Half-Life: Calculations and Applications Revision Notes | A Level AQA Physics

1. Why Half-Life Matters

Radioactive nuclei decay at random — you cannot predict when any single nucleus will go. Yet, take a billion of them together, and something remarkable emerges: the fraction that decays per second stays perfectly constant. That constancy is what gives us half-life, and it turns out to be one of the most powerful tools in modern science.

Half-life underpins everything from dating ancient artefacts to deciding how long nuclear waste must be stored. Before you can apply it, though, you need to understand exactly what it measures and how it connects to the decay constant $\lambda$ — the quantity that sits at the heart of all radioactive decay mathematics.

2. The Half-Life Equation

The decay constant $\lambda$ tells you the probability per unit time that any one nucleus will decay. A large $\lambda$ means nuclei decay quickly; a small $\lambda$ means they linger. Half-life and decay constant are two ways of expressing the same underlying property of a nuclide — and they are linked by a precise equation.

This equation comes directly from the exponential decay law $N = N_0 e^{-\lambda t}$. Setting $N = \frac{N_0}{2}$ and solving for $t$ gives $t = \frac{\ln 2}{\lambda}$, which is exactly $T_{\frac{1}{2}}$. The derivation is worth knowing because it shows why half-life is constant: $\lambda$ is fixed for a given nuclide, so $T_{\frac{1}{2}}$ never changes regardless of how many nuclei remain.

Notice that $T_{\frac{1}{2}}$ and $\lambda$ are inversely proportional. Double the decay constant and you halve the half-life. This inverse relationship is the first thing to check whenever a question gives you one quantity and asks for the other.

Worked example: Calculating Decay Constant and Half-Life

A radioactive source has an initial activity of 800 Bq. After 40 s its activity has fallen to 100 Bq. Calculate (a) the decay constant $\lambda$ and (b) the half-life $T_{\frac{1}{2}}$.

1. $A = A_0 e^{-\lambda t}$ — This is the standard exponential decay law for activity. [M1]
2. $100 = 800 \times e^{-\lambda \times 40}$ — $A_0 = 800$ Bq, $A = 100$ Bq, $t = 40$ s.
3. $e^{-40\lambda} = \frac{100}{800} = 0.125$ — Divide both sides by 800. Note that $0.125 = \frac{1}{8}$, which is $\left(\frac{1}{2}\right)^3$ — a useful check that three half-lives have elapsed.
4. $-40\lambda = \ln(0.125) = -2.079$ — $\ln(0.125) = \ln\left(\frac{1}{8}\right) = -\ln 8 = -3\ln 2 \approx -2.079$. [M1]
5. $\lambda = \frac{2.079}{40} = 0.0520 \text{ s}^{-1}$ — Divide both sides by $-40$, noting the negatives cancel. [A1]
6. $T_{\frac{1}{2}} = \frac{0.6931}{0.0520} = 13.3 \text{ s}$ — Sanity check: three half-lives = $3 \times 13.3 = 39.9 \approx 40$ s, consistent with activity dropping to $\frac{1}{8}$ of its initial value. [A1]

Answer: $\lambda = 0.0520 \text{ s}^{-1}$; $T_{\frac{1}{2}} = 13.3 \text{ s}$

3. Reading Decay Graphs

Exam questions frequently give you a graph of activity (or count rate) against time and ask you to extract the half-life. The approach depends on whether the graph is plotted on a linear or logarithmic scale — and knowing which method to use is itself a mark-winning skill.

On a linear decay curve, the activity falls steeply at first and then levels off, producing the characteristic exponential shape. To find the half-life, read off the initial activity $A_0$, then find the time at which the activity equals $\frac{A_0}{2}$. Repeat from that point to $\frac{A_0}{4}$ and check that the same time interval is obtained — this confirms the decay is truly exponential.

The curve above shows activity halving every 20 s: from 800 Bq at $t = 0$ to 400 Bq at $t = 20$ s, then to 200 Bq at $t = 40$ s. Each equal time interval produces the same fractional decrease — the defining signature of exponential decay. Note: this is a separate hypothetical source from the worked example in Section 2, which used a different source with $T_{\frac{1}{2}} = 13.3$ s.

3a. Using a Log Graph to Find $\lambda$

Taking the natural logarithm of $A = A_0 e^{-\lambda t}$ gives $\ln A = \ln A_0 - \lambda t$. This is the equation of a straight line: plot $\ln A$ on the $y$-axis against $t$ on the $x$-axis and you get a line with gradient $-\lambda$ and $y$-intercept $\ln A_0$.

The log graph is more powerful than the raw decay curve for two reasons. First, it lets you verify that the decay really is exponential — a straight line confirms it. Second, you can extract $\lambda$ precisely from the gradient rather than relying on reading half-life intervals off a curved graph, which introduces larger uncertainty.

The gradient of this line is $\frac{\Delta(\ln A)}{\Delta t} = \frac{3.912 - 6.685}{80 - 0} = \frac{-2.773}{80} = -0.0347 \text{ s}^{-1}$, giving $\lambda = 0.0347 \text{ s}^{-1}$ and therefore $T_{\frac{1}{2}} = \frac{0.6931}{0.0347} \approx 20 \text{ s}$. This matches the half-life read directly from the decay curve, confirming the method.

4. Applications of Half-Life

With the mathematics established, the real power of half-life becomes clear when you apply it to genuine scientific and engineering problems. Two applications appear repeatedly in AQA exam questions: radioactive dating and the management of radioactive waste. Both require you to reason quantitatively about how activity changes over time — and to evaluate the suitability of a particular nuclide for a given purpose.

4a. Radioactive Dating

Carbon-14 ($^{14}$C) is produced continuously in the upper atmosphere and is absorbed by all living organisms. When an organism dies, it stops exchanging carbon with its environment, so the $^{14}$C in its remains decays without replacement. Measuring the current activity and comparing it to the known initial activity allows archaeologists to calculate how long ago the organism died.

The half-life of $^{14}$C is approximately 5730 years — long enough to date organic material up to around 50 000 years old, but short enough that measurable changes in activity occur over archaeologically relevant timescales. For much older geological samples, geologists use potassium-40 ($^{40}$K, $T_{\frac{1}{2}} \approx 1.25 \times 10^9$ years) or uranium-238 ($^{238}$U, $T_{\frac{1}{2}} \approx 4.5 \times 10^9$ years), matching the half-life to the age range being investigated.

Worked example: Radiocarbon Dating Calculation

A wooden artefact has a $^{14}$C activity of 0.0625 Bq g$^{-1}$. Fresh wood has an activity of 0.250 Bq g$^{-1}$. The half-life of $^{14}$C is 5730 years. Estimate the age of the artefact.

1. $\frac{A}{A_0} = \frac{0.0625}{0.250} = 0.25 = \frac{1}{4}$ — The activity has fallen to one quarter of its initial value. [M1]
2. $\frac{1}{4} = \left(\frac{1}{2}\right)^2$, so 2 half-lives have elapsed. — Each half-life halves the activity: after 1 half-life, $\frac{1}{2}$; after 2 half-lives, $\frac{1}{4}$.
3. $t = 2 \times T_{\frac{1}{2}} = 2 \times 5730 = 11\,460 \text{ years}$ — Multiply the number of half-lives by the half-life value. [A1]
4. $A = A_0 e^{-\lambda t}$, $\lambda = \frac{\ln 2}{5730} = 1.210 \times 10^{-4} \text{ yr}^{-1}$\n$A = 0.250 \times e^{-1.210 \times 10^{-4} \times 11460}$\n$= 0.250 \times e^{-1.386} = 0.250 \times 0.250 = 0.0625 \text{ Bq g}^{-1}$ ✓ — The verification confirms the answer is consistent. In an exam, the half-life counting method is faster; the exponential equation is used when the number of half-lives is not a whole number.

Answer: The artefact is approximately 11 460 years old.

4b. Radioactive Waste Storage

Nuclear power stations produce waste containing nuclides with a wide range of half-lives. High-level waste contains fission products such as caesium-137 ($T_{\frac{1}{2}} = 30$ years) and strontium-90 ($T_{\frac{1}{2}} = 29$ years). These are dangerous for hundreds of years — roughly ten half-lives are needed before activity falls to around $\frac{1}{1000}$ of its initial value (since $\left(\frac{1}{2}\right)^{10} = \frac{1}{1024} \approx \frac{1}{1000}$), which means secure storage for approximately 300 years.

Some actinides in spent nuclear fuel, such as plutonium-239 ($T_{\frac{1}{2}} = 24 100$ years), present a far longer-term hazard. The very long half-life means activity decreases only slowly, requiring geological-timescale containment — deep underground repositories in stable rock formations are the proposed solution. Evaluating a storage strategy therefore requires quantitative reasoning about how many half-lives must elapse before the material reaches a safe activity level.

This connects directly back to the inverse relationship between $T_{\frac{1}{2}}$ and $\lambda$ established in Section 2. A nuclide with a half-life of 30 years has a much larger $\lambda$ — and therefore higher activity per nucleus — than one with a half-life of 24 000 years. Both require careful management, but for different reasons: the former for its high dose rate, the latter for its persistence.

5. Exam Technique and Mark-Winning Strategies

Half-life questions on AQA papers follow predictable patterns. Recognising those patterns — and knowing the precise language that earns marks — is what separates a confident answer from a vague one. The guidance below consolidates the key exam-technique points for this topic.

For 'calculate' questions, always write the equation first, then substitute with units, then evaluate. The mark scheme awards a method mark for the correct equation and a separate accuracy mark for the correct numerical answer — so even an arithmetic slip does not cost you both marks if your method is clear.

For 'explain' questions about why half-life is constant, the key phrase is that the decay constant $\lambda$ is a fixed property of the nuclide, independent of the number of undecayed nuclei remaining. Saying 'each nucleus has the same probability of decaying per unit time' is the underlying physical reason and earns the explanation mark.

For 'evaluate' questions about suitability of a nuclide for dating or waste storage, structure your answer as: state the half-life, compare it to the relevant timescale, and reach a justified conclusion. A nuclide is suitable for dating only if its half-life is comparable to the age of the sample — too short and all the nuclide has decayed; too long and the change in activity is too small to measure accurately.

Taken together, the half-life equation, graphical analysis methods, and real-world applications form a coherent set of skills that AQA tests in a variety of question styles — from short calculations to extended evaluations. The thread connecting all of them is the exponential decay law: master that, and every half-life question becomes a variation on the same underlying idea.