Half-Life: Calculations and Applications Revision Notes | A Level AQA Physics
Half-life and decay constant are linked by T½ = ln 2/λ, where ln 2 ≈ 0.693. Free AQA A Level Physics notes with worked examples and 11 questions answered.
Updated 12 Sept 2026 · Revise Right Team
Half-Life: Calculations and Applications is part of Radioactivity (A-level only) in A Level AQA Physics.
8 min
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What this lesson covers
Half-life and decay constant are linked by T½ = ln 2/λ, where ln 2 ≈ 0.693. Free AQA A Level Physics notes with worked examples and 11 questions answered.
Calculate the decay constant from the half-life equation $T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}$ and vice versa
Determine half-life from exponential decay curves and logarithmic graphs of activity or count rate against time
Apply knowledge of half-life to evaluate real-world contexts including radioactive waste storage and radioactive dating
Explain why the half-life of a nuclide is constant and independent of the initial number of nuclei
Key ideas to keep in view
Use this lesson to stay inside Radioactivity (A-level only) while connecting the detail back to the wider Nuclear physics (A-level only) topic.
Half-life ($T_{\frac{1}{2}}$): The time taken for the number of radioactive nuclei in a sample — or its activity — to fall to exactly half its initial value. A fixed property of each nuclide, independent of sample size or external conditions.
Decay constant ($\lambda$): The probability per unit time that a given nucleus will decay. Measured in s$^{-1}$ (or min$^{-1}$, yr$^{-1}$ depending on context). Related to half-life by $T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}$.
Activity ($A$): The number of nuclear decays occurring per second in a sample. Measured in becquerels (Bq), where 1 Bq = 1 decay per second. Activity is proportional to the number of undecayed nuclei: $A = \lambda N$.
Exponential decay: The mathematical pattern describing radioactive decay, where the quantity decreases by the same fractional amount in each equal time interval. Described by $N = N_0 e^{-\lambda t}$ and $A = A_0 e^{-\lambda t}$.
Revision notes
Read the core explanations from Half-Life: Calculations and Applications before testing yourself from memory.
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 $T_{\frac{1}{2}}$.
$A = A_0 e^{-\lambda t}$ — This is the standard exponential decay law for activity. [M1]
$e^{-40\lambda} = \frac{100}{800} = 0.125$ — 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.
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.
$\frac{A}{A_0} = \frac{0.0625}{0.250} = 0.25 = \frac{1}{4}$ — The activity has fallen to one quarter of its initial value. [M1]
$\frac{1}{4} = \left(\frac{1}{2}\right)^2$, so 2 half-lives have elapsed. — Each half-life halves the activity: after 1 half-life, $\frac{1}{2}$; after 2 half-lives, $\frac{1}{4}$.
$t = 2 \times T_{\frac{1}{2}} = 2 \times 5730 = 11\,460 \text{ years}$ — Multiply the number of half-lives by the half-life value. [A1]
$A = A_0 e^{-\lambda t}$, $\lambda = \frac{\ln 2}{5730} = 1.210 \times 10^{-4} \text{ yr}^{-1}$\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.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-12.
Living organisms absorb carbon-14, but after death it decays without replacement, so comparing the current activity with that of fresh material gives the age. Wood with an activity of 0.0625 Bq g⁻¹, against 0.250 Bq g⁻¹ for fresh wood, has fallen to a quarter: two half-lives of 5730 years, so about 11 460 years old.
How do you find half-life from a graph?
On a decay curve, read the initial activity A₀, find the time at which the activity is A₀/2, then repeat from A₀/2 to A₀/4 to check the interval is the same. Or plot ln A against t: the straight line has gradient -λ, so T½ = ln 2/λ. A gradient of -0.0347 s⁻¹ gives T½ ≈ 20 s.
What is the equation linking half-life and decay constant?
Half-life and decay constant are linked by T½ = ln 2/λ, where ln 2 ≈ 0.693. It comes from setting N = N₀/2 in N = N₀ × exp(−λt) and solving for t; because λ is fixed for a given nuclide, the half-life is constant however many nuclei remain, and doubling λ halves the half-life.
Why does radioactive waste need to be stored for so long?
Activity falls to about 1/1000 of its initial value only after roughly ten half-lives, since (½)¹⁰ = 1/1024. Caesium-137 and strontium-90, with half-lives of about 30 years, need secure storage for approximately 300 years, while plutonium-239, with a half-life of 24 100 years, needs geological-timescale containment in deep underground repositories.
How do you calculate half-life from two activity readings?
Half-life is calculated from two activity readings by using A = A₀ × exp(−λt) to find λ, then T½ = ln 2/λ. If activity falls from 800 Bq to 100 Bq in 40 s, exp(−40λ) = 0.125, so λ = 2.079 ÷ 40 = 0.0520 s⁻¹ and T½ = 0.6931 ÷ 0.0520 = 13.3 s, consistent with three half-lives in 40 s.
How do you find half-life from a graph of ln A against t?
Plotting ln A against t gives a straight line with gradient −λ and y-intercept ln A₀, since ln A = ln A₀ − λt. A gradient of −0.0347 s⁻¹ gives λ = 0.0347 s⁻¹ and a half-life of 0.6931 ÷ 0.0347 ≈ 20 s, and the straight line itself confirms the decay is exponential.
How do you do a radiocarbon dating calculation?
In radiocarbon dating, an artefact with a carbon-14 activity of 0.0625 Bq g⁻¹, compared with 0.250 Bq g⁻¹ for fresh wood, has a quarter of the initial activity. That is two half-lives, so with a half-life of 5730 years its age is 2 × 5730 = 11 460 years.
Why is half-life constant?
Half-life is constant because the decay constant λ is a fixed property of the nuclide, independent of the number of undecayed nuclei remaining. Each nucleus has the same probability of decaying per unit time, so the same fraction decays in each equal time interval.
How do you choose a nuclide for radioactive dating?
A nuclide suits radioactive dating only if its half-life is comparable to the age of the sample. Carbon-14, with a half-life of about 5730 years, dates organic material up to around 50 000 years old, while potassium-40 and uranium-238, with half-lives of about 1.25 × 10⁹ and 4.5 × 10⁹ years, date much older geological samples.
How does half-life affect the hazard of radioactive waste?
A nuclide with a short half-life, such as caesium-137 at 30 years, has a large decay constant and high activity per nucleus, so it is dangerous because of its high dose rate. One with a long half-life, such as plutonium-239 at 24 100 years, decreases in activity only slowly and needs geological-timescale containment.
How do you check that decay is exponential?
Decay is confirmed as exponential by checking that equal time intervals give the same fractional decrease: the time to fall from A₀ to A₀/2 should equal the time from A₀/2 to A₀/4. A straight line on a graph of ln A against t also confirms it.