A source has a half-life of 8.0\,days. Find its decay constant in day^-1. [2 marks]

WJEC WJEC 2020-style practice: A radioactive source contains 5.0 × 10^18 undecayed nuclei and has a half-life of 15\,hours. Calculate the decay constant, the initial activity, and the activity after 45\,hours.

Decay constant: λ = ln 2t_1/2 = 0.69315 × 3600 = 0.6935.4 × 10^4 = 1.28 × 10^-5\,s^-1. Initial activity: A_0 = λ N_0 = 1.28 × 10^-5 × 5.0 × 10^18 = 6.4 × 10^13\,Bq. After 45\,hours, which is exactly three half-lives, the activity halves three times: A = A_0 × (12)^3 = 6.4 × 10^138 = 8.0 × 10^12\,Bq. Markers reward the decay constant from ln 2 / t_1/2, the activity from A = λ N, and using three half-lives (or the exponential) for the final activity.

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How do we describe random radioactive decay with the decay law and half-life?

Random and spontaneous decay, the decay law, activity, the decay constant and half-life, and the properties of alpha, beta and gamma radiation.

A focused answer to WJEC A-Level Physics Unit 3 nuclear decay, covering radioactive decay as random and spontaneous, the exponential decay law, activity, the decay constant and half-life, and the properties of alpha, beta and gamma radiation.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

WJEC wants you to describe radioactive decay as random and spontaneous, use the exponential decay law and activity, relate the decay constant to half-life, and state the properties of alpha, beta and gamma radiation. The exponential decay law shares its mathematics with capacitor discharge, so the calculation skills transfer, and the half-life questions are dependable exam marks.

The answer

Random and spontaneous decay

Although individual decays are unpredictable, a large sample contains so many nuclei that the average behaviour is highly regular, which is what gives the smooth exponential decay curve.

The decay law and activity

$N$ is the number of undecayed nuclei, $\lambda$ the decay constant (the probability of decay per unit time), and the activity $A$ is the rate of decay, measured in becquerels (one becquerel is one decay per second).

Half-life

Activity after a non-integer number of half-lives

A source has an initial activity of $8.0 \times 10^{5}\,\text{Bq}$ and a half-life of $6.0\,\text{hours}$ . Find its activity after $10\,\text{hours}$ .

step Find the decay constant

$\lambda = \dfrac{\ln 2}{t_{1/2}} = \dfrac{0.693}{6.0} = 0.1155\,\text{hour}^{-1}$ .

step Apply the decay law to the activity

$A = A_0 e^{-\lambda t} = 8.0 \times 10^{5} \times e^{-0.1155 \times 10}$ .

step Evaluate

$e^{-1.155} = 0.315$ , so $A = 8.0 \times 10^{5} \times 0.315 = 2.5 \times 10^{5}\,\text{Bq}$ .

Alpha, beta and gamma

Radiation	Nature	Ionising	Penetration
Alpha	Helium nucleus	Most	Stopped by paper
Beta	Fast electron	Medium	Stopped by aluminium
Gamma	High-energy photon	Least	Reduced by thick lead

Examples in context

Example 1. Radiocarbon dating: Living things take in carbon-14, which decays with a half-life of about $5700$ years. Once an organism dies, no new carbon-14 is absorbed, so the activity falls exponentially. Measuring the remaining activity and applying $A = A_0 e^{-\lambda t}$ lets archaeologists date a sample tens of thousands of years old.
Example 2. Medical tracers: A technetium-99m tracer used in scans has a half-life of about $6$ hours, short enough that it decays away quickly to limit the patient's dose, but long enough to complete the scan. Its activity, $A = \lambda N$ , is chosen high enough to image clearly while the rapid exponential decay keeps the total exposure low.
Example 3. Smoke detectors: A household smoke detector contains a tiny amount of americium-241, an alpha emitter with a half-life of about $430$ years. The alpha particles ionise the air in a small chamber, allowing a steady current to flow; smoke particles disrupt this current and trigger the alarm. The long half-life means the source stays nearly constant over the detector's lifetime, and the short range of alpha radiation keeps it safe inside the unit.

Try this

Q1. A source has a half-life of $8.0\,\text{days}$ . Find its decay constant in $\text{day}^{-1}$ . [2 marks]

Cue. $\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{8.0} \approx 0.087\,\text{day}^{-1}$ .

Q2. State two ways in which radioactive decay is described as random and spontaneous. [2 marks]

Cue. Random: cannot predict which nucleus decays; spontaneous: rate unaffected by external conditions.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20206 marksA radioactive source contains

5.0 \times 10^{18}

undecayed nuclei and has a half-life of

15\,\text{hours}

. Calculate the decay constant, the initial activity, and the activity after

45\,\text{hours}

Show worked answer →

Decay constant: $\lambda = \dfrac{\ln 2}{t_{1/2}} = \dfrac{0.693}{15 \times 3600} = \dfrac{0.693}{5.4 \times 10^{4}} = 1.28 \times 10^{-5}\,\text{s}^{-1}$ .

Initial activity: $A_0 = \lambda N_0 = 1.28 \times 10^{-5} \times 5.0 \times 10^{18} = 6.4 \times 10^{13}\,\text{Bq}$ .

After $45\,\text{hours}$ , which is exactly three half-lives, the activity halves three times: $A = A_0 \times \left(\tfrac{1}{2}\right)^3 = \dfrac{6.4 \times 10^{13}}{8} = 8.0 \times 10^{12}\,\text{Bq}$ .

Markers reward the decay constant from $\ln 2 / t_{1/2}$ , the activity from $A = \lambda N$ , and using three half-lives (or the exponential) for the final activity.

WJEC 20183 marksDescribe the relative penetrating powers of alpha, beta and gamma radiation, and state a suitable absorber to distinguish each.

Show worked answer →

Alpha radiation is the least penetrating: it is stopped by a few centimetres of air or a sheet of paper, so if the count rate drops to background when paper is inserted, alpha is present.

Beta radiation is more penetrating: it passes through paper but is stopped by a few millimetres of aluminium, so a drop in count rate when aluminium is added indicates beta.

Gamma radiation is the most penetrating: it is only reduced (never fully stopped) by thick lead or concrete, so any remaining count rate after aluminium that is cut down by lead indicates gamma. Markers reward the correct order of penetration and naming paper, aluminium and lead as the distinguishing absorbers.

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Sources & how we know this

WJEC A-level Physics specification — WJEC (2015)