Radioactivity & Half-Life

By the end of this workbook you should be able to:

Explain what radioactivity is, and describe the random, spontaneous nature of nuclear decay.
Describe the origin, penetrating power and ionizing ability of alpha, beta and gamma radiation, and write balanced decay equations for each.
Explain how background radiation affects count-rate measurements, and correct readings for it.
Define activity, decay constant and half-life, and relate them using the exponential radioactive decay law and T½ = ln2/λ.
Describe practical applications of radionuclides, from medical tracers and smoke detectors to radiometric dating.

Name

1. What is radioactivity?

The nucleus of an atom is normally left alone once it forms. But some nuclei are unstable: without any warning or external cause, they can spontaneously change, throwing out a particle and/or a high-energy photon. This process is called radioactivity, and when a nucleus does this we say it has decayed or transmuted — because emitting a charged particle changes the proton number, turning the atom into a different element entirely.

Historical note

A history in three names. In 1896 Henri Becquerel noticed, almost by accident, that a uranium compound could blacken a photographic plate even in complete darkness — uranium was emitting some kind of energetic radiation on its own. Marie and Pierre Curie then discovered more radioactive elements, including polonium and radium, and physicists soon realised there were three distinct types of radiation coming from these materials. Having no idea what they actually were, they simply named them after the first three letters of the Greek alphabet: alpha (α), beta (β) and gamma (γ) — names that have stuck ever since.

A few key terms are used throughout this topic:

Radioactive — describes a substance containing unstable nuclei that will emit radiation.
Radioisotope / radionuclide — an isotope or nuclide with an unstable nucleus.
Transmutation — when a nuclide changes into a different element by emitting a particle.
Daughter product — the new nuclide left behind after a "parent" radionuclide emits a particle.

Example: A nucleus of ²³⁵₉₂U is unstable. At some unpredictable moment it emits an alpha particle (⁴₂He), transmuting into ²³¹₉₀Th — the daughter product. Note that "isotope" and "nuclide" are used almost interchangeably here; "nuclide" just emphasises that we're talking about the nucleus itself.

1Explain, in your own words, the difference between "radioactive decay" and ordinary chemical or biological decay.

This is descriptive — check your wording against Section 1 and with your teacher.

2A nucleus of thorium-232 decays by emitting an alpha particle. Name the daughter product's element if the resulting proton number is 88. (You are not expected to memorise element names from proton numbers — look this one up.)

radium (Z = 88).

2. Detecting radiation & background radiation

A typical school setup for investigating radioactivity uses a small sealed source, a Geiger–Müller (GM) tube, and a ratemeter (or counter). Radiation entering the GM tube ionises the gas inside it, producing a tiny burst of current for every particle or photon detected. The ratemeter counts these bursts and displays a count rate — typically in counts per second or per minute.

Fig. 2.1 — Basic components of a radioactivity experiment: source, GM tube, and ratemeter.

Why bigger counts are more reliable: Because decay is random, repeating a count will not give exactly the same result each time. A small average count (say, 9) might genuinely vary between about 6 and 12 from one trial to the next — a single reading could easily be misleading. A large average count (say, 900) varies by a much smaller fraction of its own size, so large count rates give far more reliable data.

Almost everything around us — rocks, soil, building materials, even our own bodies — contains tiny amounts of naturally radioactive material, and cosmic rays add to the total. This is background radiation, and a GM tube will register a small background count even with no obvious source nearby (typically 0.25–0.5 s⁻¹ in a school lab). Whenever background radiation is significant compared to the count being measured, it must be subtracted from every reading.

Worked example 2.1

A count of 42 was recorded from a source over one minute. The background count rate at that location was 0.44 s⁻¹. Determine the count rate from the source alone, adjusted for background.

Answer:
Background count in that minute = 0.44 × 60 = 26.4
Adjusted count from source = 42 − 26.4 = 16 min⁻¹

3Give two reasons why it is preferable to work with larger count rates in a radioactivity experiment.

This is descriptive — see the key-box "Why bigger counts are more reliable".

4In 15 minutes, a count of 5486 was measured with the GM tube directed at a source. The background count rate at that location was 18 per minute. Calculate the count rate, in counts per second, due to the source alone.

total count in 15 min = 5486, background count in 15 min = 18 × 15 = 270, adjusted count = 5216 in 15 min = 347.7 min⁻¹ = 5.8 s⁻¹.

5At a location with a background count of 22 min⁻¹, two separate measurements gave count rates of 50 min⁻¹ and 5000 min⁻¹. Compare how significant the background count is in each case.

for 50 min⁻¹, background (22) is nearly half the total — very significant; for 5000 min⁻¹, background is under 0.5% of the total — barely significant.

Try it yourself: gamma source and the inverse-square law

The simulation below lets you move a gamma source relative to a detector and watch how the measured count rate changes with distance — use it to see the inverse-square law in action, and to think about why background subtraction matters even more once the source's own count rate becomes small.

Move the distance slider, run a 10 s count at several distances, then use “Show Graph Analysis” to plot count rate against 1/d² and find the background count rate.

Interactive simulation by Dr Jones Physics — drjonesphysics.com/gamma. Used with permission.

An interactive simulation normally appears here — view this workbook on a device to use it.

3. Alpha, beta and gamma decay

Three kinds of radiation can be emitted from a decaying nucleus, and each corresponds to a different underlying change inside it.

Alpha (α) decay: An alpha particle is identical to a helium-4 nucleus: two protons and two neutrons bound tightly together, so it carries a nucleon number of 4 and a charge of +2. Emitting one removes 2 protons and 2 neutrons from the parent nucleus:

^A_ZX → ^A−4_Z−2Y + ⁴₂α

For example: ²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂α. All alpha particles from a given radionuclide's decay are emitted with the same kinetic energy (radium-226's alpha particles all carry 4.7 MeV) — a fact that becomes important evidence for nuclear energy levels, as covered in the companion Binding Energy & Nuclear Stability workbook.

Beta-minus (β⁻) decay: Inside an unstable nucleus, a neutron can convert into a proton, releasing a fast-moving electron (the beta-minus particle) and a new, almost undetectable particle called an antineutrino, v̄:

¹₀n → ¹₁p + ⁰₋₁β⁻ + v̄

The nucleon number is unchanged, but the proton number increases by one, forming a new element:

^A_ZX → ^A_Z+1Y + ⁰₋₁β⁻ + v̄

For example: ⁹⁰₃₈Sr → ⁹⁰₃₉Y + ⁰₋₁β⁻ + v̄. Unlike alpha particles, beta-minus particles come out with a whole range of energies up to some maximum — the reason for this range is explained in the companion Binding Energy & Nuclear Stability workbook.

Beta-plus (β⁺) decay: In a similar process, a proton inside the nucleus can convert into a neutron, releasing a positively charged electron — a positron, the antiparticle of the electron — plus a neutrino, v:

¹₁p → ¹₀n + ⁰₊₁β⁺ + v

^A_ZX → ^A_Z−1Y + ⁰₊₁β⁺ + v

For example: ²³₁₂Mg → ²³₁₁Na + ⁰₊₁β⁺ + v.

Gamma (γ) decay: After an alpha or beta decay, the daughter nucleus is often left with excess internal energy — an excited state, marked with an asterisk. It settles down by emitting a high-energy photon, a gamma ray, with no change to the nucleon number or proton number at all — no transmutation occurs:

²³⁴₉₀Th* → ²³⁴₉₀Th + γ

Historical note

Antimatter isn't just in the movies. A positron is real antimatter — the film Angels & Demons famously (and dramatically) got the physics roughly right: when a particle meets its antiparticle, both are annihilated and their mass is converted entirely into energy. For an electron–positron pair (each with rest mass 9.11 × 10⁻³¹ kg), E = mc² gives about 1.64 × 10⁻¹³ J released as two gamma-ray photons — which is why beta-plus decay is always accompanied, sooner or later, by annihilation radiation once the positron meets an ordinary electron.

Worked example 3.1

Write the balanced nuclear equation for the beta-negative decay of carbon-14 (Z = 6) into nitrogen (Z = 7).

Answer:
¹⁴₆C → ¹⁴₇N + ⁰₋₁β⁻ + v̄
Check: nucleon numbers 14 = 14 + 0 ✓. Proton numbers 6 = 7 + (−1) ✓.

6Write the balanced decay equation for the alpha decay of americium-241 (Z = 95) to neptunium (Z = 93).

²⁴¹₉₅Am → ²³⁷₉₃Np + ⁴₂α.

7Write the balanced decay equation for the beta-plus decay of sodium-22 (Z = 11) to neon (Z = 10).

²²₁₁Na → ²²₁₀Ne + ⁰₊₁β⁺ + ν.

8Explain why gamma emission does not change which element a nuclide is.

a gamma photon carries no charge and no nucleons, so emitting one changes neither Z nor A — the nuclide (and element) stays exactly the same, just in a lower energy state.

9Neutrinos and antineutrinos are extremely difficult to detect. Suggest why, in terms of their charge and mass.

neutrinos/antineutrinos have no charge (so they cannot ionise matter or be detected via ionisation) and extremely small mass, so they interact only via the weak force and almost never collide with anything.

4. Penetrating power and ionizing ability

As radiation passes through matter, it knocks electrons off atoms and molecules — ionization. The more strongly a type of radiation ionizes, the more quickly it loses its own energy, and so the less far it can travel: ionizing ability and penetrating power trade off against each other.

Property	Alpha (α)	Beta-minus (β⁻)	Gamma (γ)
Relative charge	+2	−1	0
Relative mass	4	1/1840	0
Ionizing ability	very high	low	very low
Typical range in air	≈ 4 cm	≈ 30 cm	barely absorbed
Stopped by	a sheet of paper	≈ 3 mm of aluminium	intensity halved by ≈ 2 cm of lead

Fig. 4.1 — Alpha particles are stopped by paper; beta particles need a few millimetres of aluminium; gamma rays are never fully absorbed, only progressively weakened by thick, dense shielding such as lead.

Why alpha is the least penetrating despite carrying the most energy: Alpha particles are heavy and carry a +2 charge, so they interact very strongly with the electrons in the atoms they pass. As a result, they lose their (typically large) kinetic energy very quickly, causing intense ionization over a very short distance. Beta particles are much lighter and carry only a −1 or +1 charge, so they generally cause less ionization per unit distance and can travel much farther. Gamma rays are uncharged photons, so they interact only occasionally with matter. Most pass straight through without interacting, making them the most penetrating and comparatively difficult to detect.

Because alpha and beta particles are charged, a beam of either can be deflected by electric or magnetic fields (gamma rays, having no charge, cannot). Alpha particles are deflected far less than beta particles in the same field, because they are much more massive and slower.

10Explain why a source that only emits alpha radiation is considered low-risk outside the body, but a serious hazard if it gets inside the body (for example, by being inhaled or swallowed).

outside the body, skin (or even a few cm of air) stops alpha particles before they reach living tissue; if inhaled or swallowed, they ionise very heavily over a very short range directly inside sensitive tissue, causing serious localised damage.

11Explain why gamma-ray sources are considered dangerous even from outside the body, unlike alpha sources.

gamma rays penetrate deep into (or all the way through) the body from outside, so external gamma sources can still damage internal organs, unlike alpha sources which cannot get in through skin.

12A beam containing alpha, beta and gamma radiation passes into a strong magnetic field directed into the page. Sketch what happens to each of the three components.

alpha particles curve one way (they are positive) with a large radius (heavy, fast); beta particles curve the opposite way with a smaller radius (light, negative); gamma rays travel straight through, undeflected (no charge).

5. Randomness, activity and half-life

Nobody can predict when any one particular unstable nucleus will decay — each decay is random (no pattern) and spontaneous (no external cause or trigger). Yet, put a huge number of identical unstable nuclei together, and their overall behaviour becomes remarkably predictable — in exactly the way that a single coin toss is unpredictable, but the fraction of heads from a million tosses is not.

Activity: The activity, A, of a radioactive source is the number of nuclei decaying every second. Its SI unit is the becquerel, Bq, where 1 Bq = one decay per second. A GM tube's count rate is not the same thing as activity (the tube cannot detect every single decay), but count rate is usually assumed to be proportional to activity.

The activity of every radioactive source falls over time, because as nuclei decay, fewer undecayed nuclei remain to produce further decays. The pattern this follows is an exponential decrease: in equal time intervals, the number of undecayed nuclei (and hence the activity) always falls by the same fraction — conventionally one half.

Fig. 5.1 — A radioactive decay curve: the number of undecayed nuclei (or the activity, or the count rate) halves in every successive half-life, T½ — an exponential decrease that, in theory, never quite reaches zero.

Half-life: The half-life, T½, of a radionuclide is the time taken for half of its undecayed nuclei to decay — equivalently, the time taken for its activity (or count rate) to halve. Half-lives range from tiny fractions of a second to billions of years, but the value is always fixed for a given nuclide — it never speeds up or slows down.

Worked example 5.1

Radium-226 has a half-life of 1620 years. A 0.010 g source contains 30% radium-226 and no other radionuclide. Calculate the mass of radium-226 remaining after 3240 years.

Answer:
3240 years = 2 half-lives, so the fraction remaining = (1/2)² = 1/4.
Mass remaining = ¼ × 0.30 × 0.010 = 7.5 × 10⁻⁴ g

13The half-life of francium-221 is 4.8 minutes. Calculate the fraction of a sample remaining undecayed after 24.0 minutes.

24.0 minutes = 5 half-lives, fraction remaining = (1/2)⁵ = 1/32 ≈ 0.031.

14A radioactive element has a half-life of 80 minutes and an initial count rate of 1000 min⁻¹. Determine how long it takes for the count rate to fall to 250 min⁻¹.

1000 → 250 is a fall to 1/4, i.e. 2 half-lives = 160 minutes.

15Explain, in terms of undecayed nuclei, why the activity of a source keeps decreasing even though the half-life itself stays constant.

activity A = λN, and λ is constant, but N itself keeps falling as nuclei decay — so even though the half-life (which only depends on λ) never changes, the activity (which also depends on the ever-shrinking N) keeps decreasing.

Try it yourself: decay simulation

The simulation below lets you watch a sample of radioactive nuclei decay in real time, one random event at a time, and compare the resulting curve of undecayed nuclei against the theoretical exponential prediction.

Choose an isotope and starting activity, then run the simulation and watch the activity fall away exponentially on the live graph.

Interactive simulation by Hookean Physics — sites.google.com/view/hookean-physics/decaysim. Used with permission.

An interactive simulation normally appears here — view this workbook on a device to use it.

6. Practical uses of radionuclides

Choosing a radionuclide for a real application means balancing its half-life, the type of radiation it emits, and the health risk involved. A few common examples:

Medical tracers — technetium-99m (half-life 6 hours, gamma emitter) is injected or swallowed and tracked through the body with a gamma camera. Its half-life is long enough to complete a scan but short enough to minimise the patient's radiation dose.
Carbon dating — living organisms maintain a constant fraction of radioactive carbon-14 while alive; once they die, that fraction steadily decreases with carbon-14's 5700-year half-life, letting scientists estimate age from once-living material.
Smoke detectors — a tiny amount of americium-241 ionises air between two electrodes; smoke disrupts the current and triggers the alarm.
Thickness control — a beta source and detector either side of a moving sheet of metal or plastic can monitor and automatically correct its thickness during manufacture.

16Suggest why technetium-99m's half-life of 6 hours is described as "a good compromise" for medical tracer use.

long enough that it doesn't decay away before/during the scan, but short enough that the patient isn't left radioactive for long afterwards, minimising their radiation dose.

17Explain why a beta source (rather than an alpha or gamma source) is well suited to controlling the thickness of a thin metal sheet during manufacture.

beta particles have a range of a few mm in metal — enough to be sensitive to small changes in thickness, but not so penetrating (like gamma) that thickness changes barely affect the reading, and not so easily stopped (like alpha) that they can't get through the sheet at all.

7. The decay constant and the radioactive decay law

Radioactive decay is random for any individual nucleus, but a large collection of identical unstable nuclei behaves very predictably: in any short time interval, a fixed fraction of the remaining nuclei will decay, regardless of how many happen to be left. That fraction, per unit time, is the decay constant.

The decay constant, λ: λ = −(ΔN/N) / Δt
The decay constant is the probability that any one given nucleus will decay in a unit time interval (SI unit: s⁻¹). Rearranging, the rate of decay −ΔN/Δt = λN — the more undecayed nuclei there are, the faster they decay, which is exactly the exponential behaviour seen in Section 5.

An important limitation: λ only equals the probability of decay in unit time when λΔt is small enough that N barely changes during the interval Δt. Over a longer interval, N keeps falling as nuclei decay, so ΔN/N measured over that whole interval is not a reliable estimate of λ — this is why the exponential equations below, rather than simple proportion, are needed for anything beyond a short time step.

Solving the defining equation for λ gives the equations you will use for almost every calculation in this section — each has exactly the same exponential form:

number of undecayed nuclei, N = N₀e^−λt

activity, A = A₀e^−λt

count rate, C = C₀e^−λt

where the subscript 0 always means "at the start of the time interval being considered" — not necessarily the very beginning of the source's existence. Because activity is simply the rate of decay,

activity, A = λN

Worked example 7.1

The activity of a radioactive sample is 3.6 × 10⁵ Bq. Its decay constant is 2.4 × 10⁻⁶ s⁻¹. Determine the number of undecayed nuclei in the sample.

Answer:
A = λN ⟹ N = A/λ = (3.6 × 10⁵)/(2.4 × 10⁻⁶) = 1.5 × 10¹¹ nuclei

Worked example 7.2 — combining half-life reasoning with the exponential equation

The number of radioactive nuclei in a sample falls to 1/8 of its initial value after 15 days. Find the half-life, and predict the fraction remaining after 40 days.

Answer — by counting half-lives:
1/8 = (1/2)³, so 15 days = 3 half-lives ⟹ T½ = 5.0 days. After 40 days = 8 half-lives, fraction remaining = (1/2)⁸ = 1/256 ≈ 0.0039.

Answer — by finding λ directly:
1/8 = e^−15λ ⟹ λ = (ln 8)/15 = 0.139 day⁻¹
After 40 days: N/N₀ = e^{−0.139 × 40} = e^−5.55 = 0.0039 — the same answer, as it must be.

Worked example 7.3

A radioisotope has a decay constant of 0.048 y⁻¹. Its activity at the start of 2020 was 620 Bq. Calculate its activity at the start of 2025.

Answer:
A = A₀e^−λt = 620 × e^{−0.048 × 5} = 620 × e^−0.24 = 620 × 0.787 = 488 Bq

18A sample initially contains 5.0 × 10¹⁰ undecayed nuclei. In the first 8.0 s, 620 of them decay. Estimate the decay constant, and explain why this can only ever be an approximation to the true value of λ.

λ ≈ (ΔN/N)/Δt = 620/(5.0 × 10¹⁰ × 8.0) ≈ 1.55 × 10⁻⁹ s⁻¹; this is only an approximation because N (and so the true probability of decay per second) is assumed constant over the 8.0 s interval, whereas strictly N falls slightly throughout it.

19A sample has an activity of 7.4 × 10⁴ Bq and contains 9.0 × 10¹² undecayed nuclei. Calculate its decay constant.

λ = A/N = (7.4 × 10⁴)/(9.0 × 10¹²) ≈ 8.2 × 10⁻⁹ s⁻¹.

20The count rate from a source falls from 350 s⁻¹ to 90 s⁻¹ in 6.0 minutes. Determine the decay constant of the source, in s⁻¹.

6.0 minutes = 360 s; C = C₀e^−λt ⟹ λ = ln(350/90)/360 = 1.358/360 ≈ 3.77 × 10⁻³ s⁻¹.

8. Half-life and the decay constant

Section 5 introduced half-life, T½, as the time taken for the number of undecayed nuclei (or the activity, or the count rate) to fall to half its previous value. Now that we have the exponential decay equation, we can derive an exact relationship between T½ and λ.

Deriving T½ = ln2/λ: By definition, N = N₀/2 when t = T½. Substituting into N = N₀e^−λt:
N₀/2 = N₀e^−λT½ ⟹ 1/2 = e^−λT½ ⟹ 2 = e^λT½
Taking natural logarithms of both sides: ln 2 = λT½, so:

T½ = ln2 / λ = 0.693 / λ

Larger decay constants correspond to shorter half-lives — a nuclide that decays very readily (large λ) cannot remain half-undecayed for very long.

Taking natural logarithms of C = C₀e^−λt gives ln C = ln C₀ − λt — an equation of the same straight-line form as y = mx + c. This means that a graph of ln(count rate) against time should be a straight line, whose gradient is exactly −λ. This is by far the most accurate way to determine a decay constant experimentally, since it uses every data point collected rather than just two.

Fig. 8.1 — A graph of ln(count rate) against time gives a straight line of gradient −λ, since ln C = ln C₀ − λt. This is the standard experimental method for finding a decay constant.

Worked example 8.1

A detector records a count rate of 84 s⁻¹ at a certain moment, falling to 21 s⁻¹ exactly 60 s later. The background count rate is 4 s⁻¹. Calculate the half-life of the source.

Answer:
Adjusted initial count rate = 84 − 4 = 80 s⁻¹. Adjusted final count rate = 21 − 4 = 17 s⁻¹.
C = C₀e^−λt ⟹ 17 = 80 × e^−60λ ⟹ 60λ = ln(80/17) = 1.549 ⟹ λ = 0.0258 s⁻¹
T½ = 0.693/0.0258 = 27 s

Worked example 8.2 — radiometric dating

Potassium-40 decays (eventually) to stable argon-40, with a half-life of 1.25 × 10⁹ years. A rock sample is found to contain potassium-40 and argon-40 (which was trapped in the rock as it formed) in the ratio 1 : 3. Estimate the age of the rock.

Answer:
Every argon-40 atom now present came from a potassium-40 atom that has since decayed, so the original number of potassium-40 atoms was 1 + 3 = 4 "parts". The fraction of potassium-40 remaining is 1/4 = (1/2)², i.e. exactly 2 half-lives have passed.
Age = 2 × (1.25 × 10⁹) = 2.5 × 10⁹ years

21Starting from N = N₀e^−λt, show that T½ = ln2/λ.

setting N = N₀/2 at t = T½ in N = N₀e^−λt gives 1/2 = e^−λT½, so 2 = e^λT½; taking natural logs, ln2 = λT½, i.e. T½ = ln2/λ.

22A radioactive source has a decay constant of 3.2 × 10⁻³ s⁻¹. Calculate its half-life, in seconds and in minutes.

T½ = 0.693/(3.2 × 10⁻³) ≈ 216 s ≈ 3.6 minutes.

23Explain why plotting ln(count rate) against time, rather than count rate against time directly, allows the decay constant to be found from the gradient of a straight line.

taking natural logs of C = C₀e^−λt gives ln C = ln C₀ − λt, which has exactly the straight-line form y = mx + c with gradient m = −λ — so plotting ln C against t turns an exponential relationship into a straight line whose gradient directly gives −λ.

Glossary

Radioactive decay: The random, spontaneous process by which an unstable nucleus emits a particle and/or a high-energy photon, often transmuting into a different nuclide.
Radionuclide (radioisotope): An isotope or nuclide with an unstable nucleus, capable of radioactive decay.
Alpha particle (α): A helium-4 nucleus (two protons and two neutrons) emitted from a decaying nucleus; highly ionizing but easily stopped, even by paper.
Beta particle (β): A fast-moving electron (β⁻) or positron (β⁺) emitted from a decaying nucleus, accompanied by an antineutrino or neutrino respectively.
Gamma ray (γ): A high-energy photon emitted when a nucleus drops from an excited state to a lower energy state, with no change to its proton or nucleon number.
Ionization: The process by which radiation knocks electrons off atoms or molecules as it passes through matter.
Background radiation: The small, ever-present count rate registered by a detector due to naturally occurring radioactive materials and cosmic rays, which must be subtracted from source readings.
Activity: The number of nuclei decaying per second in a radioactive source, measured in becquerel (Bq).
Half-life: The time taken for half of the undecayed nuclei in a sample (or its activity, or count rate) to decay away.
Decay constant, λ: The probability that any one given nucleus will decay in a unit time interval, with SI unit s⁻¹.
Becquerel (Bq): The SI unit of activity, equal to one nuclear decay per second.
Count rate: The number of decay events registered by a detector such as a GM tube per unit time, usually assumed proportional to activity.