A source has a decay constant of 0.020\ s^-1. Find its half-life. Take ln 2 = 0.693.

OCR OCR 2019-style practice: A radioactive source has a half-life of 6.0 hours. A sample initially contains 4.0 × 10^16 undecayed nuclei. Calculate the decay constant and the number of nuclei remaining after 15 hours. Take ln 2 = 0.693.

Decay constant: λ = ln 2T_1/2 = (0.693)/(6.0 × 3600) = 3.21 × 10^-5\ s^-1. Time in seconds: t = 15 × 3600 = 5.4 × 10^4\ s. Decay law: N = N_0 e^-λ t = (4.0 × 10^16) e^-(3.21 × 10^-5)(5.4 × 10^4) = (4.0 × 10^16) e^-1.73. e^-1.73 = 0.177, so N = (4.0 × 10^16)(0.177) = 7.1 × 10^15 nuclei. Markers reward λ = ln 2T_1/2, the decay law, and the value about 7.1 × 10^15 (consistent with 2.5 half-lives elapsed).

EnglandPhysicsSyllabus dot point

How does radioactive decay follow an exponential law, and how are activity and half-life related?

Radioactive decay: the random and spontaneous nature of decay, the decay constant and activity, the exponential decay law, half-life and its relation to the decay constant, and radioactive dating.

A focused answer to the OCR H556 radioactive decay content, covering the random and spontaneous nature of decay, the decay constant and activity, the exponential decay law for the number of nuclei and activity, the relationship between half-life and the decay constant, and applications such as radioactive dating.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

OCR wants you to describe radioactive decay as random and spontaneous, define the decay constant and activity, use the exponential decay law for the number of nuclei and the activity, relate the half-life to the decay constant, and apply these ideas to radioactive dating.

The answer

The nature of radioactive decay

The decay constant and activity

The exponential decay law

Half-life and dating

Activity from the number of nuclei

A sample contains $6.0 \times 10^{18}$ undecayed nuclei of an isotope with a half-life of $2.0$ years. Find its activity. Take $1$ year $= 3.16 \times 10^{7}\ \text{s}$ and $\ln 2 = 0.693$ .

step 1: Decay constant

$\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{2.0 \times 3.16 \times 10^{7}} = \frac{0.693}{6.32 \times 10^{7}} = 1.10 \times 10^{-8}\ \text{s}^{-1}$ .

step 2: Activity

$A = \lambda N = (1.10 \times 10^{-8})(6.0 \times 10^{18})$ .

step 3: Evaluate

$A = 6.6 \times 10^{10}\ \text{Bq}$ .

Examples in context

Carbon-14 dating, with its half-life of about 5730 years, dates organic material such as wood, bone and cloth up to tens of thousands of years old. The decay of uranium and potassium isotopes, with half-lives of billions of years, dates rocks and meteorites and gives the age of the Earth. Medical tracers use short-half-life isotopes so the patient's exposure falls quickly, and smoke detectors, sterilisation and industrial gauging all rely on the predictable exponential decay of activity.

Try this

Q1. State what is meant by the activity of a radioactive source and its unit. [2 marks]

Cue. The number of nuclear decays per second, $A = \lambda N$ , in becquerels.

Q2. A source has a decay constant of $0.020\ \text{s}^{-1}$ . Find its half-life. Take $\ln 2 = 0.693$ . [2 marks]

Cue. $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.020} = 35\ \text{s}$ .

Q3. Explain what is meant by saying radioactive decay is random and spontaneous. [2 marks]

Cue. Random: you cannot predict which nucleus decays or when. Spontaneous: the decay is unaffected by external conditions such as temperature or chemical state.

Exam-style practice questions

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

OCR 20194 marksA radioactive source has a half-life of

6.0

hours. A sample initially contains

4.0 \times 10^{16}

undecayed nuclei. Calculate the decay constant and the number of nuclei remaining after

15

hours. Take

\ln 2 = 0.693

Show worked answer →

Decay constant: $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{6.0 \times 3600} = 3.21 \times 10^{-5}\ \text{s}^{-1}$ .

Time in seconds: $t = 15 \times 3600 = 5.4 \times 10^{4}\ \text{s}$ .

Decay law: $N = N_0 e^{-\lambda t} = (4.0 \times 10^{16}) e^{-(3.21 \times 10^{-5})(5.4 \times 10^{4})} = (4.0 \times 10^{16}) e^{-1.73}$ .

$e^{-1.73} = 0.177$ , so $N = (4.0 \times 10^{16})(0.177) = 7.1 \times 10^{15}$ nuclei.

Markers reward $\lambda = \frac{\ln 2}{T_{1/2}}$ , the decay law, and the value about $7.1 \times 10^{15}$ (consistent with 2.5 half-lives elapsed).

OCR 20214 marksA sample of carbon from an ancient artefact has an activity of

0.25\ \text{Bq}

per gram, while living material has an activity of

0.50\ \text{Bq}

per gram. The half-life of carbon-14 is

5730

years. Estimate the age of the artefact. Take

\ln 2 = 0.693

Show worked answer →

The activity has fallen to half, $A = \frac{1}{2}A_0$ , which is exactly one half-life.

Check with the decay law: $\frac{A}{A_0} = e^{-\lambda t}$ , so $\frac{0.25}{0.50} = 0.5 = e^{-\lambda t}$ , giving $\lambda t = \ln 2$ , so $t = \frac{\ln 2}{\lambda} = T_{1/2} = 5730$ years.

Markers reward recognising the activity has halved, linking this to one half-life, and the age $5730$ years.

Related dot points

Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)