Random decay, radiation, activity and half-life

Key idea: A reviewed, static H2 Physics learning chain for all official Nuclear Physics outcomes, from Rutherford evidence to fusion and fission.

  • GCE A-Level H2 Physics 2027

Learn the idea

Big question: How can random individual decays produce a predictable population law?

Each unstable nucleus decays spontaneously and randomly with constant probability per unit time. For a large sample, N = N₀e⁻λt and activity A = λN, with half-life t½ = ln2/λ. Alpha, beta and gamma radiation differ in charge, ionising power, penetration and behaviour in fields.

Separate unpredictable events from a predictable population

Nuclear decay is spontaneous and random: no external trigger is required, and the decay time of one nucleus cannot be predicted. Repeated counts fluctuate even when the average activity is steady.

Background radiation comes from sources such as cosmic rays, rocks and building materials. Measure it separately and subtract its mean count rate before inferring source behaviour.

Check your understanding: Do fluctuating counts prove the source activity is changing?

No. Random fluctuations are expected; look for a sustained change beyond the scatter.

Connect decay constant, activity and the exponential

The decay constant λ is the probability per unit time that one nucleus decays. For a large population, N = N₀e⁻λt and activity A = λN, so activity follows the same exponential form.

Half-life satisfies t½ = ln2/λ. Equal half-life intervals multiply N or A by one half; they do not subtract equal amounts. Activity is measured in becquerels, where 1 Bq means one decay per second.

Check your understanding: What fraction remains after three half-lives?

(1/2)³ = 1/8.

Distinguish alpha, beta and gamma radiation

Alpha radiation is a helium nucleus with charge +2e, beta-minus is a fast electron with charge −e and gamma is an electromagnetic photon with no charge. Alpha is strongly ionising and weakly penetrating; gamma is less densely ionising and highly penetrating; beta is intermediate.

Electric and magnetic fields deflect charged alpha and beta emissions in opposite directions, with beta normally curving more because of its much smaller mass. Gamma is not deflected.

Check your understanding: Which emission is undeflected by an electric field and why?

Gamma, because it has no electric charge.

Random individual decays and predictable population half-lifeThree groups contain sixteen, eight and four undecayed nuclei at zero, one and two half-lives, illustrating statistical decay of a large population.t = 0N/N₀ = 16/16t = t½N/N₀ = 8/16t = 2t½N/N₀ = 4/16which nucleus?unpredictablepopulation trend?predictable
Scroll diagram horizontally to read all labels.
Individual decay times are random, but a large population follows a predictable exponential law: the expected number remaining halves after each half-life.
Comparing the penetration of alpha, beta and gamma radiationThree beams travel from a source toward paper, aluminium and thick lead. Alpha ends at paper, beta passes paper but ends at aluminium, and gamma passes both before being attenuated by lead.Source☢PaperThin aluminiumThick leadαabsorbedβ⁻absorbedγintensity reduced
Paper absorbs alpha, thin aluminium absorbs beta, and thick lead reduces gamma intensity. Shielding reduces exposure; gamma is not completely stopped by a single stated thickness.
Alpha, beta-minus and gamma in an electric fieldParallel alpha, beta-minus and gamma beams enter between a positive upper plate and negative lower plate. Alpha curves slightly downward, beta curves strongly upward, and gamma continues straight.+−positive platenegative plateSource☢αγβ⁻
In an electric field, positive alpha bends toward the negative plate, negative beta-minus bends toward the positive plate, and uncharged gamma is not deflected. Beta bends more because its mass is much smaller.

Key ideas to keep

  • Activity is measured in becquerels; a measured count rate may include background and is not automatically the source activity.
  • Half-life is independent of the initial number in the ideal model.
  • Random decay does not mean the population curve is unpredictable.

Worked example

Move from activity data to half-life, decay constant and nuclei

Question: A sample's activity falls from 800 Bq to 200 Bq in 600 s. Find its half-life, decay constant and initial number of undecayed nuclei.

  1. Step 1: Count the halvings

    Why: 800 to 200 is a factor of four.

    Working: Two half-lives occur in 600 s, so t½ = 300 s.

  2. Step 2: Find the decay constant

    Why: Half-life and decay constant describe the same exponential.

    Working: λ = ln2/300 = 2.31 × 10⁻³ s⁻¹.

  3. Step 3: Use activity as decay rate

    Why: A = λN links the macroscopic activity to population size.

    Working: N₀ = 800/(2.31×10⁻³) = 3.46 × 10⁵ nuclei.

Answer: t½ = 300 s, λ = 2.31 × 10⁻³ s⁻¹ and N₀ = 3.46 × 10⁵ nuclei.

Check: A smaller λ would require more nuclei to produce the same activity, which matches A = λN.

Question

Net activity falls from 960 to 240 Bq in 12 min. Find half-life and λ.

Check the worked solution

Two halvings occur, so t½ = 6 min = 360 s and λ = ln2/360 = 1.93 × 10⁻³ s⁻¹.

Practise with support

Try this

Three half-lives pass. State the remaining fraction.

Hint: Apply one factor 1/2 per half-life.

Check your answer

1/8.

Practise independently

Your turn

Explain randomness, count fluctuations, background, radiation properties and the equations A = λN and x = x₀e⁻λt.

Check your answer

Individual decay time is unpredictable but population probability is constant. Counts fluctuate statistically and include environmental background. Alpha, beta and gamma differ in nature, ionisation and penetration. λ links activity to undecayed nuclei; exponential decay gives t½ = ln2/λ.

Common mistakes

Common mistake

Half-life predicts when one nucleus decays.

What is wrong with this reasoning?

Show better thinking

It describes a population; individual decay is random.

Common mistake

Measured count rate is automatically source activity.

What is wrong with this reasoning?

Show better thinking

Subtract background and account for detection efficiency before inference.

Exam guidance

Subtract background count rate before using measurements and retain the exponential, rather than subtracting equal amounts each half-life.

Exam-style practice [7 marks]

A detector records 920 counts min⁻¹ initially and 150 counts min⁻¹ after 30 min. Background is 40 counts min⁻¹. Find the half-life and decay constant. The initial source activity is 58.0 Bq; estimate the initial number of undecayed nuclei.

Plan before you answer

  • Subtract background from both count rates.
  • Use the corrected ratio to count halvings.
  • Use the stated activity in A = λN.
Mark your answer and compare the model

Marking points

Tick each point only if your answer states it clearly.

Model answer

Corrected rates are 920 − 40 = 880 and 150 − 40 = 110 counts min⁻¹. The factor 880/110 = 8 represents three half-lives in 30 min, so t½ = 10 min = 600 s and λ = ln2/600 = 1.16 × 10⁻³ s⁻¹. From A = λN, N = 58.0/(1.16×10⁻³) ≈ 5.0 × 10⁴ nuclei.

Exam-style practice [6 marks]

Compare alpha, beta-minus and gamma radiation by nature, charge, ionising effect and penetration. State how each behaves in an electric field.

Plan before you answer

  • Identify what each radiation consists of.
  • Connect charge to field deflection.
  • Compare ionisation and penetration oppositely.
Mark your answer and compare the model

Marking points

Tick each point only if your answer states it clearly.

Model answer

Alpha is a helium nucleus with charge +2e, is strongly ionising and weakly penetrating, and deflects towards the negative plate. Beta-minus is an electron with charge −e, has intermediate penetration and ionisation, and deflects oppositely and more strongly. Gamma is an uncharged electromagnetic photon, is highly penetrating and less densely ionising, and is not deflected.

Check what stayed with you

Recall question 1

Define decay constant.

Check the answer

The probability per unit time that one undecayed nucleus decays.

Recall question 2

State the activity relation.

Check the answer

A = λN.

Recall question 3

Why subtract background count rate?

Check the answer

It is not caused by the source and would distort activity and half-life inferences.

Try this next

Continue to the next lesson in this topic.

Applications and hazards

Syllabus and review details

This lesson covers the listed H2 Physics 9478 outcomes. Topic 20 excludes knowledge of positron emission in 20(g) and detailed knowledge of the antineutrino and particle zoo in 20(o). Nuclide equations conserve nucleon number, charge, mass-energy and momentum. Count data require background correction before population-law inference. Applications must relate half-life, penetration and ionisation to benefit and hazard. The binding-energy-per-nucleon curve, not a claim that mass disappears, explains fusion and fission energy release.

  • GCE A-Level H2 PhysicsTopic 20(d) / Topic 20(e) / Topic 20(f) / Topic 20(g) / Topic 20(h) / Topic 20(i) / Topic 20(j) / Topic 20(k) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027