Skip to main content
EnglandPhysicsSyllabus dot point

If radioactive decay is random, how can we predict the behaviour of a large sample so reliably?

Radioactive decay as a random process, the decay constant, the activity of a source, the exponential decay law, half-life and applications such as radioactive dating.

A focused answer to AQA A-Level Physics 3.8.1.4, covering radioactive decay as a random process, the decay constant, activity, the exponential decay law, half-life and its link to the decay constant, and radioactive dating.

Generated by Claude Opus 4.810 min answer

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

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

Jump to a section
  1. What this dot point is asking
  2. Decay as a random process
  3. The decay constant and activity
  4. The exponential decay law
  5. Half-life
  6. Radioactive dating
  7. Try this

What this dot point is asking

AQA specification point 3.8.1.4 wants you to treat radioactive decay as a random process, define and use the decay constant and activity, apply the exponential decay law, relate half-life to the decay constant, and apply these ideas to radioactive dating.

Decay as a random process

The randomness shows in the irregular, fluctuating count rate from a small source; only when very many nuclei are present does the smooth exponential law emerge, in the same way that individual coin tosses are unpredictable but the fraction of heads in a million tosses is reliably one half.

The decay constant and activity

The relation A=λNA = \lambda N makes sense: a larger sample (more NN) or a more unstable isotope (larger λ\lambda) gives more decays per second. Because the activity is proportional to NN, and NN falls with time, the activity itself falls exponentially.

The exponential decay law

The number of undecayed nuclei, the activity and the measured count rate all fall exponentially:

This follows because the rate of decay is proportional to the number of nuclei present, dNdt=−λN\dfrac{\mathrm{d}N}{\mathrm{d}t} = -\lambda N, the same mathematical form as capacitor discharge. Taking logarithms gives ln⁡N=ln⁡N0−λt\ln N = \ln N_0 - \lambda t, a straight line whose gradient is −λ-\lambda.

Half-life

The half-life is the average time for the number of undecayed nuclei (or the activity) to halve. It is constant for a given isotope, regardless of how much is left, which is the signature of exponential decay. Half-lives range from fractions of a second to billions of years.

Radioactive dating

Try this

Q1. Define the activity of a radioactive source. [1 mark]

  • Cue. The number of nuclei that decay per second, measured in becquerels.

Q2. An isotope has a decay constant of 0.025 s−10.025 \text{ s}^{-1}. Calculate its half-life. [2 marks]

  • Cue. T1/2=ln⁡2λ=0.6930.025=28 sT_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{0.025} = 28 \text{ s}.

Q3. State the two properties that describe radioactive decay as a process. [1 mark]

  • Cue. Random and spontaneous.

Exam-style practice questions

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

AQA 20184 marksA radioactive source contains 5.0×10185.0 \times 10^{18} undecayed nuclei and has a decay constant of 3.2×10−9 s−13.2 \times 10^{-9} \text{ s}^{-1}. Calculate the activity of the source and its half-life.
Show worked answer →

The activity is A=λN=(3.2×10−9)(5.0×1018)=1.6×1010 BqA = \lambda N = (3.2 \times 10^{-9})(5.0 \times 10^{18}) = 1.6 \times 10^{10} \text{ Bq}.

The half-life is T1/2=ln⁡2λ=0.6933.2×10−9=2.2×108 sT_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{3.2 \times 10^{-9}} = 2.2 \times 10^{8} \text{ s} (about 6.96.9 years).

Markers reward using A=λNA = \lambda N for the activity and T1/2=ln⁡2λT_{1/2} = \dfrac{\ln 2}{\lambda} for the half-life, with correct units.

AQA 20214 marksA sample of carbon-14 in a piece of ancient wood has an activity that is 0.250.25 times that of an equal mass of living wood. The half-life of carbon-14 is 5730 years5730 \text{ years}. Estimate the age of the wood.
Show worked answer →

An activity of 0.250.25 of the living value is one quarter, which is two halvings: 1→12→141 \to \tfrac{1}{2} \to \tfrac{1}{4}.

Each halving takes one half-life, so the age is 2×5730=1.15×104 years2 \times 5730 = 1.15 \times 10^4 \text{ years}.

Equivalently, using A=A0e−λtA = A_0 e^{-\lambda t} with λ=ln⁡25730\lambda = \dfrac{\ln 2}{5730} and AA0=0.25\dfrac{A}{A_0} = 0.25 gives t=ln⁡4λ=1.15×104 yearst = \dfrac{\ln 4}{\lambda} = 1.15 \times 10^4 \text{ years}.

Markers reward recognising a quarter as two half-lives (or using the exponential law) and the correct age.

Related dot points

Sources & how we know this