Decay Constant and Half-Life

In the last lesson we met the four modes of radioactive decay: alpha, beta-minus, beta-plus and gamma. We said that decay is spontaneous and random — any individual nucleus in a sample may or may not decay in the next second, and there is no way of predicting which. What saves us from complete ignorance is that, in a sufficiently large sample, the statistics are extraordinarily reliable. This lesson is about those statistics: the decay constant λ, the activity A, the exponential decay law, and the half-life t_{1/2}.

This material is the core of section 6.4.2 of the OCR A-Level Physics A specification (H556). You will need it for every subsequent topic in the nuclear module — and for all of the medical physics at the end of the course, where half-life considerations determine which radioisotopes are suitable for clinical use.

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The Decay Constant

Imagine a sample containing N identical radioactive nuclei. We cannot predict which ones will decay in the next second, but we can ask: what is the probability that any given nucleus will decay in a short time interval Δt? Experiment shows that this probability is proportional to Δt (for small enough Δt), and that the constant of proportionality is the same for every nucleus of that species. We call this constant the decay constant, denoted λ.

Definition. The decay constant λ of a radioactive species is the probability per unit time that a single nucleus will decay. Its units are s⁻¹ (or equivalently min⁻¹, h⁻¹, yr⁻¹).

A large λ corresponds to a very unstable isotope that decays quickly; a small λ corresponds to a long-lived one. For example:

• Polonium-214: λ ≈ 4.2 × 10³ s⁻¹ (very unstable, half-life 164 μs).
• Technetium-99m: λ ≈ 3.2 × 10⁻⁵ s⁻¹ (half-life 6 hours).
• Carbon-14: λ ≈ 3.8 × 10⁻¹² s⁻¹ (half-life 5730 years).
• Uranium-238: λ ≈ 4.9 × 10⁻¹⁸ s⁻¹ (half-life 4.5 billion years).

The decay constant is a fundamental property of the isotope, determined entirely by the quantum-mechanical structure of its nucleus. It cannot be changed by heating, cooling, compressing, dissolving, or applying electric or magnetic fields. The decay constant of carbon-14 is the same whether the atom is in a tree trunk, a fossil, or a laboratory in Geneva.

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Activity

The activity A of a radioactive sample is the number of decays per unit time. If the sample contains N nuclei, each with decay probability λ per unit time, then on average:

A = λN

The activity is measured in becquerels (Bq), where 1 Bq = 1 decay per second. A sample with activity 1 MBq is experiencing one million nuclear decays every second. (An older unit, the curie, is still occasionally used: 1 Ci = 3.7 × 10¹⁰ Bq, which was roughly the activity of 1 g of pure radium-226.)

Note that the activity depends on both the decay constant and the number of nuclei present. A small sample of a short-lived isotope can have the same activity as a large sample of a long-lived one. For example:

• 10¹⁰ atoms of I-131 (λ ≈ 10⁻⁶ s⁻¹) gives A ≈ 10⁴ Bq.
• 10¹⁶ atoms of U-238 (λ ≈ 5 × 10⁻¹⁸ s⁻¹) gives A ≈ 5 × 10⁻² Bq.

To get significant activity from a long-lived isotope, you need enormous quantities of material; a short-lived isotope gives high activity from tiny samples.

Worked Example 1: Activity of a small sample

A sample of technetium-99m contains 5.0 × 10¹⁴ nuclei. The decay constant is λ = 3.2 × 10⁻⁵ s⁻¹. What is the activity?

Solution.

A = λN = (3.2 × 10⁻⁵)(5.0 × 10¹⁴) = 1.6 × 10¹⁰ Bq = 16 GBq

This is a substantial activity — 16 billion decays per second — but a typical clinical dose of Tc-99m delivered to a patient is much smaller, of the order of a few hundred MBq.

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The Exponential Decay Law

Because each nucleus has a fixed decay probability λ per unit time, the rate at which the number of nuclei decreases is proportional to the number present:

dN/dt = -λN

The minus sign indicates that N decreases with time. This is the defining equation of exponential decay, familiar from any A-Level maths course. Its solution is:

N = N₀ e^(-λt)

where N₀ is the number of nuclei at t = 0. Multiplying both sides by λ gives the same exponential law for the activity:

A = A₀ e^(-λt)

Both N and A decrease exponentially with time, with the same exponent -λt. The ratio A/N = λ is constant.

A plot of N or A against t is a decaying exponential curve. A plot of ln N or ln A against t is a straight line with gradient -λ. This is the standard experimental technique for measuring λ: record activity over time and fit a straight line to the log-plot.

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The Half-Life

The half-life t_{1/2} is the time taken for the number of radioactive nuclei in a sample (or equivalently, the activity) to fall to half its initial value. It is a more intuitive measure of decay speed than λ.

Setting N = N₀/2 in the exponential decay law:

N₀/2 = N₀ e^(-λ t_{1/2})
1/2 = e^(-λ t_{1/2})
ln(1/2) = -λ t_{1/2}
-ln 2 = -λ t_{1/2}

Hence:

t_{1/2} = ln 2 / λ     (equivalently λ = ln 2 / t_{1/2})

with ln 2 ≈ 0.693. This is one of the most important relations in the nuclear module and you should know it by heart.

After one half-life, N = N₀/2. After two half-lives, N = N₀/4. After three, N = N₀/8. After n half-lives, N = N₀/2ⁿ. For example, after 10 half-lives the sample has fallen to about N₀/1024 ≈ 0.1% of its original activity — a useful rule of thumb for estimating when a sample has "effectively decayed away".

Worked Example 2: From half-life to decay constant

Iodine-131, used in thyroid treatment, has a half-life of 8.02 days. What is its decay constant in s⁻¹?

Solution. First convert the half-life to seconds:

t_{1/2} = 8.02 × 24 × 3600 s = 6.93 × 10⁵ s

Then:

λ = ln 2 / t_{1/2} = 0.693 / (6.93 × 10⁵) ≈ 1.00 × 10⁻⁶ s⁻¹

Worked Example 3: From activity to mass

A sample of carbon-14 has an activity of 1.0 × 10³ Bq. Given that the half-life is 5730 years and the molar mass is 14 g mol⁻¹, find the mass of C-14 in the sample.

Solution.

Decay constant:

t_{1/2} = 5730 × 365.25 × 24 × 3600 ≈ 1.81 × 10¹¹ s
λ = ln 2 / t_{1/2} ≈ 3.83 × 10⁻¹² s⁻¹

Number of C-14 nuclei:

N = A / λ = 1.0 × 10³ / 3.83 × 10⁻¹² ≈ 2.6 × 10¹⁴ nuclei

Mass: