Free Pearson Edexcel International A Level Physics lesson

Random decay and predictable populations

Use probability, half-life and logarithmic evidence.

What you’ll learn

  • Use constant individual probability to explain a falling population rate.
  • Use repeated fractional changes or the exponential law with consistent units.
  • Identify the log base, fit the corrected signal and inspect evidence of model failure.

An unpredictable nucleus, a statistical law

Spontaneous decay needs no external trigger. Random decay means that the time of one nucleus’s transformation cannot be predicted. Nevertheless, a large collection follows a stable statistical pattern. For a given isotope in the ordinary constant-probability model, every surviving nucleus has the same chance per short time interval of decaying.

The decay constant λ specifies that probability rate. The expected population change is dN/dt = −λN, where N counts undecayed nuclei. Activity A = λN is positive and measured in becquerels, one transformation per second. A detector count rate additionally depends on geometry, absorption and detector efficiency.

Twice as many surviving nuclei gives twice the expected activity at the same λ. Equal time intervals remove the same fraction, rather than the same number, so the rate falls as the sample decays. Small count totals fluctuate visibly; this is compatible with a constant half-life and does not show that an individual old nucleus is becoming more likely to decay.

dN/dt = −λN A = λN 1 Bq = 1 transformation s⁻¹

Symbols and units

  • N is the number of undecayed nuclei, a dimensionless count; t is time in s.
  • λ is the decay constant in s⁻¹: the constant probability rate for each surviving nucleus in the stated model.
  • dN/dt is the rate of decrease of the model population in s⁻¹; the minus sign distinguishes depletion from positive activity.
  • A is source activity in Bq = transformations s⁻¹, not nucleon number or oscillation amplitude. 1 kBq = 1000 Bq.

How the equation explains the idea

  1. In a sufficiently short interval Δt, each nucleus has decay probability approximately λΔt. The expected loss is therefore λNΔt, giving dN/dt = −λN in the continuous population model.
  2. Define positive activity as the expected transformations per second: A = −dN/dt = λN. Substitute the example values: A = (2.0 × 10⁻⁵)(3.0 × 10¹⁰) = 6.0 × 10⁵ Bq = 600 kBq.
  3. As N decreases, A decreases even though λ stays fixed. Individual survivors do not become more likely to decay just because they have survived longer.
  4. The detector records only a geometry- and efficiency-dependent share of the emissions, with random fluctuations and possible background, so 600 kBq does not imply 600000 detected counts every second.

Worked example

Activity from survivors

  1. λ = 2.0 × 10⁻⁵ s⁻¹ and N = 3.0 × 10¹⁰.
  2. A = λN = 6.0 × 10⁵ Bq = 600 kBq.
  3. A detector need not count all 600000 transformations each second.

Activity belongs to the source; count rate also depends on detection.

Keep this idea. Use constant individual probability to explain a falling population rate.

Half-life describes repeated fractions

The constant-probability law gives N = N₀ exp(−λt), where exp(y) means e raised to the power y. Multiplying by λ gives the identical time dependence A = A₀ exp(−λt). At one half-life, N/N0 = 1/2; taking logarithms yields λT½ = ln2. Thus half-life is ln2/λ, with the time unit matching the inverse-time unit of λ.

After n half-lives the expected fraction is 2⁻ⁿ. Two half-lives leave one quarter: the second interval removes half of what survived the first. The continuous model approaches zero rather than reaching it in finite time, although a finite real collection eventually has a last transformation.

A graphical half-life should be read from background-corrected activity or a proportional corrected count rate. Compare several intervals where the signal is adequate. Dating requires a justified initial isotope abundance and a sufficiently closed sample history. A neat exponential calculation does not by itself verify either assumption.

N = N₀ exp(−λt) A = A₀ exp(−λt) T½ = ln2/λ

Symbols and units

  • N₀ and N are initial and surviving nuclear populations; A₀ and A are initial and later source activities in Bq.
  • λ is decay constant in s⁻¹ when t and half-life T½ are in s. T½ is a duration, not absolute temperature.
  • exp(y) means eʸ; ln is the natural logarithm; λt and ln 2 are dimensionless.
  • n = t/T½ is the number of elapsed half-lives, a dimensionless ratio that need not be an integer.

How the equation explains the idea

  1. At one half-life, set N/N₀ = 1/2 = exp(−λT½). Taking natural logs gives −ln 2 = −λT½, so T½ = ln 2/λ and λ = ln 2/T½.
  2. Convert 20.0 min to 1200 s before requesting λ in s⁻¹: λ = 0.693147/1200 = 5.78 × 10⁻⁴ s⁻¹.
  3. At 60 min, n = 60/20 = 3, so N/N₀ = exp(−n ln 2) = 2⁻³ = 1/8. Since A = λN with fixed λ, activity has the same remaining fraction.
  4. Each interval halves the survivors, not the original total. Apply this to a single isotope’s corrected activity or a count rate proportional to it under unchanged detection conditions.
Equal time intervals remove equal fractions

Conditions. Computed source activity A = 640 exp(−λt) Bq, with λ = ln(2)/600 s⁻¹ and half-life 600 s. One isotope, no added nuclei, and constant decay probability per surviving nucleus. This is an expected population curve, not scheduled individual decays.

Time / s; Expected activity / Bq.

Activities at 0, 600, 1200 and 1800 s are 640, 320, 160 and 80 Bq. Each interval halves what remains, so the absolute number lost decreases. The exponential approaches zero without reaching it at a finite plotted time.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Convert time units

  1. T½ = 20.0 min = 1200 s.
  2. λ = 0.693147/1200 = 5.78 × 10⁻⁴ s⁻¹.
  3. After 60 min, three half-lives leave one eighth.

Match the units of λ and time.

Keep this idea. Use repeated fractional changes or the exponential law with consistent units.

Use a logarithmic plot critically

Taking natural logarithms gives ln(A/Aref) = ln(A0/Aref) − λt, where a fixed reference activity makes the logarithm dimensionless. The gradient of this graph is −λ. The intercept describes activity at the chosen time origin, and changing the activity unit shifts that intercept without changing the gradient.

Use a large triangle on a fitted line rather than two adjacent raw points. Base-ten logarithms instead give gradient −λ/ln10, so the log base matters. Background must be subtracted before taking logarithms; its additive effect is not removed by a vertical intercept adjustment.

Curvature may indicate multiple isotopes, changing geometry or incorrect background, rather than a new decay law. Relative uncertainty increases as corrected rates fall. A spreadsheet update Nnext ≈ N(1 − λΔt) is only a small-step approximation; require λΔt much less than one. The exact exponential update stays positive for any positive step and makes the approximation’s limitation visible.

ln(A/Aref) = ln(A₀/Aref) − λt small step: Nnext ≈ N(1 − λΔt)

Symbols and units

  • A and A₀ are later and initial activity in Bq; Aref is a fixed reference activity in the same unit, making A/Aref dimensionless.
  • t is time in s; λ is decay constant in s⁻¹; T½ is half-life in s. ln is the natural logarithm.
  • A log-activity versus time gradient has units s⁻¹. A negative slope corresponds to a positive λ.
  • N and Nnext are current and next expected populations; Δt is the numerical time step in s, and λΔt is dimensionless.

How the equation explains the idea

  1. Divide A = A₀ exp(−λt) by Aref and take natural logs: ln(A/Aref) = ln(A₀/Aref) − λt. Comparing with a straight-line equation identifies gradient = −λ.
  2. Use the fitted values, not adjacent noisy points: gradient = (6.50 − 8.00)/500 = −0.00300 s⁻¹, so λ = +0.00300 s⁻¹.
  3. Convert that rate to a duration: T½ = ln 2/λ = 0.693147/0.00300 = 231 s. If the plot used log₁₀ instead, its gradient would be −λ/ln 10.
  4. For numerical modelling, the exact update is Nnext = N exp(−λΔt). Expanding for λΔt much smaller than one gives Nnext ≈ N(1 − λΔt); large steps invalidate that linear approximation.
Correct background before taking the logarithm

Conditions. Expected source count rate R = 80 exp(−λt) s⁻¹, λ = ln(2)/600 s⁻¹, and fixed background 10 s⁻¹. Detection efficiency and geometry remain fixed. Natural logarithms use the stated reference rate to make their arguments dimensionless.

Time / s; ln[R / (1 s⁻¹)].

The corrected line has gradient −λ = −0.001155 s⁻¹. Adding background before taking logs produces curvature and a progressively shallower slope, not merely a shifted intercept. Real low-count data also require uncertainty analysis.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Read a fitted gradient

  1. The fitted log value falls from 8.00 to 6.50 over 500 s.
  2. Gradient = −1.50/500 = −0.00300 s⁻¹, so λ = 0.00300 s⁻¹.
  3. T½ = 0.693147/0.00300 = 231 s.

The negative slope gives a positive decay constant.

Keep this idea. Identify the log base, fit the corrected signal and inspect evidence of model failure.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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