Activity, Specific Activity and Mean Lifetime

Introduction

Detectors do not count nuclei; they count decays. The quantity that matters for measurement, dosimetry and regulation is therefore the activity — the number of decays per second — rather than the number of radioactive atoms. This page connects the two through A = λN, introduces the units used on every radioactive source label, and shows why a microgram of a short-lived nuclide can be far more active than a kilogram of a long-lived one.

Core explanation

Activity. From the decay law, the rate of decay is −dN/dt = λN. This rate is the activity:

A = λN

Since N falls exponentially, so does A: A = A₀e^(−λt). Activity and number of nuclei have the same half-life.

Units. The SI unit is the becquerel , 1 Bq = 1 decay per second (s⁻¹). Laboratory sources are typically kBq to MBq; medical doses of technetium-99m are hundreds of MBq; spent reactor fuel is measured in PBq (10¹⁵ Bq). The older curie was defined as approximately the activity of 1 g of radium-226 and is now fixed at 1 Ci = 3.7 × 10¹⁰ Bq = 37 GBq.

Activity is not radiation output. One becquerel means one nuclear decay, but a single decay may emit several photons (cobalt-60 emits two gamma rays per decay) or none that escape. Converting activity into detector counts or dose requires emission probabilities and detection efficiencies.

Specific activity. For a pure radionuclide of molar mass M, one gram contains N = N A/M nuclei, so the specific activity is

SA = λN A/M = (ln 2)N A/(M t½)

Short half-lives and low molar masses give enormous specific activities. Pure carbon-14 has about 1.65 × 10¹¹ Bq g⁻¹, while uranium-238 (t½ = 4.47 × 10⁹ years) has only about 1.24 × 10⁴ Bq g⁻¹.

Carrier-free versus carrier-added. In practice a radionuclide is often mixed with stable isotopes of the same element (the carrier ). The specific activity of the element is then much lower than the theoretical maximum. Radiopharmaceuticals and tracers for receptor studies need high specific activity so that the chemical amount injected is tiny and does not disturb the biology.

Mean lifetime. The average survival time of a nucleus is τ = 1/λ. Since t½ = (ln 2)/λ, τ = t½/ln 2 ≈ 1.443 t½. After one mean lifetime, a fraction 1/e ≈ 0.368 remains. Physicists often quote τ; chemists more often quote t½.

Total decays. Integrating the activity over all time gives ∫A dt = N₀: every nucleus decays once. Equivalently, N₀ = A₀τ, a quick way to find how many atoms a source of known activity contains.

Formulae

A = λN = (ln 2/t½)N; A = A₀e^(−λt); SA = (ln 2)N A/(M t½); τ = 1/λ = t½/ln 2; 1 Ci = 3.7 × 10¹⁰ Bq.

Step-by-step reasoning

To find the activity of a given mass of a pure radionuclide:

1. Convert the mass to moles using the molar mass of the nuclide. 2. Multiply by N A to obtain N. 3. Convert the half-life to seconds and calculate λ = 0.693/t½. 4. Multiply: A = λN, giving becquerels. 5. Convert to curies if needed by dividing by 3.7 × 10¹⁰.

Visual explanation

Picture two jars of glowing beads. In one jar each bead flashes rarely but there are billions; in the other there are few beads but each flashes often. The total flash rate — the activity — depends on the product of the number of beads and the flashing probability, exactly as A = λN.

Real-world analogy

A town's daily number of birthdays is its population multiplied by 1/365 per day. Activity is the same idea: the number of nuclei times the probability per second that any one of them decays. A small village with an unusually high "rate" can outpace a larger town.

Real-world example

Smoke detectors of the ionisation type contain americium-241 of roughly 30–40 kBq, less than 1 µCi. Because americium-241 has t½ = 432 years, this activity corresponds to only about 0.3 µg of the metal. By contrast, hospital cobalt-60 teletherapy sources historically held hundreds of TBq.

Why?

Why does a shorter half-life give a higher specific activity? A = λN, and λ is inversely proportional to t½. For the same number of atoms, a nuclide that decays a million times faster produces a million times more decays per second. The price is that it also disappears a million times faster.

Common misconception

"A source with a higher activity always contains more radioactive material." Activity depends on λ as well as N. A few nanograms of a short-lived nuclide can be more active than kilograms of uranium-238.

Worked example

Question: Calculate the activity of 1.00 µg of pure iodine-131 (M = 131 g mol⁻¹, t½ = 8.02 days).

Reasoning: n = 1.00 × 10⁻⁶ ÷ 131 = 7.63 × 10⁻⁹ mol, so N = 7.63 × 10⁻⁹ × 6.022 × 10²³ = 4.60 × 10¹⁵. t½ = 8.02 × 86 400 = 6.93 × 10⁵ s, so λ = 0.693 ÷ 6.93 × 10⁵ = 1.00 × 10⁻⁶ s⁻¹. A = λN = 4.60 × 10⁹ Bq.

Answer: About 4.6 GBq (roughly 0.12 Ci) from a single microgram.

Quick check

1. A nuclide has a half-life of 10.0 hours. What is its mean lifetime in hours? Answer: τ = t½/ln 2 = 10.0/0.693 = 14.4 hours.

Exam focus

Keep λ in s⁻¹ when you want activity in Bq. Examiners commonly test A = λN, specific activity from half-life and molar mass, and conversions between Bq and Ci. Distinguish activity (decays per second) from count rate (detected events per second).

Advanced insight

For a nuclide with several decay modes, λ is the sum of partial decay constants: λ = λ₁ + λ₂ + …, and the branching fraction for mode i is λᵢ/λ. The partial activity for one mode is λᵢN, but all modes share the single observed half-life. Potassium-40, for example, decays about 89% by β⁻ to calcium-40 and about 11% by electron capture to argon-40.

Summary

Activity is the decay rate, A = λN, measured in becquerels (1 Bq = 1 s⁻¹) or curies (1 Ci = 3.7 × 10¹⁰ Bq). Specific activity, (ln 2)N A/(M t½) for a pure nuclide, is largest for short-lived, light nuclides and is reduced by carrier. The mean lifetime τ = 1/λ = 1.443 t½, and N₀ = A₀τ gives the number of atoms in a source.

Practice questions

1. Calculate the specific activity of pure cobalt-60 (M = 60.0 g mol⁻¹, t½ = 5.27 years; 1 year = 3.156 × 10⁷ s). Answer: λ = 0.693 ÷ (1.663 × 10⁸ s) = 4.17 × 10⁻⁹ s⁻¹; N = 1.00 × 10²² g⁻¹; SA ≈ 4.2 × 10¹³ Bq g⁻¹ (about 1100 Ci g⁻¹). 2. Convert 370 MBq into millicuries. Answer: 370 × 10⁶ ÷ 3.7 × 10¹⁰ = 0.0100 Ci = 10.0 mCi. 3. A source of technetium-99m (t½ = 6.01 h) has an activity of 500 MBq. How many ⁹⁹ᵐTc nuclei does it contain? Answer: λ = 0.693 ÷ 21 636 s = 3.20 × 10⁻⁵ s⁻¹; N = A/λ = 5.00 × 10⁸ ÷ 3.20 × 10⁻⁵ = 1.56 × 10¹³ nuclei. 4. Why does a carrier-added preparation have a lower specific activity than a carrier-free one? Answer: The stable carrier atoms add mass but no decays, so the activity is spread over a larger mass of the element.