Mastering Decay Equations Revision Notes | A Level AQA Physics

1. Why Exponential Decay?

Picture a large crowd of unstable nuclei. Each one has no memory of how long it has been waiting — at any instant, every nucleus has the same fixed probability of decaying. That single idea forces the mathematics into one specific shape: exponential decay.

The rate at which nuclei disappear is proportional to how many are still present. Halve the number of nuclei and you halve the rate of decay. This self-similar behaviour is the hallmark of exponential processes, and it is why the same equation governs everything from carbon-14 dating to hospital radioisotope dosing.

The mathematical statement of this proportionality is $\frac{dN}{dt} = -\lambda N$, where the minus sign confirms that $N$ is decreasing. Solving this differential equation gives the core result of the entire topic.

Notice that when $t = 0$, $e^{0} = 1$ so $N = N_0$ — the equation is self-consistent at the start. As $t$ increases, the exponent $-\lambda t$ becomes more negative, so $e^{-\lambda t}$ shrinks towards zero. The number of undecayed nuclei never quite reaches zero, which reflects the probabilistic nature of radioactive decay.

The graph confirms the key feature: each half-life interval reduces $N$ by exactly half, regardless of when you start counting. That constant fractional decrease is what distinguishes exponential decay from linear decay.

2. Activity and the Link to Nuclei

Knowing how many nuclei remain is useful, but in practice you measure activity — the number of decays per second detected by a Geiger counter or scintillation detector. Activity connects directly to $N$ through the decay constant.

This equation is not just a definition — it follows directly from the decay rate equation. Since $A = \left|\frac{dN}{dt}\right| = \lambda N$, activity and the number of nuclei are always proportional to each other through the same constant $\lambda$.

Because $N$ decays exponentially, $A$ must also decay exponentially with the same decay constant. Substituting $N = N_0 e^{-\lambda t}$ into $A = \lambda N$ gives the activity version of the decay equation.

The two equations $N = N_0 e^{-\lambda t}$ and $A = A_0 e^{-\lambda t}$ are structurally identical. Any technique you apply to one — rearranging for $t$, finding $\lambda$, or reading off a graph — works equally for the other. The only difference is what you measure: nuclei or decays per second.

3. Finding the Number of Nuclei from Mass

Exam questions frequently give you a mass of radioactive material rather than a number of nuclei. To use the decay equations, you need to convert mass into $N$. This is where the Avogadro constant and molar mass come in.

One mole of any substance contains $N_A = 6.02 \times 10^{23}$ particles. The molar mass $M$ (in g mol$^{-1}$) tells you the mass of one mole. So if you have a mass $m$ of a nuclide with molar mass $M$, the number of nuclei is:

For a pure isotope, the molar mass in g mol$^{-1}$ equals the nucleon number (mass number). For example, carbon-14 has a molar mass of 14 g mol$^{-1}$, and iodine-131 has a molar mass of 131 g mol$^{-1}$. Always check whether the question gives mass in grams or kilograms and convert before substituting.

4. Worked Examples: Putting the Equations Together

The real challenge in exam questions is chaining these equations together: convert mass to $N_0$, find $A_0$ using $A = \lambda N$, then track how $N$ or $A$ changes over time. The worked examples below model this chain explicitly.

Worked example: Calculating Activity from Mass and Decay Constant

A sample of iodine-131 has a mass of $1.31 \times 10^{-6}$ g. The decay constant of iodine-131 is $1.00 \times 10^{-6}$ s$^{-1}$. Calculate the initial activity of the sample. ($N_A = 6.02 \times 10^{23}$ mol$^{-1}$, molar mass of I-131 = 131 g mol$^{-1}$)

1. $n = \frac{m}{M} = \frac{1.31 \times 10^{-6}}{131} = 1.00 \times 10^{-8}$ mol — Divide the mass in grams by the molar mass. [M1] for correct method.
2. $N_0 = n \times N_A = 1.00 \times 10^{-8} \times 6.02 \times 10^{23} = 6.02 \times 10^{15}$ — Multiply moles by Avogadro's constant to get the number of individual nuclei. [M1]
3. $A_0 = 1.00 \times 10^{-6} \times 6.02 \times 10^{15}$ — Activity equals decay constant multiplied by number of nuclei. [M1]
4. $A_0 = 6.02 \times 10^{9}$ Bq — Units are becquerels (Bq = decays per second). [A1] for correct value and unit.

Answer: $A_0 = 6.02 \times 10^{9}$ Bq

Now extend this: what if the question asks for the activity after a certain time has passed? You use $A = A_0 e^{-\lambda t}$, substituting the $A_0$ you just calculated. The chain continues naturally.

Worked example: Finding Remaining Nuclei After a Given Time

A radioactive source initially contains $N_0 = 4.80 \times 10^{20}$ undecayed nuclei. The decay constant is $\lambda = 2.00 \times 10^{-4}$ s$^{-1}$. Calculate the number of undecayed nuclei remaining after $t = 5000$ s.

1. $N = N_0 \, e^{-\lambda t}$ — State the equation first — this earns [M1] in an exam even if arithmetic goes wrong later.
2. $\lambda t = 2.00 \times 10^{-4} \times 5000 = 1.00$ — Multiply decay constant by time. Check units: s$^{-1}$ $\times$ s = dimensionless. [M1]
3. $e^{-1.00} = 0.3679$ — Use your calculator: $e^{-1} \approx 0.368$. This is the fraction of nuclei remaining.
4. $N = 4.80 \times 10^{20} \times 0.3679 = 1.77 \times 10^{20}$ — Carry the full calculator value of $e^{-1}$ through to avoid rounding error. [A1]
5. $\lambda t = 1.00$, so roughly one $e$-folding has occurred; $N$ should be about $37\%$ of $N_0$. $\frac{1.77}{4.80} \approx 0.369$ ✓ — Always verify the fraction makes physical sense. After one half-life, $N$ would be $50\%$; after one $e$-folding ($\lambda t = 1$), it is $\approx 37\%$, which is consistent.

Answer: $N = 1.77 \times 10^{20}$ undecayed nuclei

The sanity check in step 5 is worth building into every calculation. If $\lambda t$ is small (much less than 1), very few nuclei have decayed and $N \approx N_0$. If $\lambda t$ is large, most nuclei have decayed and $N \ll N_0$. A quick order-of-magnitude check catches arithmetic slips before they cost marks.

5. Rearranging for Time or Decay Constant

So far the examples have calculated $N$ or $A$ at a known time. Exam questions also ask you to work backwards: given a measured activity ratio, find the time elapsed, or find $\lambda$ from a half-life. Both require taking the natural logarithm of both sides.

Starting from $A = A_0 e^{-\lambda t}$, divide both sides by $A_0$ to get $\frac{A}{A_0} = e^{-\lambda t}$. Taking $\ln$ of both sides: $\ln\left(\frac{A}{A_0}\right) = -\lambda t$. Rearranging gives $t = -\frac{1}{\lambda} \ln\left(\frac{A}{A_0}\right)$, or equivalently $t = \frac{1}{\lambda} \ln\left(\frac{A_0}{A}\right)$.

The connection between $\lambda$ and the half-life $t_{1/2}$ follows the same logic. At $t = t_{1/2}$, $N = \frac{N_0}{2}$, so $\frac{1}{2} = e^{-\lambda t_{1/2}}$. Taking $\ln$: $\ln\left(\frac{1}{2}\right) = -\lambda t_{1/2}$, giving $\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{t_{1/2}}$. This relationship lets you convert freely between $\lambda$ and $t_{1/2}$.

Worked example: Finding Time Elapsed from an Activity Ratio

The activity of a cobalt-60 source falls to $30\%$ of its initial value. The half-life of cobalt-60 is 5.27 years. Calculate the time elapsed. Give your answer in years.

1. $\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.6931}{5.27} = 0.1315$ yr$^{-1}$ — Keep $t_{1/2}$ in years so that $\lambda$ is in yr$^{-1}$ and the final answer comes out in years. [M1]
2. $\frac{A}{A_0} = 0.30$, so $\ln\left(\frac{A_0}{A}\right) = \ln\left(\frac{1}{0.30}\right) = \ln(3.333) = 1.204$ — Note: $\ln\left(\frac{A_0}{A}\right) = \ln\left(\frac{1}{0.30}\right)$ because $A = 0.30 \, A_0$. [M1]
3. $t = \frac{\ln(A_0/A)}{\lambda} = \frac{1.204}{0.1315} = 9.16$ yr — Divide the logarithm by $\lambda$. Units: $\frac{\text{dimensionless}}{\text{yr}^{-1}} = \text{yr}$. [A1]
4. After 1 half-life (5.27 yr): $A = 50\%$. After 2 half-lives (10.54 yr): $A = 25\%$. Our answer of 9.16 yr gives $30\%$, which lies between 50% and 25% — consistent. ✓ — The answer falls between one and two half-lives, which matches the activity being between 50% and 25% of $A_0$.

Answer: $t \approx 9.16$ years

6. Exam Technique and Common Errors

Decay equation questions are reliable mark-earners once you know the traps. The errors below account for the majority of lost marks in this topic at A-level, so it is worth being deliberate about each one.

One more examiner favourite: questions that give activity in kBq or MBq rather than Bq. Convert to Bq (multiply by $10^3$ or $10^6$) before substituting into $A = \lambda N$, otherwise $N$ will be out by the same factor.

7. Bringing It All Together: A Full Multi-Step Problem

The most demanding exam questions combine all four spec requirements in a single problem: convert mass to nuclei using $N_A$, find initial activity using $A = \lambda N$, then track the decay over time using $A = A_0 e^{-\lambda t}$. The worked example below models the complete chain.

Worked example: Full Multi-Step Decay Problem

A hospital physics department prepares a sample of technetium-99m (Tc-99m) with a mass of $2.00 \times 10^{-9}$ g. The half-life of Tc-99m is $2.16 \times 10^{4}$ s and its molar mass is 99 g mol$^{-1}$. (a) Calculate the initial number of nuclei $N_0$. (b) Calculate the initial activity $A_0$. (c) Calculate the activity 6 hours after preparation. ($N_A = 6.02 \times 10^{23}$ mol$^{-1}$)

1. $N_0 = \frac{m}{M} \times N_A = \frac{2.00 \times 10^{-9}}{99} \times 6.02 \times 10^{23}$ — Mass is already in grams; molar mass is 99 g mol$^{-1}$. [M1]
2. $n = \frac{2.00 \times 10^{-9}}{99} = 2.020 \times 10^{-11}$ mol $N_0 = 2.020 \times 10^{-11} \times 6.02 \times 10^{23} = 1.216 \times 10^{13}$ — Keep extra significant figures here to avoid rounding errors in later steps. [A1]
3. $\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.6931}{2.16 \times 10^{4}} = 3.209 \times 10^{-5}$ s$^{-1}$ — Half-life is already in seconds. [M1]
4. $A_0 = \lambda N_0 = 3.209 \times 10^{-5} \times 1.216 \times 10^{13} = 3.90 \times 10^{8}$ Bq — Units: s$^{-1}$ $\times$ (dimensionless count) = s$^{-1}$ = Bq. [A1]
5. $t = 6 \times 3600 = 21600$ s — Always convert time to seconds when $\lambda$ is in s$^{-1}$. This is a common unit trap.
6. $\lambda t = 3.209 \times 10^{-5} \times 21600 = 0.6931$ — Notice that $\lambda t = \ln 2 \approx 0.693$, which means exactly one half-life has elapsed (6 hours $\approx t_{1/2}$). This is a useful cross-check.
7. $A = 3.90 \times 10^{8} \times e^{-0.6931} = 3.90 \times 10^{8} \times 0.5000 = 1.95 \times 10^{8}$ Bq — $e^{-0.6931} = 0.5000$ exactly (since $e^{-\ln 2} = 1/2$). After one half-life, activity halves. [A1]

Answer: (a) $N_0 = 1.22 \times 10^{13}$ nuclei; (b) $A_0 = 3.90 \times 10^{8}$ Bq; (c) $A = 1.95 \times 10^{8}$ Bq after 6 hours

Step 6 revealed something elegant: 6 hours is almost exactly one half-life of Tc-99m, so the activity halved precisely. Spotting this kind of relationship in an exam saves calculator time and provides a built-in check. When $\lambda t = \ln 2$, you know one half-life has elapsed and $N$ or $A$ is exactly half its initial value.

Every spec requirement has now appeared in both teaching and application: $N = N_0 e^{-\lambda t}$ and $A = A_0 e^{-\lambda t}$ were used in the worked examples; $A = \lambda N$ linked activity to nuclei; and the Avogadro constant with molar mass converted mass to $N_0$. These four tools, used in sequence, solve every standard decay equation question at A-level.