First-Order Decay Kinetics

Introduction

Earlier courses introduced half-life as the time for half of a radioactive sample to decay. At this level we derive the decay law from a single physical assumption: every nucleus of a given nuclide has the same, constant probability of decaying per unit time, independent of its age and its surroundings. From that one statement come the exponential law, the logarithmic plots used in the laboratory, and the link between decay constant and half-life that underpins every later calculation in radiochemistry.

Core explanation

The starting assumption. In a short time interval dt, each nucleus has a probability λ dt of decaying, where λ is the decay constant . The nucleus has no memory: an atom of iodine-131 made yesterday is exactly as likely to decay in the next second as one made last year.

The rate law. If N undecayed nuclei are present, the expected number decaying in dt is Nλ dt, so

dN/dt = −λN

This is a first-order rate law, identical in form to that for a unimolecular chemical reaction, with λ playing the role of the rate constant k.

Integration. Separating variables: dN/N = −λ dt. Integrating from N₀ at t = 0 to N at time t gives ln(N/N₀) = −λt, so

N = N₀e^(−λt)

Half-life. Setting N = N₀/2 gives ln(1/2) = −λt½, hence

t½ = ln 2/λ ≈ 0.693/λ

The half-life does not depend on N₀: it takes the same time to go from 1000 nuclei to 500 as from 10²⁰ to 5 × 10¹⁹. After n half-lives the fraction remaining is (1/2)ⁿ, so an equivalent form is N = N₀(1/2)^(t/t½).

Linear plots. Taking logarithms, ln N = ln N₀ − λt. A plot of ln N (or ln of the count rate) against time is a straight line of gradient −λ. Curvature in such a plot is diagnostic: it reveals a mixture of nuclides with different half-lives, or the growth of a radioactive daughter.

Range of half-lives. Known half-lives span more than 50 orders of magnitude, from below 10⁻²⁰ s for some very unstable light nuclei to about 10²⁴ years for tellurium-128. The same law applies throughout.

Environmental independence. Because decay is governed by nuclear forces deep inside the atom, λ is essentially unaffected by temperature, pressure or chemical bonding. Tiny exceptions occur for electron-capture decays (e.g. beryllium-7), whose rate depends slightly on electron density at the nucleus, changing λ by no more than about 1%.

Formulae

dN/dt = −λN; N = N₀e^(−λt); ln(N/N₀) = −λt; t½ = ln 2/λ; N/N₀ = (1/2)^(t/t½).

Step-by-step reasoning

To find the fraction of a nuclide remaining after time t:

1. Convert the half-life and the elapsed time into the same unit. 2. Calculate λ = 0.693/t½. 3. Calculate the exponent λt (it must be dimensionless). 4. Evaluate N/N₀ = e^(−λt). 5. Check against a half-life estimate: t/t½ half-lives should give roughly (1/2)^(t/t½).

Visual explanation

Plot N against t: a smooth curve falling by half every t½, never quite reaching zero. Beside it, plot ln N against t: a straight line of slope −λ. In the simulation, individual nuclei blink out at random, yet the total count tracks the smooth curve ever more closely as the starting number increases.

Real-world analogy

Imagine a large bowl of popcorn kernels where each unpopped kernel, every second, has the same small chance of popping, regardless of how long it has been heated. The number popping per second is proportional to the number still unpopped, so the unpopped count falls exponentially. (Real popcorn is not quite like this, which is why nuclei are the cleaner example.)

Real-world example

Iodine-131 (t½ = 8.02 days) is released in nuclear accidents and used in thyroid therapy. Because 80 days is about ten half-lives, its activity falls to roughly 1/1000 of the initial value in under three months. This is why food restrictions for iodine-131 contamination can be lifted comparatively quickly, whereas caesium-137 (t½ ≈ 30 years) remains a concern for decades.

Why?

Why is the decay exponential rather than linear? Because the number of decays per second is proportional to the number of nuclei still present. As the sample shrinks, the decay rate shrinks in proportion, so each equal time interval removes the same fraction, not the same number. Constant fractional loss is the defining signature of an exponential.

Common misconception

"After two half-lives, all of the sample has decayed." After one half-life half remains; after two, a quarter remains; after ten, about 0.1% remains. The amount approaches zero but a finite sample only disappears completely through the randomness of the last few decays.

Worked example

Question: A sample contains iodine-131 (t½ = 8.02 days). What percentage remains after 20.0 days?

Reasoning: λ = 0.693/8.02 = 0.0864 day⁻¹. λt = 0.0864 × 20.0 = 1.728. N/N₀ = e^(−1.728) = 0.178. Check: 20.0/8.02 = 2.49 half-lives, and (1/2)^2.49 ≈ 0.18.

Answer: About 17.8% remains.

Quick check

1. A nuclide has a decay constant of 0.0231 min⁻¹. What is its half-life in minutes? Answer: t½ = 0.693/0.0231 = 30.0 minutes.

Exam focus

Examiners expect the derivation from dN/dt = −λN, the relation t½ = ln 2/λ, and consistent units in λt. Show the half-life check alongside exponential calculations, and use a ln(count rate) versus time graph to extract λ from its gradient.

Advanced insight

Exponential decay is not exact at all times. Quantum mechanics predicts tiny deviations at extremely short times (the quantum Zeno regime) and at extremely long times, when the decay follows a power law. For nuclear decays these deviations occur far outside measurable ranges, so the exponential law is, in practice, perfectly obeyed. Its statistical nature, however, matters greatly when counting few events.

Summary

A constant decay probability per unit time, λ, gives the first-order law dN/dt = −λN and hence N = N₀e^(−λt). The half-life t½ = ln 2/λ is independent of the amount present. A plot of ln N against t is a straight line of gradient −λ. Decay constants are essentially unaffected by temperature, pressure and chemistry, with only very small effects for electron capture.

Practice questions

1. Show that t½ = ln 2/λ starting from N = N₀e^(−λt). Answer: At t = t½, N = N₀/2, so 1/2 = e^(−λt½); taking natural logarithms, −ln 2 = −λt½, so t½ = ln 2/λ. 2. Carbon-14 has t½ = 5730 years. Calculate λ in year⁻¹. Answer: λ = 0.693/5730 = 1.21 × 10⁻⁴ year⁻¹. 3. What fraction of a phosphorus-32 sample (t½ = 14.3 days) remains after 50.0 days? Answer: λ = 0.0485 day⁻¹; λt = 2.42; e^(−2.42) = 0.089, so about 8.9% remains. 4. A plot of ln(count rate) against time is curved, flattening at long times. Suggest an explanation. Answer: The sample contains two (or more) nuclides; the short-lived one dies away first, leaving the slower decay of the longer-lived one.