Radioactive Decay: Randomness and Rate Revision Notes | A Level AQA Physics

1. Why Radioactive Decay Is Fundamentally Unpredictable

Picture a large hall filled with thousands of unstable nuclei. At any moment, one of them will suddenly emit radiation and transform — but you cannot point to which one, or say exactly when. That unpredictability is not a gap in our knowledge; it is a fundamental feature of radioactive decay.

Radioactive decay is described as random and spontaneous. Random means the decay of any individual nucleus cannot be predicted — not even in principle. Spontaneous means no external trigger causes it: temperature, pressure, chemical bonding, and electric fields have no effect on when a nucleus decays.

Here is the key insight that makes everything else in this topic click: although individual decays are unpredictable, the probability of a given nucleus decaying in the next second is constant. That probability is called the decay constant, symbol $\lambda$, with units of s$^{-1}$.

Think of it like rolling a fair die. You cannot predict whether the next roll will be a six, but you know the probability is always $\frac{1}{6}$ — it never changes based on previous rolls. Each nucleus is like a die being rolled every second: the chance of it decaying is always $\lambda$, no matter how long it has already survived.

This section establishes the foundation for the entire topic: randomness at the individual level, combined with a constant probability, produces a beautifully predictable pattern at the population level — exponential decay. That connection is what the next section unpacks.

2. From Constant Probability to the Decay Equation

Because each nucleus has the same constant probability $\lambda$ of decaying per second, the number of decays happening in a short time interval $\Delta t$ depends directly on how many undecayed nuclei $N$ are currently present. More nuclei means more decays — proportionally.

Combining these two ideas — constant probability per nucleus, and a large number of nuclei — gives the central equation of radioactive decay. The rate of change of the number of undecayed nuclei is proportional to the number currently present, with $\lambda$ as the constant of proportionality.

The negative sign is not cosmetic — it tells you that $N$ is decreasing over time. As nuclei decay, there are fewer left to decay, so the rate of decay itself falls. This is the self-reinforcing logic that produces exponential behaviour.

The left-hand side, $\frac{\Delta N}{\Delta t}$, is the magnitude of the activity $A$ of the source. Activity is the number of decays per second, measured in becquerels (Bq). So the equation is often written as $A = \lambda N$, where $A = \left|\frac{\Delta N}{\Delta t}\right|$.

Notice what the equation $\frac{\Delta N}{\Delta t} = -\lambda N$ is really saying: the rate of decay is proportional to the number of nuclei remaining. This is the mathematical signature of exponential decay, and it arises directly from the constant-probability model established in Section 1.

3. Modelling with Constant Decay Probability

The equation $\frac{\Delta N}{\Delta t} = -\lambda N$ is a model. It rests on one core assumption: that every undecayed nucleus has the same, unchanging probability $\lambda$ of decaying per unit time. This assumption is well-supported experimentally and leads to a specific, testable prediction — exponential decay.

You can build this model from scratch using a simple analogy. Suppose you start with 1000 dice, each representing a nucleus. Every second, you roll all the dice and remove any that show a six (probability $\frac{1}{6}$ per roll). After each round, you have fewer dice, so fewer are removed in the next round — but the probability per die never changes. The number remaining follows an exponential curve.

This dice model is a direct analogue of radioactive decay. The fraction removed per round is constant (like $\lambda$), the number removed each round decreases as the total falls (like activity decreasing as $N$ falls), and the overall pattern is exponential. Crucially, you cannot predict which die will show a six on any given roll — randomness is built in from the start.

The model also has a clear limitation worth noting at A-level: it applies to large numbers of nuclei. For very small samples, statistical fluctuations become significant and the smooth exponential prediction breaks down. The model is a population-level description, not a statement about individual nuclei.

4. Applying the Decay Equation: Worked Calculations

The equation $\frac{\Delta N}{\Delta t} = -\lambda N$ connects three quantities: activity, decay constant, and number of undecayed nuclei. Exam questions will give you two and ask for the third. The worked examples below show the method for each direction.

Worked example: Calculating activity from decay constant and number of nuclei

A radioactive sample contains $4.0 \times 10^{15}$ undecayed nuclei. The decay constant of the isotope is $3.5 \times 10^{-4}$ s$^{-1}$. Calculate the activity of the sample. [3 marks]

1. $A = \lambda N$ — Activity equals the magnitude of the decay rate. [M1] for correct formula.
2. $\lambda = 3.5 \times 10^{-4}$ s$^{-1}$, $N = 4.0 \times 10^{15}$ — Both values are already in SI units, so no conversion is needed. [M1] for correct substitution.
3. $A = 3.5 \times 10^{-4} \times 4.0 \times 10^{15}$ — Multiply the coefficients: $3.5 \times 4.0 = 14.0$. Combine powers: $10^{-4} \times 10^{15} = 10^{11}$.
4. $A = 14.0 \times 10^{11}$ Bq $= 1.4 \times 10^{12}$ Bq — Express in standard form. Units are becquerels (Bq). [A1] for correct answer with units.

Answer: $A = 1.4 \times 10^{12}$ Bq

Now consider the reverse: you measure the activity of a sample and count the nuclei present, and you want to find $\lambda$. Rearranging $A = \lambda N$ gives $\lambda = \frac{A}{N}$. The method is identical — the algebra is the only change.

Worked example: Calculating the decay constant from activity and number of nuclei

A sample has an activity of $6.0 \times 10^{8}$ Bq and contains $2.4 \times 10^{12}$ undecayed nuclei. Calculate the decay constant. [3 marks]

1. $\lambda = \frac{A}{N}$ — Starting from $A = \lambda N$, divide both sides by $N$. [M1] for correct rearrangement.
2. $A = 6.0 \times 10^{8}$ Bq, $N = 2.4 \times 10^{12}$ — Both values are in SI units. No conversion required.
3. $\lambda = \frac{6.0 \times 10^{8}}{2.4 \times 10^{12}}$ — Divide coefficients: $6.0 \div 2.4 = 2.5$. Divide powers: $10^{8} \div 10^{12} = 10^{8-12} = 10^{-4}$. [M1] for correct substitution.
4. $\lambda = 2.5 \times 10^{-4}$ s$^{-1}$ — Units of decay constant are always s$^{-1}$ (per second). [A1] for correct answer with units.

Answer: $\lambda = 2.5 \times 10^{-4}$ s$^{-1}$

5. The Exponential Decay Curve and What It Tells You

Because activity is proportional to $N$, and $N$ decreases exponentially, activity also decreases exponentially. The graph of $N$ against time — or equivalently $A$ against time — is a characteristic curve that starts steep and flattens out, never quite reaching zero.

The shape of this curve is a direct consequence of the constant-probability model. At any point on the curve, the gradient (rate of decrease) equals $-\lambda N$ — steeper when $N$ is large, shallower when $N$ is small. The curve never touches the time axis because there is always a non-zero probability that some nuclei remain undecayed.

Earlier, in Section 1, the constant decay probability was introduced as a property of individual nuclei. Here you can see the population-level consequence: that same constant probability, applied to a large group, produces the smooth exponential curve. The randomness of individuals averages out into a predictable pattern for the whole sample.

6. Exam Technique and Common Mistakes

The calculation questions on this topic are straightforward once you know the equation, but the explanation questions are where marks are most often dropped. Examiners are looking for precise language that demonstrates understanding of the underlying physics, not just the formula.

When asked to explain why activity decreases over time, a weak answer says 'because there are fewer nuclei'. A mark-winning answer says: 'As nuclei decay, the number of undecayed nuclei $N$ decreases. Since $A = \lambda N$ and $\lambda$ is constant, activity is proportional to $N$, so activity also decreases.' The chain of reasoning — linking $N$, $\lambda$, and $A$ — is what earns the marks.

For longer 'evaluate' or 'discuss' questions about the constant-probability model, structure your answer as: state the assumption (constant $\lambda$ per nucleus), explain the consequence (exponential decay of $N$ and $A$), then assess the limitation (the model is statistical and breaks down for very small $N$). That three-part structure covers the marks available.

Every spec requirement in this lesson — the random and spontaneous nature of decay, the constant decay probability, the equation $\frac{\Delta N}{\Delta t} = -\lambda N$, and the constant-probability model — connects back to one central idea: a fixed $\lambda$ per nucleus, applied to a large population, produces exponential decay. Keep that thread visible in every exam answer you write.

The decay law can also be written in differential form as ΔN/Δt = −λN. This states that the rate of decrease of the number of undecayed nuclei is proportional to the number still present. The minus sign matters because N is falling with time.