Exponential Decay and Decay Constant

Introduction

Repeated halving is convenient when time is an exact multiple of a half-life. Radioactive decay occurs continuously, however, so a formula is useful for times such as 1.5 half-lives or 7.2 hours. The exponential decay law describes the expected surviving parent population at any time. Its decay constant λ connects microscopic probability to the measurable activity and half-life of a nuclide.

Core explanation

For one nuclide in a closed sample without replenishment, the expected number of undecayed parent nuclei is N(t) = N₀e^(−λt). Here N₀ is the initial parent count, t is elapsed time and λ is the decay constant. The product λt has no unit, so if t is in seconds, λ must be in s⁻¹. The negative sign makes the population decrease as positive time passes. The equation gives an expectation for a large population, not a precise prediction for a tiny collection of nuclei.

The decay constant represents a rate of decay probability per remaining nucleus. For a short interval Δt, the probability that a particular surviving nucleus decays is approximately λΔt when λΔt is small. The expected decays per second in a population are therefore A = λN, where A is activity. As N decreases, activity decreases proportionally while λ for that nuclide remains the same under the ordinary model.

Half-life is the time when N/N₀ = 1/2. Substitute this fraction into the exponential law: 1/2 = e^(−λt₁⁄₂). Taking natural logarithms gives ln 2 = λt₁⁄₂, so t₁⁄₂ = ln 2/λ. Because ln 2 is about 0.693, a nuclide with λ = 0.0693 day⁻¹ has half-life about 10.0 days. Conversely, λ = ln 2/t₁⁄₂. The result shows that a larger decay constant corresponds to a shorter half-life.

The exponential form and repeated-halving form are equivalent: N(t) = N₀(1/2)^(t/t₁⁄₂). This follows because λ = ln 2/t₁⁄₂. At t = 2t₁⁄₂, the equation gives (1/2)² = 1/4, exactly matching the table from the previous lesson. At t = 0.5t₁⁄₂, the expected fraction is (1/2)^0.5 ≈ 0.707, not 0.5. A half of a half-life does not itself halve the sample.

Activity follows the same exponential form for a pure isotope: A(t) = A₀e^(−λt). This works because A(t) = λN(t) and λ is fixed. It does not automatically describe a gross detector rate, which may contain background. If a detector's net efficiency and geometry remain constant, background-subtracted count rate can follow a similar shape, but its scale is set by detection fraction. In a decay chain, a daughter's activity can initially rise as it is produced, so applying a single parent exponential to the total activity can be wrong.

The equation is not a statement that individual nuclei get “more likely” to decay as they age. In the simple model, a surviving nucleus has the same probability per unit time at any time. The population falls because fewer nuclei remain. This memoryless property distinguishes nuclear decay from many everyday aging processes and explains the exponential form.

Use the logarithmic inverse when time is unknown: t = (1/λ) ln(N₀/N) = t₁⁄₂ log₂(N₀/N). If one-eighth remains, log₂(8) = 3, so three half-lives have elapsed. For more awkward fractions, a calculator is appropriate. Always check that N < N₀ gives positive elapsed time; a negative answer suggests the ratio was inverted.

Step-by-step reasoning

1. Identify whether the problem gives half-life, λ, initial amount or activity. 2. Convert all time units so λ and t are reciprocal units. 3. Find λ = ln 2/t₁⁄₂ if necessary. 4. Use N/N₀ = e^(−λt) or (1/2)^(t/t₁⁄₂) for remaining fraction. 5. Apply the fraction to parent amount or pure-parent activity, then assess assumptions and units.

Visual explanation

Draw a smooth curve starting at N₀ and approaching the horizontal axis without crossing it. Mark heights N₀/2 at t₁⁄₂, N₀/4 at 2t₁⁄₂ and N₀/8 at 3t₁⁄₂. At half a half-life, mark a point near 0.707N₀. A tangent arrow near the start is steeper than one later, showing that absolute loss rate decreases while fractional behavior stays constant.

Real-world analogy

If each remaining item in a large stock has the same small chance of leaving each minute, the number leaving per minute shrinks as the stock shrinks. The total can follow an exponential curve. Nuclear decay is more precise in its statistical law than this analogy, and individual nuclei do not decide when to leave.

Real-world example

Environmental scientists may monitor a radioactive tracer whose activity changes during a study. Knowing its half-life lets them correct for the decay expected during transport or measurement, before interpreting concentration changes. Movement and dilution may also alter a measured signal, so decay correction alone cannot explain every change.

Why?

Why is ln 2 in the half-life formula? The exponential law uses base e. At half-life the surviving fraction is one-half; solving e^(−λt) = 1/2 requires a logarithm, and ln(1/2) = −ln 2. Thus λt₁⁄₂ = ln 2.

Common misconception

“The decay constant is the number of nuclei lost each second.” λ is a per-nucleus rate parameter with inverse-time units. The total activity A = λN depends on how many parent nuclei remain. Two samples of the same isotope share λ but can have different activities.

Worked example

A nuclide has half-life 8 hours. Find the expected remaining fraction after 12 hours and λ in h⁻¹. First, λ = ln 2/8 ≈ 0.0866 h⁻¹. The elapsed time is 12/8 = 1.5 half-lives, so N/N₀ = (1/2)^1.5 ≈ 0.354. Equivalently, e^(−0.0866×12) ≈ 0.354. A 200-Bq pure source would then have expected activity about 200×0.354 = 70.8 Bq. The answer lies between 100 Bq at 8 hours and 50 Bq at 16 hours, a useful reasonableness check.

Quick check

1. If a nuclide's λ doubles, what happens to its half-life? Answer: Since t₁⁄₂ = ln 2/λ, its half-life is halved.

Exam focus

Write λ and time in compatible units, use a negative exponent for remaining amount and report a fraction between zero and one for positive elapsed time. Show the relation t₁⁄₂ = ln 2/λ and distinguish λ from total activity.

Advanced insight

The rate equation dN/dt = −λN says the instantaneous population decline is proportional to the current population. Integrating it gives N = N₀e^(−λt). The number of decays in a short interval fluctuates statistically, but averaging many comparable samples approaches this smooth law.

Summary

Exponential decay gives N(t) = N₀e^(−λt), activity A = λN and half-life t₁⁄₂ = ln 2/λ. Repeated halving is the same model evaluated at whole-number half-lives. The equations describe expected population behavior for one nuclide under simple conditions.

Practice questions

1. A nuclide has a 5-day half-life. What is λ in day⁻¹? Answer: λ = ln 2/5 ≈ 0.1386 day⁻¹. 2. What fraction remains after 1.5 half-lives? Answer: (1/2)^1.5 ≈ 0.354 remains on average. 3. Why can two samples of the same nuclide have different activities while sharing one λ? Answer: Activity is λN, so different numbers of undecayed nuclei give different decays per second even with the same decay constant.