An isotope has a decay constant of 0.030\ s^-1. Find its half-life. [2 marks]

WJEC Eduqas Eduqas 2018-style practice: A radioactive isotope has a half-life of 8.0\ days. A sample initially contains 6.0 × 10^20\ atoms. Calculate the decay constant and the number of atoms remaining after 20\ days.

Decay constant from the half-life: λ = ln 2T_1/2 = 0.6938.0 = 0.0866\ day^-1. Number remaining from N = N_0 e^-λ t: N = (6.0 × 10^20)e^-0.0866 × 20 = (6.0 × 10^20)e^-1.733. e^-1.733 = 0.177, so N = (6.0 × 10^20)(0.177) = 1.06 × 10^20\ atoms. Markers reward λ = ln 2T_1/2 = 0.087\ day^-1, using N = N_0 e^-λ t, and about 1.1 × 10^20 atoms remaining.

EnglandPhysicsSyllabus dot point

How do unstable nuclei decay, and how do we describe the random process mathematically?

Nuclear decay: alpha, beta and gamma radiation and their properties, the random nature of decay, activity and the decay constant, the exponential decay law, and half-life.

A focused answer to the Eduqas A-Level Physics Component 3 nuclear decay content, covering the properties of alpha, beta and gamma radiation, the random nature of radioactive decay, activity and the decay constant, the exponential decay law N = N0 e^(-lambda t), and half-life.

Generated by Claude Opus 4.813 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

Eduqas wants you to describe the nature and properties of alpha, beta and gamma radiation, explain that radioactive decay is random and spontaneous, define activity and the decay constant, use the exponential decay law $N = N_0 e^{-\lambda t}$ , and use the half-life with $\lambda = \frac{\ln 2}{T_{1/2}}$ .

The answer

Alpha, beta and gamma radiation

Randomness and spontaneity

Activity and the decay constant

The decay law and half-life

Activity of a sample

A sample contains $2.5 \times 10^{18}$ atoms of an isotope with a half-life of $5.0\ \text{hours}$ . Find its initial activity in becquerel.

step 1: Find the decay constant

$\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{5.0 \times 3600\ \text{s}} = \dfrac{0.693}{1.8 \times 10^{4}} = 3.85 \times 10^{-5}\ \text{s}^{-1}$ (the half-life converted to seconds).

step 2: Apply A = lambda N

$A = \lambda N = (3.85 \times 10^{-5})(2.5 \times 10^{18})$ .

step 3: Evaluate

$A = 9.6 \times 10^{13}\ \text{Bq}$ . The activity is high because the sample contains a vast number of nuclei.

Examples in context

Radioactive decay underpins radiometric dating (carbon-14 for archaeology, uranium isotopes for rock ages), medical tracers and cancer treatment, smoke detectors (which use a weak alpha source), and the monitoring of nuclear reactors. The penetrating powers determine radiation shielding and safety: alpha emitters are dangerous if inhaled or ingested, while gamma emitters require heavy external shielding.

Try this

Q1. State what alpha radiation is and one of its properties. [2 marks]

Cue. A helium nucleus (2 protons, 2 neutrons); highly ionising but weakly penetrating (stopped by paper).

Q2. An isotope has a decay constant of $0.030\ \text{s}^{-1}$ . Find its half-life. [2 marks]

Cue. $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.030} = 23\ \text{s}$ .

Q3. State the fraction of a radioactive sample remaining after three half-lives. [1 mark]

Cue. $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$ .

Exam-style practice questions

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

Eduqas 20184 marksA radioactive isotope has a half-life of

8.0\ \text{days}

. A sample initially contains

6.0 \times 10^{20}\ \text{atoms}

. Calculate the decay constant and the number of atoms remaining after

20\ \text{days}

Show worked answer →

Decay constant from the half-life: $\lambda = \dfrac{\ln 2}{T_{1/2}} = \dfrac{0.693}{8.0} = 0.0866\ \text{day}^{-1}$ .

Number remaining from $N = N_0 e^{-\lambda t}$ : $N = (6.0 \times 10^{20})e^{-0.0866 \times 20} = (6.0 \times 10^{20})e^{-1.733}$ .

$e^{-1.733} = 0.177$ , so $N = (6.0 \times 10^{20})(0.177) = 1.06 \times 10^{20}\ \text{atoms}$ .

Markers reward $\lambda = \frac{\ln 2}{T_{1/2}} = 0.087\ \text{day}^{-1}$ , using $N = N_0 e^{-\lambda t}$ , and about $1.1 \times 10^{20}$ atoms remaining.

Eduqas 20214 marksCompare the penetrating power and ionising ability of alpha, beta and gamma radiation, and state what each is.

Show worked answer →

Alpha is a helium nucleus (2 protons, 2 neutrons): the most ionising but the least penetrating, stopped by a few centimetres of air or a sheet of paper.

Beta-minus is a fast electron: moderately ionising and moderately penetrating, stopped by a few millimetres of aluminium.

Gamma is a high-energy electromagnetic photon: the least ionising but the most penetrating, reduced (not fully stopped) by thick lead or concrete.

Markers reward identifying each radiation (helium nucleus, electron, photon), and the inverse relationship between ionising ability and penetrating power (alpha most ionising/least penetrating, gamma least ionising/most penetrating).

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Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)