5.4 Radioactive decay
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A2 content, including Core Practical 15. Mass deficit and decay energies are in 5.3; the quark picture of beta decay is in Unit 4.
What this page covers
7 points
- Background radiation: sources and how to correct a count rate for it
- The nature, ionising ability, penetration and range of alpha, beta and gamma radiation
- Nuclear equations for α, β⁻, β⁺ and γ emission
- The spontaneous and random nature of decay, activity and the decay constant
- Half-life, \(A = \lambda N\), \(N = N_0e^{-\lambda t}\), \(\lambda = \ln 2/t_{1/2}\) and the log forms
- Carbon dating and medical tracers as contexts
- Core Practical 15: the absorption of gamma radiation by lead
In one minute
6 points
- Every calculation on this page comes from one statement, that each unstable nucleus has the same constant probability of decaying per second, which gives \(A = \lambda N\), a constant half-life and exponential decay.
- Always subtract the background count rate (measured with the source removed) before using a count rate.
- Alpha: strongly ionising, a few cm of air, stopped by paper. Beta: moderately ionising, stopped by a few mm of aluminium. Gamma: weakly ionising, only reduced by several cm of lead.
- In every nuclear equation nucleon numbers and proton numbers balance. α: A − 4, Z − 2. β⁻: Z + 1. β⁺: Z − 1. γ: no change.
- Decay is random (no way to tell which nucleus decays next) and spontaneous (unaffected by temperature, pressure or chemistry).
- \(N = N_0e^{-\lambda t}\) with \(t_{1/2} = \ln 2/\lambda\); a graph of \(\ln A\) against \(t\) is a straight line of gradient \(-\lambda\).
Contents
14 sections
- What is background radiation, and how do you correct for it?
- What are alpha, beta and gamma radiation?
- How do you write a nuclear equation?
- What does it mean that decay is random and spontaneous?
- What is half-life, and how do you use the decay equations?
- How do you find the decay constant from a log graph?
- How are carbon dating and medical tracers based on decay?
- How do you investigate the absorption of gamma radiation by lead?
- Exam tips for 5.4
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is background radiation, and how do you correct for it?
Natural sources: radon gas from rocks and soil (often the largest single contribution indoors), rocks and building materials such as granite, cosmic rays, and radioactive isotopes in food, drink and our bodies (potassium-40, carbon-14). Artificial sources: medical X-rays and treatments, and small contributions from past weapons tests and the nuclear industry.
A Geiger–Müller (GM) tube counts background events as well as those from your source. The background fluctuates randomly too, so measure it for several minutes and use the mean rate. Then
\[\text{corrected count rate} = \text{measured count rate} - \text{background count rate}\]Not in the formulae list (learn it). Both rates must be in the same unit (s⁻¹ or min⁻¹). Correct every reading before you halve it, plot it or take its log; it matters most when the source count rate is small.
With no source present, a GM tube records 486 counts in 20.0 minutes. With a source in front of the tube, 1840 counts are recorded in 2.00 minutes. (a) Calculate the corrected count rate due to the source, in s⁻¹. (b) Explain why the background was measured for much longer.
(a) Background \(= 486 \div 20.0 = 24.3\) min⁻¹; with source \(= 1840 \div 2.00 = 920\) min⁻¹. Corrected: \(920 - 24.3 = 896\) min⁻¹: 14.9 s⁻¹.
(b) Background counts fluctuate randomly; a longer count averages out the fluctuations and gives a more reliable mean rate.
What are alpha, beta and gamma radiation?
Unstable nuclei become more stable by emitting radiation. What each kind is decides how strongly it ionises and how far it travels.
| Radiation | What it is | Charge | Ionising | Range in air | Absorbed by |
|---|---|---|---|---|---|
| alpha, α | helium nucleus: 2 protons + 2 neutrons | +2e | strong | a few cm | a sheet of paper, or the outer layer of skin |
| beta-minus, β⁻ | high-speed electron created in the nucleus | −e | moderate | up to about 1 m (varies with energy) | a few mm of aluminium |
| beta-plus, β⁺ | positron (antielectron) created in the nucleus | +e | moderate | very short: it meets an electron and annihilates into two gamma photons | — |
| gamma, γ | high-frequency electromagnetic radiation: photons | 0 | weak | no fixed range; intensity falls roughly as 1/r² | reduced, never stopped completely, by several cm of lead or about a metre of concrete |
- Ionising vs penetrating: each ionisation takes energy. An α particle (large charge, slow, massive) ionises thousands of atoms per mm of air and runs out of energy within a few cm; an uncharged γ photon interacts only occasionally, so it travels much further.
- Hazard: outside the body α cannot reach living cells, so β and γ are the concern. Inside the body an α source is the most dangerous, because all its energy is deposited in a small volume of tissue.
- Identifying a source: add absorbers between it and a detector. A large drop with paper means α; a further drop with a few mm of aluminium means β; what remains is γ.
Saying a β⁻ particle is one of the atom's orbiting electrons. It is created in the nucleus at the moment of decay, when a neutron changes into a proton.
How do you write a nuclear equation?
A nuclide is written \({}^{A}_{Z}\text{X}\): \(A\) is the nucleon (mass) number, \(Z\) the proton (atomic) number. In every nuclear equation the total nucleon number and the total proton number (charge) are the same on both sides.
| Decay | General equation | What happens in the nucleus | Change |
|---|---|---|---|
| α | \({}^{A}_{Z}\text{X} \rightarrow {}^{A-4}_{Z-2}\text{Y} + {}^{4}_{2}\alpha\) | 2 protons and 2 neutrons leave together | A − 4, Z − 2 |
| β⁻ | \({}^{A}_{Z}\text{X} \rightarrow {}^{A}_{Z+1}\text{Y} + {}^{0}_{-1}\beta + \bar{\nu}_e\) | a neutron becomes a proton; an electron and an electron antineutrino are emitted | A same, Z + 1 |
| β⁺ | \({}^{A}_{Z}\text{X} \rightarrow {}^{A}_{Z-1}\text{Y} + {}^{0}_{+1}\beta + \nu_e\) | a proton becomes a neutron; a positron and an electron neutrino are emitted | A same, Z − 1 |
| γ | \({}^{A}_{Z}\text{X}^{*} \rightarrow {}^{A}_{Z}\text{X} + \gamma\) | an excited nucleus drops to a lower energy state and emits a photon; often follows α or β decay | no change |
The (anti)neutrino has no charge and almost no mass, so it does not affect the balance, but include it: it is needed to conserve energy, momentum and lepton number, and it explains why β particles from one isotope have a range of energies.
(a) Polonium-210 (\({}^{210}_{84}\text{Po}\)) is an α emitter. (b) Strontium-90 (\({}^{90}_{38}\text{Sr}\)) is a β⁻ emitter. (c) Fluorine-18 (\({}^{18}_{9}\text{F}\)) is a β⁺ emitter. Write each decay equation. (d) Thorium-232 (\({}^{232}_{90}\text{Th}\)) decays through α and β⁻ emissions to stable lead-208 (\({}^{208}_{82}\text{Pb}\)). How many of each are emitted?
(a) \({}^{210}_{84}\text{Po} \rightarrow {}^{206}_{82}\text{Pb} + {}^{4}_{2}\alpha\)
(b) Element 39 is yttrium: \({}^{90}_{38}\text{Sr} \rightarrow {}^{90}_{39}\text{Y} + {}^{0}_{-1}\beta + \bar{\nu}_e\)
(c) Element 8 is oxygen: \({}^{18}_{9}\text{F} \rightarrow {}^{18}_{8}\text{O} + {}^{0}_{+1}\beta + \nu_e\)
(d) Only α changes A: fall in A = 232 − 208 = 24, so 6 α decays, which alone would take Z to 90 − 12 = 78. Each β⁻ raises Z by 1: 82 − 78 = 4. 6 α and 4 β⁻.
After writing a nuclear equation, add the top numbers and the bottom numbers on each side. Write β⁻ as \({}^{0}_{-1}\beta\) and β⁺ as \({}^{0}_{+1}\beta\) so the charge balance is visible.
What does it mean that decay is random and spontaneous?
- Random: it is impossible to predict which nucleus will decay next, or when a particular nucleus will decay. Every nucleus of the same isotope has the same probability of decaying in the next second, however long it has existed.
- Spontaneous: decay is not triggered by anything outside the nucleus; its rate is unaffected by temperature, pressure, chemical combination or physical state.
- Evidence of randomness: repeated equal-time counts from a long-lived source scatter about a mean.
If each of \(N\) nuclei has probability \(\lambda\Delta t\) of decaying in a short time \(\Delta t\), the expected number decaying is \(\lambda N\Delta t\): the rate of decay is proportional to the number left. For large \(N\) the fluctuations average out and the sample follows a smooth exponential; for a few nuclei the pattern is ragged. A detector catches only a fraction of the emissions, so the count rate is smaller than the activity but proportional to it for a fixed arrangement.
What is half-life, and how do you use the decay equations?
Both are in the formulae list; the minus sign shows that \(N\) decreases. With \(A\) in Bq, \(\lambda\) must be in s⁻¹, so convert a half-life in hours, days or years to seconds first.
\[N = N_0e^{-\lambda t} \qquad A = A_0e^{-\lambda t} \qquad \lambda = \frac{\ln 2}{t_{1/2}}\]These are in the formulae list. Put \(N = N_0/2\) at \(t = t_{1/2}\): \(\tfrac{1}{2} = e^{-\lambda t_{1/2}}\), so \(\lambda t_{1/2} = \ln 2\). After \(n\) half-lives the fraction left is \((\tfrac{1}{2})^n\) (not in the list: learn it). The same equations work for \(N\), \(A\), the corrected count rate and the mass of isotope left, since all are proportional to \(N\).
To find a half-life from a graph, read the time for the activity to fall from any value to half of it; repeat from two or three starting points and average.
A patient is injected with technetium-99m, a gamma emitter with a half-life of 6.0 hours; the activity at injection is 400 MBq. (a) Calculate the decay constant in s⁻¹. (b) Calculate the number of nuclei injected. (c) Calculate the activity 24 hours later. (d) Calculate the time for the activity to fall to 1.0 MBq.
(a) \(\lambda = 0.693 \div (6.0 \times 3600)\): \(3.21 \times 10^{-5}\) s⁻¹.
(b) \(N = A/\lambda = 4.00 \times 10^{8} \div 3.21 \times 10^{-5}\): \(1.2 \times 10^{13}\) (1.25 × 10¹³).
(c) 24 h is exactly 4 half-lives: \(400 \times (\tfrac{1}{2})^4\): 25 MBq.
(d) \(t = \ln(A_0/A)/\lambda = \ln(400 \div 1.0) \div 3.21 \times 10^{-5} = 1.87 \times 10^{5}\) s: 52 hours. Shortcut: \(t = t_{1/2} \times \ln(A_0/A)/\ln 2 = 6.0 \times 8.64\) h.
A sample contains 2.0 µg of iodine-131 (atom mass 131 u, half-life 8.0 days; 1 u = 1.66 × 10⁻²⁷ kg). (a) Calculate the activity. (b) Calculate how long it takes for the activity to fall to 5.0% of this value.
(a) \(N = 2.0 \times 10^{-9} \div (131 \times 1.66 \times 10^{-27}) = 9.20 \times 10^{15}\); \(\lambda = 0.693 \div (8.0 \times 24 \times 3600) = 1.00 \times 10^{-6}\) s⁻¹; \(A = \lambda N\): \(9.2 \times 10^{9}\) Bq (mass given to 2 s.f.).
(b) \(t = t_{1/2} \times \ln(1/0.050)/\ln 2 = 8.0 \times \ln 20 \div 0.693\): 35 days.
Using a half-life in days or years in \(A = \lambda N\). The becquerel is decays per second, so \(\lambda\) must be in s⁻¹. In \(N = N_0e^{-\lambda t}\) any time unit works, as long as \(\lambda\) and \(t\) share it. Remember µg → kg is × 10⁻⁹.
How do you find the decay constant from a log graph?
An exponential curve is hard to judge by eye. Taking natural logs turns it into a straight line:
\[\ln N = \ln N_0 - \lambda t \qquad \ln A = \ln A_0 - \lambda t\]These are in the formulae list too (they are the natural logs of the exponential forms). Compared with \(y = mx + c\), plot \(\ln A\) (or \(\ln C\)) against \(t\): gradient \(= -\lambda\), intercept \(= \ln A_0\). A straight line confirms exponential decay. Label the axis ln (C / s⁻¹), since a log has no unit.
Corrected count rates from a short-lived sample, every 40 s:
| t / s | 0 | 40 | 80 | 120 | 160 | 200 | 240 |
|---|---|---|---|---|---|---|---|
| C / s⁻¹ | 64.0 | 45.5 | 30.6 | 22.1 | 15.0 | 10.3 | 7.57 |
| ln (C / s⁻¹) | 4.16 | 3.82 | 3.42 | 3.10 | 2.71 | 2.33 | 2.02 |
(a) Explain why ln C against t should be a straight line. (b) Use the graph to determine the decay constant and half-life. (c) Determine the initial count rate.
(a) Taking logs of \(C = C_0e^{-\lambda t}\) gives \(\ln C = -\lambda t + \ln C_0\), the form \(y = mx + c\) with constant \(m = -\lambda\).
(b) Gradient \(= -1.80 \div 200 = -9.0 \times 10^{-3}\) s⁻¹, so \(\lambda = 9.0 \times 10^{-3}\) s⁻¹; \(t_{1/2} = 0.693 \div 9.0 \times 10^{-3}\): 77 s.
(c) Intercept \(\ln C_0 = 4.16\), so \(C_0 = e^{4.16}\): 64 s⁻¹. Check: the table halves from 64.0 to about 32 s⁻¹ in about 80 s, consistent with 77 s.
Taking ln of the measured count rate before subtracting background. The background does not decay or get absorbed, so leaving it in makes the ln graph curve upwards at long times (or large thicknesses) and gives a gradient that is too shallow.
How are carbon dating and medical tracers based on decay?
Carbon dating
Cosmic rays produce carbon-14 in the upper atmosphere (β⁻, half-life 5730 years), which mixes into atmospheric carbon dioxide. Living things keep exchanging carbon with their surroundings, so the activity per gram of their carbon stays steady. After death no new carbon-14 is taken in and the activity per gram falls exponentially; comparing an old sample with living material gives the time since death.
Limitations: it assumes past atmospheric carbon-14 levels were the same (in practice corrected with samples of known age); very old samples have so little activity that background and random fluctuations swamp it, limiting the method to roughly 50 000 years; contamination with modern carbon gives wrong ages.
Carbon from living wood has an activity of 0.25 Bq per gram; carbon from a wooden tool has 0.084 Bq per gram. Half-life of carbon-14 = 5730 years. (a) Calculate the age of the tool. (b) The sample contains 5.0 g of carbon. Explain why measuring its activity accurately is difficult.
(a) \(\lambda = 0.693 \div 5730 = 1.21 \times 10^{-4}\) year⁻¹ (no need for seconds). \(t = \ln(0.25 \div 0.084) \div 1.21 \times 10^{-4} = 1.091 \div 1.21 \times 10^{-4}\): \(9.0 \times 10^{3}\) years.
(b) Activity \(= 5.0 \times 0.084 = 0.42\) Bq: fewer than one count per second even if every emission were detected, no more than a typical background rate. The corrected count is small with a large random uncertainty, so long counting times and shielding are needed.
Medical tracers
- A tracer is a radioactive isotope attached to a chemical the body takes to the organ of interest; a camera outside the body builds an image from the radiation.
- Gamma emitter: γ leaves the body and, being weakly ionising, does least damage on the way. An α emitter would be absorbed inside and do most harm.
- Half-life of a few hours: long enough to reach the organ and complete the scan, short enough that the activity soon becomes negligible. Technetium-99m (6.0 h) is the standard example.
- The daughter should be stable or very long-lived and the chemical non-toxic. Positron emitters such as fluorine-18 are used in PET scans: the two gamma photons from each annihilation travel in opposite directions and locate the source.
How do you investigate the absorption of gamma radiation by lead?
Core Practical 15 finds how the gamma count rate through lead depends on the thickness of lead.
- With the source locked away, record the background over several minutes (e.g. 10 min).
- Clamp the GM tube and set the sealed source a fixed distance \(d\) from its window. Record the count over a fixed time (e.g. 100 s) with no lead.
- Measure each sheet's thickness with a micrometer at several points and average. Add sheets one at a time, recording the count for each total thickness \(x\); repeat and average.
- Correct every count rate for background, then plot \(\ln C\) against \(x\).
- Control variables: same source, distance (gamma intensity also falls with distance), counting time and GM tube settings. Longer counts give more reliable rates; steepest and least steep acceptable lines give the gradient's uncertainty.
- Safety: handle the source with tongs, pointing away from people, and return it to its lead-lined box promptly; keep exposure short and distant; lead is toxic, so wear gloves or wash hands.
If each millimetre of lead absorbs the same fraction of the photons reaching it, the corrected count rate falls exponentially: \(C = C_0e^{-\mu x}\), with \(\mu\) a constant for lead and this gamma energy. It is the same mathematics as decay with time: \(\ln C\) against \(x\) has gradient \(-\mu\), and the half-thickness \(\ln 2/\mu\) plays the part of the half-life.
Background = 0.45 s⁻¹; counts taken for 100 s at each thickness.
| x / mm | counts in 100 s | C / s⁻¹ | ln (C / s⁻¹) |
|---|---|---|---|
| 0 | 4287 | 42.4 | 3.75 |
| 4 | 3325 | 32.8 | 3.49 |
| 8 | 2718 | 26.7 | 3.29 |
| 12 | 2122 | 20.8 | 3.03 |
| 16 | 1733 | 16.9 | 2.83 |
| 20 | 1337 | 12.9 | 2.56 |
| 24 | 1104 | 10.6 | 2.36 |
(a) Show how C for 12 mm was calculated. (b) Use the ln graph to determine μ. (c) Calculate the thickness that halves the count rate. (d) Estimate the thickness needed to reduce C to 1.0% of its value with no lead.
(a) \(C = 2122 \div 100 - 0.45\): 20.8 s⁻¹.
(b) Gradient \(= -1.16 \div 20 = -0.058\) mm⁻¹, so \(\mu = 0.058\) mm⁻¹.
(c) \(x_{1/2} = 0.693 \div 0.058\): 12 mm.
(d) \(x = \ln(100)/\mu = 4.61 \div 0.058\): 79 mm (about 8 cm). The table agrees: every 12 mm roughly halves C (42.4 → 20.8 → 10.6 s⁻¹).
Exam tips for 5.4
Make "subtract background" your first line whenever a count rate appears, with both rates in the same unit.
In \(A = \lambda N\), convert the half-life to seconds. In \(A = A_0e^{-\lambda t}\) you may stay in hours, days or years, provided \(\lambda\) and \(t\) share the unit.
For a whole number of half-lives, halve repeatedly; otherwise use the exponential. Check your answer lies between the values for the whole numbers either side.
"Random": you cannot predict which nucleus will decay or when, but each has a constant probability per unit time. "Spontaneous": unaffected by external conditions. Do not swap them.
Use a triangle covering more than half the line, quote the gradient with its sign, then say λ = −gradient. Scatter about the line is expected, because decay is random.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| After two half-lives the sample has completely decayed. | Each half-life halves what is left. | After two half-lives ¼ remains, after three ⅛, and so on. |
| Heating a radioactive source makes it decay faster. | Decay is nuclear, unaffected by the energies of heating or chemical change. | Decay is spontaneous: λ and the half-life are fixed for each isotope. |
| A nucleus that has existed for a long time is "due" to decay. | Nuclei have no memory. | Every nucleus has the same probability of decaying per second, whatever its age. |
| The count rate equals the activity. | A detector catches only some emissions and also counts background. | Corrected count rate is proportional to activity for a fixed arrangement, but smaller. |
| Gamma emission turns the nucleus into a different element. | A photon has no charge and no nucleons. | Gamma emission leaves A and Z unchanged; the nucleus just loses energy. |
| Alpha is the most dangerous radiation in every situation. | Outside the body, α is stopped by the dead outer layer of skin. | α is most dangerous inside the body; outside, β and γ are the bigger hazards. |
How this topic connects
5.3 Binding energy, fission and fusion, and the standard model (Unit 4)
Why nuclei are unstable and the energy released in a decay come from mass deficit and \(\Delta E = c^2\Delta m\). In β⁻ decay a down quark becomes an up quark (udd → uud); neutrinos conserve lepton number; a positron annihilates with an electron to give two gamma photons.
Capacitors (Unit 4)
Discharge follows the same mathematics: \(Q = Q_0e^{-t/RC}\), time to halve \(RC\ln 2\), and \(\ln Q\) against \(t\) is a straight line, just as \(t_{1/2} = \ln 2/\lambda\) here.
Photons (Unit 2) and practical skills (Unit 6)
Gamma photon energies follow \(E = hf\) and are quoted in MeV; the log graphs, gradients and uncertainties of Core Practical 15 are exactly the skills assessed in practical work.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| corrected count rate = measured − background | removing background before any analysis | no — learn it |
| \(A = \lambda N\) | activity from the number of nuclei (λ in s⁻¹ for Bq) | yes |
| \(\dfrac{dN}{dt} = -\lambda N\) | rate of decay proportional to nuclei left | yes |
| \(\lambda = \ln 2/t_{1/2}\) | decay constant from half-life and back | yes |
| \(N = N_0e^{-\lambda t}\), \(A = A_0e^{-\lambda t}\) | number or activity after any time | yes |
| \(\ln N = \ln N_0 - \lambda t\), \(\ln A = \ln A_0 - \lambda t\) | straight-line graph: gradient −λ; solving for t | yes |
| fraction left = \((\tfrac{1}{2})^n\) | after a whole number n of half-lives | no — learn it |
| N = mass of sample ÷ mass of one atom | number of nuclei from a mass (atom mass ≈ nucleon number × u) | no — derive it |
| \(C = C_0e^{-\mu x}\) | gamma absorption in lead (Core Practical 15) | no — learn the shape; ln C against x has gradient −μ |
Check your understanding
1. What fraction of a radioactive isotope remains after three half-lives?
\((\tfrac{1}{2})^3 = \tfrac{1}{8} = 0.125\) of the original nuclei.
2. An isotope has a half-life of 30 minutes. Calculate its decay constant in s⁻¹.
\(\lambda = 0.693 \div (30 \times 60) = 3.85 \times 10^{-4}\) s⁻¹.
3. A nucleus emits a β⁺ particle. What happens to its nucleon number and its proton number?
The nucleon number is unchanged and the proton number decreases by 1: a proton has changed into a neutron, emitting a positron and an electron neutrino.
4. A source has an activity of 3.0 × 10⁵ Bq and a half-life of 2.5 days. What is its activity after 10 days?
10 days = 4 half-lives, so \(A = 3.0 \times 10^{5} \times (\tfrac{1}{2})^4 = 1.9 \times 10^{4}\) Bq (1.88 × 10⁴ Bq).
5. Why does the count rate from a long-lived source vary from one 10 s count to the next?
Decay is random: the number of nuclei decaying in each interval fluctuates about the mean, even though the probability of decay per nucleus is constant. The background fluctuates in the same way.
Specification checklist
Wording is ours; numbers refer to the specification points.
- (137) I can name the main sources of background radiation and correct a measured count rate by subtracting the background count rate.
- (138) I can describe what alpha, beta and gamma radiation are, and compare their ionising ability, penetration and range.
- (139) I can write and complete balanced nuclear equations for α, β⁻, β⁺ and γ emission, including the neutrino or antineutrino.
- (140) I can describe how to investigate the absorption of gamma radiation by lead (Core Practical 15), including safety, control variables and the background correction, and analyse the results with a log graph.
- (141) I can explain what is meant by the spontaneous and random nature of radioactive decay.
- (142) I can find half-life from a decay graph and use \(A = \lambda N\), \(dN/dt = -\lambda N\), \(\lambda = \ln 2/t_{1/2}\), \(N = N_0e^{-\lambda t}\), \(A = A_0e^{-\lambda t}\) and their log forms, including in contexts such as carbon dating and medical tracers.