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How does the random, spontaneous nature of radioactive decay lead to the exponential decay law and the idea of half-life?

Describe radioactive decay as a random spontaneous process, apply the exponential decay law and the decay constant, and relate it to half-life and activity

A focused answer to the H2 Physics learning outcome on radioactive decay. Decay as a random spontaneous process, the decay constant and activity, the exponential decay law, and the link between half-life and the decay constant.

Generated by Claude Opus 4.88 min answer

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

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  1. What this dot point is asking
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  3. Examples in context
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What this dot point is asking

SEAB wants you to describe radioactive decay as a random and spontaneous process, to apply the exponential decay law with the decay constant, and to relate the decay constant to half-life and activity. The randomness at the level of individual nuclei produces a precise statistical law for large numbers.

The answer

Random and spontaneous decay

Radioactive decay is random: it is impossible to predict which nucleus will decay next or exactly when a given nucleus will decay. Every undecayed nucleus has the same constant probability of decaying per unit time. Decay is also spontaneous: it is unaffected by external conditions such as temperature, pressure or chemical state. Despite this randomness, the behaviour of a very large number of nuclei is highly predictable on average.

The decay constant and activity

The decay constant λ\lambda is the probability per unit time that a given nucleus decays. The activity AA of a source is the rate at which nuclei decay, measured in becquerels (Bq), one decay per second. Because each nucleus has the same probability of decaying:

A=λNA = \lambda N

where NN is the number of undecayed nuclei. Since the rate of decay is proportional to the number remaining, the population falls exponentially.

The exponential decay law

The number of undecayed nuclei follows:

N=N0eλtN = N_0 e^{-\lambda t}

and the activity, being proportional to NN, follows the same form:

A=A0eλtA = A_0 e^{-\lambda t}

A graph of lnN\ln N (or lnA\ln A) against time is a straight line of gradient λ-\lambda, which is the standard way to measure the decay constant from data.

Half-life

The half-life t1/2t_{1/2} is the time for half the undecayed nuclei (and so half the activity) to decay. Setting N=12N0N = \tfrac{1}{2}N_0 in the decay law gives:

t1/2=ln2λ=0.693λt_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}

Each half-life reduces the quantity to half its previous value, so after nn half-lives a fraction (12)n(\tfrac{1}{2})^n remains. The half-life is constant for a given isotope, independent of how much material is present.

Examples in context

Example 1. Carbon dating. Living things maintain a fixed proportion of radioactive carbon-14. After death, the carbon-14 decays with a half-life of about 57005700 years. Measuring the remaining activity and applying A=A0eλtA = A_0 e^{-\lambda t} gives the age of organic remains, a direct use of the exponential decay law.

Example 2. Medical tracers. A radioactive tracer used in diagnosis is chosen with a short half-life so it decays quickly and limits the patient's dose. Technetium-99m, with a half-life of about 66 hours, gives a strong signal during a scan but falls to a negligible activity within a day, a deliberate use of the half-life relation.

Try this

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

  • Cue. The rate at which nuclei decay, A=λNA = \lambda N; unit becquerel (Bq), one decay per second.

Q2. An isotope has a decay constant of 0.050 s10.050\ \text{s}^{-1}. Find its half-life. [2 marks]

  • Cue. t1/2=ln2λ=0.6930.050=13.9 st_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{0.050} = 13.9\ \text{s}.

Q3. A source has an initial activity of 1.6×105 Bq1.6 \times 10^5\ \text{Bq} and a half-life of 2.02.0 hours. Find its activity after 6.06.0 hours. [2 marks]

  • Cue. 6.06.0 hours is 33 half-lives, so A=1.6×105×(12)3=2.0×104 BqA = 1.6 \times 10^5 \times (\tfrac{1}{2})^3 = 2.0 \times 10^4\ \text{Bq}.

Exam-style practice questions

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

Original5 marksA radioactive source has a half-life of 8.08.0 days and an initial activity of 4.0×106 Bq4.0 \times 10^6\ \text{Bq}. (a) Find the decay constant. (b) Find the activity after 2424 days. (c) Find the activity after 2020 days.
Show worked answer →

(a) Decay constant: λ=ln2t1/2=0.6938.0=0.0866 day1\lambda = \dfrac{\ln 2}{t_{1/2}} = \dfrac{0.693}{8.0} = 0.0866\ \text{day}^{-1}.

(b) 2424 days is exactly 33 half-lives, so the activity halves three times: A=4.0×106×(12)3=4.0×106×18=5.0×105 BqA = 4.0 \times 10^6 \times (\tfrac{1}{2})^3 = 4.0 \times 10^6 \times \tfrac{1}{8} = 5.0 \times 10^5\ \text{Bq}.

(c) For 2020 days use A=A0eλt=4.0×106×e0.0866×20=4.0×106×e1.732=4.0×106×0.177=7.1×105 BqA = A_0 e^{-\lambda t} = 4.0 \times 10^6 \times e^{-0.0866 \times 20} = 4.0 \times 10^6 \times e^{-1.732} = 4.0 \times 10^6 \times 0.177 = 7.1 \times 10^5\ \text{Bq}.

Markers reward λ=ln2/t1/2\lambda = \ln 2 / t_{1/2}, the use of whole half-lives for part (b), and the exponential law for the non-integer time in part (c).

Original4 marks(a) Explain what is meant by saying radioactive decay is random and spontaneous. (b) State what is meant by the activity of a source and how it relates to the number of undecayed nuclei.
Show worked answer →

(a) Random means it is impossible to predict which nucleus will decay next or exactly when any individual nucleus will decay; every undecayed nucleus has the same constant probability of decaying per unit time. Spontaneous means the decay is unaffected by external conditions such as temperature, pressure or chemical state.

(b) The activity is the rate at which nuclei in the source decay, A=ΔNΔtA = \dfrac{\Delta N}{\Delta t} (decays per second, the becquerel). It is proportional to the number of undecayed nuclei present: A=λNA = \lambda N, so as NN falls the activity falls in proportion.

Markers reward random as unpredictable per-nucleus with constant probability, spontaneous as independent of external conditions, activity as decays per second, and the relation A=λNA = \lambda N.

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