Half-life is the time for the number of undecayed nuclei to halve

Radioactive Decay & Half-Life — AQA A-Level Physics

Key Definition

Half-life (t₁/₂) — The average time taken for a given number of nuclei of a particular isotope to halve. Equivalently, the time for the activity to halve.

Deriving the half-life equation

Start with N = N₀$e^{−λt}$. At t = t₁/₂, $N = ½N_{0}$.

\frac{1}{2}N_0 = N_0 e^{-\lambda t_{1/2}}

Divide both sides by N₀: ½ = $e^{−λt₁/₂}$.
Take the natural log: ln(½) = −λt₁/₂.
Since ln(½) = −ln 2: $\lambda t_{1}/_{2} = \ln 2.$

t_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}

$t₁/₂$: half-lifeThe time taken for half the number of radioactive nuclei in a sample to decay, or for the activityThe number of nuclear decays per unit time. Measured in becquerels (Bq), where 1 Bq = 1 decay per second. to halve. (s)
$λ$: decay constantThe probability of decay of a nucleus per unit time. Measured in s⁻¹. (s⁻¹)

Half-lifeThe time taken for half the number of radioactive nuclei in a sample to decay, or for the activityThe number of nuclear decays per unit time. Measured in becquerels (Bq), where 1 Bq = 1 decay per second. to halve. and decay constantThe probability of decay of a nucleus per unit time. Measured in s⁻¹. are inversely proportional.
Short half-$life = large decay constantThe probability of decay of a nucleus per unit time. Measured in s⁻¹. = fast decay$.
Long half-$life = small decay constant = slow decay$.
After 1 half-life: 50% remains. After 2: 25%. After 3: 12.5%. After n half-lives: (½)ⁿ remains.

Worked Example

Strontium-90 has a half-life of 28.0 years. Calculate the decay constant in s⁻¹.

Show Solution

Convert half-life to seconds

$$t_{1/2} = 28 \times 365 \times 24 \times 3600 = 8.83 \times 10^8 \text{ s}$$

Apply the half-life equation

$$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{8.83 \times 10^8} = 7.85 \times 10^{-10} \text{ s}^{-1}$$

Answer

$\lambda = 7.85 \times 10^{-10}$ s⁻¹