Radioactive Decay and Half-Life

Radioactive decay is a random process, but when we have a large number of nuclei the statistical behaviour becomes highly predictable. This lesson covers the exponential decay law, the decay constant, activity, half-life, and their applications including carbon dating. This material is central to AQA section 3.8.1 and is examined in Paper 2.

---

The Exponential Decay Law

If we start with N₀ undecayed nuclei at time t = 0, the number remaining undecayed at time t is:

N = N₀ e^(−λt)

where λ (lambda) is the decay constant, measured in s⁻¹ (or sometimes min⁻¹ or yr⁻¹). The decay constant represents the probability that any given nucleus will decay per unit time.

This equation can also be written:

N/N₀ = e^(−λt)

The graph of N against t is a smooth exponential curve that starts at N₀ and asymptotically approaches zero.

---

Activity

The activity (A) of a radioactive source is the number of decays per unit time. It is measured in becquerels (Bq), where 1 Bq = 1 decay per second.

Activity is directly proportional to the number of undecayed nuclei:

A = λN

Since N decreases exponentially, so does the activity:

A = A₀ e^(−λt)

where A₀ = λN₀ is the initial activity.

Similarly, the count rate (C) measured by a detector (which is proportional to activity) follows the same exponential law:

C = C₀ e^(−λt)

Key Point: Activity, the number of undecayed nuclei, and the count rate all follow the same exponential decay pattern. Any of them can be used to determine the half-life or decay constant.

---

Half-Life

The half-life (t½) is the time taken for:

• The number of undecayed nuclei to halve, OR
• The activity to halve, OR
• The count rate to halve.

Deriving the Half-Life Formula

Starting from N = N₀ e^(−λt), at time t = t½ we have N = N₀/2:

N₀/2 = N₀ e^(−λt½)

1/2 = e^(−λt½)

Taking natural logarithms: ln(1/2) = −λt½

−ln 2 = −λt½

t½ = ln 2 / λ ≈ 0.693 / λ

This is one of the most important equations in nuclear physics. It relates the measurable quantity (half-life) to the fundamental constant of the decay (decay constant).

---

Worked Examples

Worked Example 1 — Finding the decay constant from half-life

Cobalt-60 has a half-life of 5.27 years. Find the decay constant in s⁻¹.

First convert half-life to seconds: t½ = 5.27 × 365.25 × 24 × 3600 = 1.663 × 10⁸ s

λ = ln 2 / t½ = 0.6931 / 1.663 × 10⁸

λ = 4.17 × 10⁻⁹ s⁻¹

Worked Example 2 — Finding activity

A sample initially contains 5.00 × 10²⁰ atoms of cobalt-60. What is the initial activity?

A₀ = λN₀ = 4.17 × 10⁻⁹ × 5.00 × 10²⁰

A₀ = 2.08 × 10¹² Bq (or 2.08 TBq)

Worked Example 3 — Number remaining after a given time

How many cobalt-60 atoms remain after 15.0 years?

Convert time: t = 15.0 × 365.25 × 24 × 3600 = 4.734 × 10⁸ s

N = N₀ e^(−λt) = 5.00 × 10²⁰ × e^(−4.17 × 10⁻⁹ × 4.734 × 10⁸)

N = 5.00 × 10²⁰ × e^(−1.974)

N = 5.00 × 10²⁰ × 0.1390

N = 6.95 × 10¹⁹ atoms

Alternatively, using half-lives: 15.0 years / 5.27 years ≈ 2.847 half-lives. N = 5.00 × 10²⁰ × (1/2)^2.847 = 5.00 × 10²⁰ × 0.1390 ≈ 6.95 × 10¹⁹ ✓

Worked Example 4 — Finding half-life from measurements

A detector records a count rate of 480 counts per minute from a radioactive source. After 6.0 hours, the count rate has fallen to 60 counts per minute. Find the half-life.

C = C₀ e^(−λt)

60 = 480 e^(−λ × 6.0)

60/480 = e^(−6.0λ)

0.125 = e^(−6.0λ)

ln(0.125) = −6.0λ

−2.079 = −6.0λ