3.8.1.4
Half-life can be found from decay curves and log graphs
Radioactive Decay & Half-Life — AQA A-Level Physics
From a decay curve
- Read the initial activityThe number of nuclear decays per unit time. Measured in becquerels (Bq), where 1 Bq = 1 decay per second. (or count rate) from the y-axis.
- Draw a horizontal line at half this value across to the curve.
- Draw a vertical line down to the time axis. This time is t₁/₂.
- You can verify by checking: at 2t₁/₂ the activityThe number of nuclear decays per unit time. Measured in becquerels (Bq), where 1 Bq = 1 decay per second. should be ¼ of the initial value.
From a logarithmic graph
- Taking ln of both sides of N = N₀\(e^{−λt}\): ln $N = \ln N_{0} - \lambda t$.
- Compare with y = mx + c: plot ln N (y) against t (x).
- $Gradient = -\lambda. Y-intercept = \ln N_{0}$.
- On a log graph, half-lives appear as equal intervals on the time axis.
Worked Example
A technetium sample has initial activityThe number of nuclear decays per unit time. Measured in becquerels (Bq), where 1 Bq = 1 decay per second. 8.0 × 10⁷ Bq. From the decay curve, the half-lifeThe time taken for half the number of radioactive nuclei in a sample to decay, or for the activity to halve. is 6 hours. Find (a) the decay constantThe probability of decay of a nucleus per unit time. Measured in s⁻¹. and (b) the number of atoms remaining after 24 hours.
Show Solution
1
(a) Decay constantThe probability of decay of a nucleus per unit time. Measured in s⁻¹.
$$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{6 \times 3600} = 3.2 \times 10^{-5} \text{ s}^{-1}$$
2
(b) Activity after 24 hours
24 hours = 4 half-lives. Activity = $8.0 \times 10^7 \times (\frac{1}{2})^4 = 5.0 \times 10^6$ Bq.
3
Number of atoms from A = λN
$$N = \frac{A}{\lambda} = \frac{5.0 \times 10^6}{3.2 \times 10^{-5}} = 1.56 \times 10^{11}$$
Answer
(a) λ = 3.2 × 10⁻⁵ s⁻¹, (b) $N = 1.56 \times 10¹¹$ atoms