | Decay Constant | Use / Context |
|---|---|---|---|
| Polonium-214 | 164 microseconds | 4.23 × 10³ s⁻¹ | Alpha decay chain |
| Radon-220 | 55.6 seconds | 1.25 × 10⁻² s⁻¹ | Thorium decay product |
| Iodine-131 | 8.02 days | 1.00 × 10⁻⁶ s⁻¹ | Thyroid treatment |
| Cobalt-60 | 5.27 years | 4.17 × 10⁻⁹ s⁻¹ | Radiotherapy |
| Strontium-90 | 28.8 years | 7.63 × 10⁻¹⁰ s⁻¹ | Environmental contaminant |
| Carbon-14 | 5,730 years | 3.84 × 10⁻¹² s⁻¹ | Archaeological dating |
| Uranium-238 | 4.47 billion years | 4.92 × 10⁻¹⁸ s⁻¹ | Geological dating |
Half-life can be determined from a graph of activity (or count rate, or number of nuclei) against time:
On a graph of ln(A) against t, the decay appears as a straight line with gradient −λ. This is a useful technique because:
For problems involving whole numbers of half-lives, you can use repeated halving without the exponential equation:
After 1 half-life: N = N₀/2 After 2 half-lives: N = N₀/4 After 3 half-lives: N = N₀/8 After n half-lives: N = N₀/2ⁿ
Example: A sample has an initial activity of 6400 Bq and a half-life of 3 hours. What is its activity after 12 hours?
12 hours = 4 half-lives Activity = 6400 / 2⁴ = 6400 / 16 = 400 Bq
Problem: A sample of phosphorus-32 (t½ = 14.3 days) has an initial activity of 2.4 MBq. What is the activity after 20 days?
20 days is not a whole number of half-lives, so we must use the exponential equation.
Step 1: Find λ. λ = ln 2 / t½ = 0.693 / (14.3 × 24 × 3600) = 0.693 / 1,235,520 = 5.61 × 10⁻⁷ s⁻¹
Step 2: Convert 20 days to seconds. t = 20 × 24 × 3600 = 1,728,000 s
Step 3: Calculate activity. A = A₀ e^(−λt) = 2.4 × 10⁶ × e^(−5.61 × 10⁻⁷ × 1,728,000) A = 2.4 × 10⁶ × e^(−0.9695) A = 2.4 × 10⁶ × 0.3796 A = 9.11 × 10⁵ Bq = 911 kBq
Problem: A sample of iodine-131 (t½ = 8.0 days) has an activity of 5.0 GBq. How long until the activity falls to 0.1 GBq?
Method: A = A₀ e^(−λt), so t = −(1/λ) ln(A/A₀)
λ = ln 2 / (8.0 × 86400) = 1.003 × 10⁻⁶ s⁻¹
t = −(1/1.003 × 10⁻⁶) × ln(0.1/5.0) t = −(9.97 × 10⁵) × ln(0.02) t = −(9.97 × 10⁵) × (−3.912) t = 3.90 × 10⁶ s = 45.1 days
Check: 45.1 / 8.0 = 5.64 half-lives. After 5 half-lives the ratio would be 1/32 = 0.03125, after 6 half-lives 1/64 = 0.015625. Since 0.02 is between these, ~5.6 half-lives is reasonable. ✔
Carbon-14 is continuously produced in the upper atmosphere by cosmic ray bombardment of nitrogen:
¹⁴₇N + ¹₀n → ¹⁴₆C + ¹₁p
Living organisms constantly exchange carbon with the environment through respiration, eating, and photosynthesis, maintaining a roughly constant ratio of C-14 to C-12 in their tissues. When an organism dies, it stops taking in new carbon. The C-14 already present decays with a half-life of 5,730 years while the stable C-12 remains unchanged.
By measuring the ratio of C-14 to C-12 in a sample and comparing it to the ratio in living organisms, the time since the organism died can be calculated using:
N = N₀ e^(−λt) → t = −(1/λ) ln(N/N₀)
Carbon dating is reliable for ages up to about 50,000 years (roughly 9 half-lives). Beyond that, the remaining C-14 activity is too low to measure accurately above background.
Problem: A wooden tool found at an archaeological site has a C-14 activity of 0.18 Bq per gram of carbon. Living wood has an activity of 0.23 Bq per gram. Estimate the age of the tool.
λ = ln 2 / 5730 = 1.21 × 10⁻⁴ yr⁻¹
t = −(1/λ) ln(A/A₀) = −(1/1.21 × 10⁻⁴) × ln(0.18/0.23)
t = −(8264) × ln(0.7826) = −(8264) × (−0.2451) = 2025 years
Activity (A = λN) is the rate of decay in becquerels. The decay constant λ represents the probability of decay per unit time for each nucleus. Both N and A follow exponential decay curves. The half-life t½ = ln 2/λ is the time for half the nuclei to decay. Half-life can be determined graphically and is used in applications ranging from medical treatment to archaeological dating. For non-integer half-lives, use the full exponential equation; for whole-number half-lives, repeated halving is faster and less error-prone.
Edexcel 9PH0 specification Topic 11 — Nuclear and Particle Physics covers the radioactive decay law, the relationship A = λN, the use of the exponential equation N = N₀e^(-λt), the definition of half-life, and the connection t½ = ln 2 / λ (refer to the official specification document for exact wording). This material is examined principally on 9PH0-02 Paper 2 (Physics on the Move), but exponential and logarithmic manipulation is reused on Topic 5 (Capacitor discharge), Topic 13 (Oscillations — exponential damping), and Topic 14 (Astrophysics — luminosity decay of variable stars). Practical handling of count-rate data underpins Core Practical 15 on the inverse-square law and absorption of gamma radiation, and the algebra here also feeds the uniform mathematical-skills appendix of the specification, which expects fluent use of natural logarithms in the form ln(A/A₀) = -λt. The Edexcel data and formulae booklet supplies A = λN and N = N₀e^(-λt); it does not supply t½ = ln 2 / λ as a stand-alone line, so candidates are expected to derive it.
Question (8 marks):
A sealed sample of caesium-137 has an initial activity of 6.4 × 10⁴ Bq. The half-life of caesium-137 is approximately 30.0 years.
(a) Determine the decay constant λ in s⁻¹. (2)
(b) Calculate the number N₀ of caesium-137 nuclei initially present. (2)
(c) Predict the activity of the sample after 75 years. (4)
Solution with mark scheme:
(a) Step 1 — convert the half-life to seconds.
t1/2=30.0×365.25×24×3600≈9.47×108s
Step 2 — apply λ=ln2/t1/2.
λ=0.693/(9.47×108)≈7.32×10−10s−1
M1 — correct rearrangement and substitution into λ = ln 2 / t½. A1 — answer in s⁻¹ to 2–3 sf. Common slip: leaving the half-life in years, which generates a λ that is dimensionally inconsistent with A measured in Bq.
(b) Step 1 — rearrange A = λN.
N0=A0/λ=6.4×104/7.32×10−10
M1 — correct rearrangement of A = λN to give N = A / λ. A1 — N0≈8.7×1013 nuclei.
(c) Step 1 — convert 75 years to seconds.
t=75×365.25×24×3600≈2.37×109s
Step 2 — apply the activity form of the decay law.
Because A ∝ N, the activity satisfies A=A0e−λt.
λt=7.32×10−10×2.37×109≈1.735
M1 — substituting consistent units into λt. M1 — recognising A and N follow the same exponential form.
Step 3 — evaluate.
A=6.4×104×e−1.735≈6.4×104×0.1764≈1.13×104Bq
A1 — numerical answer to 2–3 sf. A1 — units (Bq) and a sensible comment that 75 years ≈ 2.5 half-lives, so the activity should fall to roughly (1/2)2.5≈0.177 of its initial value, in agreement with the calculation.
Total: 8 marks (M4 A4 as shown).
Question (6 marks): A school sample of an unknown beta emitter is monitored with a Geiger–Müller tube. After background subtraction, the corrected count rate is 480 s⁻¹ at the start of the experiment and 132 s⁻¹ exactly four hours later.
(a) Determine the decay constant λ for this isotope, stating an assumption you have made. (3)
(b) Hence estimate the half-life and comment on whether the isotope is suitable for a school experiment lasting one lesson (about one hour). (3)
Mark scheme decomposition by AO:
(a)
(b)
Total: 6 marks split AO1 = 3, AO2 = 1, AO3 = 2. This is a typical 9PH0-02 short-data-handling item: most marks are AO1/AO2 procedural, with the AO3 marks reserved for the contextual judgement about practicality.
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