The number of undecayed nuclei falls exponentially: $N = N_{0}e^{-\lambdat}$

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

Radioactive decayThe spontaneous and random disintegration of an unstable nucleus, emitting radiation (alpha, beta, or gamma) to become more stable. follows an exponential pattern — the number of undecayed nuclei falls rapidly but never reaches zero.
The steeper the curve, the larger the decay constantThe probability of decay of a nucleus per unit time. Measured in s⁻¹. λ.

N = N_0 e^{-\lambda t}

$N$: number of undecayed nuclei at time t
$N₀$: initial number of undecayed nuclei (at t = 0)
$λ$: decay constantThe probability of decay of a nucleus per unit time. Measured in s⁻¹. (s⁻¹)
$t$: time (s)

Since activityThe number of nuclear decays per unit time. Measured in becquerels (Bq), where 1 Bq = 1 decay per second. A is proportional to N, activityThe number of nuclear decays per unit time. Measured in becquerels (Bq), where 1 Bq = 1 decay per second. also decays exponentially: $A = A_{0} e^{-\lambda t}$.
Count rate C is proportional to activityThe number of nuclear decays per unit time. Measured in becquerels (Bq), where 1 Bq = 1 decay per second., so: $C = C_{0} e^{-\lambda t}$.
The exponential constant e ≈ 2.718. On a calculator, use the eˣ button. The inverse function is ln.

Worked Example

Strontium-90 has a decay constant of 0.025 year⁻¹. Express the activity after 5.0 years as a fraction of the initial activity.

Show Solution

Write the exponential decay equation for activity

$$\frac{A}{A_0} = e^{-\lambda t}$$

Substitute values (units match: both in years)

$$\frac{A}{A_0} = e^{-(0.025 \times 5.0)} = e^{-0.125} = 0.88$$

Answer

The activity decreases to 0.88 (88%) of its initial value — a 12% reduction after 5 years.