Detailed notes on Nuclear and quantum physics for IB DP Physics, covering key concepts, explanations, examples, and exam-focused revision points.
E.3 Radioactive decay — IB Physics SL Study Notes (Theme E: Nuclear and Quantum Physics)
Why some nuclei are unstable and decay at random: nuclear stability and the neutron–proton curve, the strong nuclear force, the three types of radiation (alpha, beta, gamma), balancing decay equations by conserving nucleon and proton number, the (anti)neutrino and the continuous beta spectrum, and half-life from decay curves — always remembering to subtract background radiation first.
At a glance
Radioactivity is spontaneous and random: an unstable nucleus decays by itself, and you cannot predict when any single nucleus will decay — only the probability.
The strong nuclear force is very short-ranged (~10−15 m), attractive between nucleons, and holds the nucleus together against the electrostatic repulsion of the protons.
Stability depends on the neutron–proton ratio: light stable nuclei have N≈Z; heavier stable nuclei need more neutrons (N>Z) to stay bound.
Three radiations, in order of increasing penetration: alpha (α, a 24He nucleus) < beta (β, a fast electron or positron) < gamma (γ, a high-energy photon). Ionising power runs the opposite way: α > β > γ.
Balancing a decay equation: both the nucleon number A (top) and the proton number Z (bottom) must be conserved — add them up on each side.
Beta decay always emits a third particle: β⁻ releases an antineutrino, β⁺ releases a neutrino. The continuous range of beta energies is the evidence for it.
Half-lifet1/2 is the time for half the radioactive nuclei (or the activity) to decay. After n whole half-lives the fraction remaining is (21)n.
Always subtract the background count rate before analysing count-rate data — this is the single most common Paper 2 slip.
What you’ll learn
Mapped to the 100452 subject guide (2025-onwards).
Describe radioactive decay as the spontaneous and random disintegration of an unstable nucleus, and explain nuclear stability in terms of the neutron–proton ratio and the strong nuclear force.
State the nature, charge, ionising power and penetrating power of alpha, beta-minus, beta-plus and gamma radiation, and predict their deflection in electric and magnetic fields.
Write and balance nuclear equations for α, β⁻, β⁺ and γ decay, conserving both nucleon number and proton number.
Explain the role of the (anti)neutrino and interpret the continuous energy spectrum of beta particles as evidence for a third emitted particle.
Define half-life, determine it from a decay curve, and calculate the fraction of nuclei (or activity) remaining after a whole number of half-lives, correcting count-rate data for background radiation.
Unstable nuclei: random, spontaneous decay and the strong force
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Decay is spontaneous and random; stability is set by the N–Z balance and the strong nuclear force.
A nucleus is a tiny, dense bundle of protons and neutrons (together called nucleons). The protons all carry positive charge, so they repel one another electrostatically — yet most nuclei hold together. What glues them?
The strong nuclear force. Between any two nucleons that are very close together (separations of about 10−15m, the size of a nucleus) there acts an attractive strong nuclear force. It is much stronger than the electrostatic repulsion at these tiny distances, so it binds protons and neutrons into a nucleus. Its defining feature is its very short range: beyond a few femtometres it essentially vanishes. (At even smaller separations it becomes repulsive, which stops the nucleus from collapsing.)
Why some nuclei are unstable. Because the strong force is short-ranged but electrostatic repulsion reaches right across the nucleus, adding more protons piles up long-range repulsion faster than the short-range glue can compensate. To stay bound, larger nuclei need extra neutrons — neutrons add strong-force attraction without adding any repulsion. If the balance of neutrons to protons is wrong, the nucleus is unstable and will eventually decay, emitting radiation to move towards a more stable configuration.
The neutron–proton (N–Z) curve.
Light stable nuclei sit close to the line N=Z (roughly equal numbers of neutrons and protons — e.g. carbon-12 has 6 of each).
Heavier stable nuclei lie above that line: they need more neutrons than protons (N>Z) to remain bound (e.g. lead-206 has 82 protons but 124 neutrons).
There is no stable nucleus with Z>83 — every element heavier than bismuth is radioactive.
Two words that earn marks: spontaneous and random.
Spontaneous — the decay is not triggered by anything outside the nucleus. It is unaffected by temperature, pressure or chemical state.
Random — you cannot predict when a particular nucleus will decay; every undecayed nucleus of a given type has the same probability of decaying in the next second. We can only describe the behaviour of a large number of nuclei statistically (which is what half-life does).
Strong nuclear force: short-range (~10⁻¹⁵ m), attractive between nucleons, holds the nucleus together against proton repulsion.
Stability needs the right N–Z balance: light nuclei N ≈ Z; heavier nuclei need N > Z; nothing stable above Z = 83.
Decay is spontaneous (unaffected by external conditions) and random (only the probability, not the timing, is known).
The three types of radiation: alpha, beta and gamma
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α (helium nucleus), β (fast electron/positron) and γ (photon) differ in charge, ionising power and penetration.
Unstable nuclei emit three kinds of ionising radiation. You must know the nature, charge, ionising power and penetration of each.
Radiation
What it is
Symbol
Charge
Ionising power
Penetration (stopped by)
Alpha (α)
A helium nucleus (2 protons + 2 neutrons)
24He
+2
Most ionising
Least penetrating — a sheet of paper or a few cm of air
Beta-minus (β⁻)
A fast-moving electron from the nucleus
−10e
−1
Moderate
A few mm of aluminium
Beta-plus (β⁺)
A positron (anti-electron) from the nucleus
+10e
+1
Moderate
(Annihilates with an electron almost at once)
Gamma (γ)
A high-energy electromagnetic photon
00γ
0
Least ionising
Most penetrating — reduced (never fully stopped) by thick lead or concrete
The key inverse relationship (a favourite of examiners): the more ionising a radiation is, the less penetrating it is, and vice versa.
Alpha is heavy, slow and doubly charged, so it interacts strongly with matter — it ionises heavily but loses its energy over a very short distance (least penetrating).
Gamma has no charge or mass, so it interacts weakly — it ionises very little but travels a long way (most penetrating).
Penetrating power increases from alpha (stopped by paper) to beta (stopped by a few mm of aluminium) to gamma (only reduced by thick lead or concrete). Ionising power runs in the opposite order: α > β > γ.
Deflection in electric and magnetic fields confirms the charges:
Alpha (+2) and beta-minus (−1) are deflected in opposite directions by a field, because their charges have opposite sign.
Gamma (0) is not deflected at all — it has no charge.
In the same field, beta is deflected much more than alpha, because a beta particle (an electron) has a far smaller mass than an alpha particle, and beta-plus curves the opposite way to beta-minus.
These properties explain their uses and hazards: alpha is very dangerous inside the body (it ionises intensely) but is easily stopped outside; gamma is a whole-body external hazard because it penetrates deeply.
α = ⁴₂He nucleus, +2, most ionising, least penetrating (paper/skin).
β⁻ = fast electron ⁰₋₁e, −1; β⁺ = positron ⁰₊₁e, +1; moderate, stopped by ~mm of aluminium.
γ = uncharged EM photon, least ionising, most penetrating (thick lead/concrete); undeflected by fields.
Conserve BOTH nucleon number A (top) and proton number Z (bottom) on each side.
A nuclear equation is written with each species as ZAX, where A is the nucleon (mass) number and Z is the proton (atomic) number. The golden rule:
In every nuclear decay, the total nucleon number A and the total proton number Z are each conserved. Add up the top numbers on both sides (they must match) and the bottom numbers on both sides (they must match too).
Alpha decay — the nucleus loses a 24He nucleus, so A drops by 4 and Z drops by 2:
ZAX→Z−2A−4Y+24α
Example (uranium-238):
92238U→90234Th+24α(238=234+4;92=90+2)
Beta-minus (β⁻) decay — inside the nucleus a neutron changes into a proton, emitting an electron and an antineutrino (νˉ):
n→p+−10e+νˉ
So A stays the same but Zincreases by 1:
ZAX→Z+1AY+−10e+νˉ
Example (carbon-14):
614C→714N+−10e+νˉ(14=14;6=7+(−1))
Beta-plus (β⁺) decay — a proton changes into a neutron, emitting a positron and a neutrino (ν):
p→n++10e+ν
So A stays the same but Zdecreases by 1:
ZAX→Z−1AY++10e+ν
Example (sodium-22):
1122Na→1022Ne++10e+ν(22=22;11=10+1)
Gamma (γ) emission — the nucleus loses energy (it was left in an excited state after an earlier decay) by emitting a gamma photon. Neither A nor Z changes; the nucleus simply drops to a lower energy state:
ZAX∗→ZAX+γ
The reliable method for any "complete the equation" question:
Write the top numbers (A) on both sides and make them balance.
Write the bottom numbers (Z) on both sides and make them balance.
Use Z to name the daughter element from the periodic table.
α decay: A − 4, Z − 2 (emits ⁴₂He).
β⁻ decay: A unchanged, Z + 1 (n → p + e⁻ + antineutrino).
β⁺ decay: A unchanged, Z − 1 (p → n + e⁺ + neutrino); γ emission: A and Z unchanged.
First principles: why beta decay needs the (anti)neutrino — the continuous spectrum
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A two-body decay would give beta particles one fixed energy; the observed continuous spread proves a third particle shares the energy.
This section shows how physicists reasoned from evidence to a particle they could not yet detect — a beautiful example of first-principles thinking that the IB loves to examine.
The puzzle. When alpha particles are emitted, they come out with a single, definite energy (or a few sharp values). That is exactly what conservation of energy and momentum predict for a two-body split: the parent nucleus divides into a daughter nucleus and an alpha particle, and the fixed energy release must be shared between just those two in one fixed way.
So physicists expected beta particles to behave the same: if ZAX simply split into a daughter nucleus and one electron, the electron should always carry away the same kinetic energy.
The evidence. Instead, beta particles are emitted with a continuous range of kinetic energies, from almost zero up to a definite maximum, Emax:
Beta particles emerge with a *continuous* range of energies up to a maximum E_max. A two-body decay would give a single fixed energy (the dashed line). The spread is the evidence that a third particle is also emitted.
The reasoning. If only the daughter nucleus and the electron were produced, energy conservation would force every beta particle to have the same fixed energy — a single spike, not a spread. The continuous spectrum means the fixed energy released by the decay is being shared among more than two particles. Since only the electron and daughter are detected, there must be a third, undetected particle carrying off the missing energy (and momentum).
The conclusion. Wolfgang Pauli proposed exactly such a particle in 1930; it is now called the (anti)neutrino. It:
has no charge (so it does not disturb the balance of Z in the decay equation),
has a very small mass (essentially negligible at SL), and
interacts extremely weakly with matter, which is why it went undetected for decades.
In β⁻ decay the third particle is the antineutrino (νˉ); in β⁺ decay it is the neutrino (ν). Its job is to carry away the balance of energy and momentum — which is why, when the neutrino happens to take almost none, the electron gets nearly Emax, and when the neutrino takes most of it, the electron gets almost nothing. The whole spread of the spectrum is the neutrino's fingerprint.
Alpha particles have a single fixed energy — the signature of a two-body decay.
Beta particles have a continuous spectrum from ~0 up to E_max, which a two-body decay cannot explain.
A third particle — the (anti)neutrino — shares the energy and momentum, explaining the spread; β⁻ emits an antineutrino, β⁺ a neutrino.
Half-life is the time for half the nuclei (or activity) to decay; after n half-lives a fraction (½)ⁿ remains.
Because decay is random, we describe a large collection of nuclei with a single statistical quantity.
Definition. The half-lifet1/2 of a radioactive isotope is the time taken for half of the radioactive nuclei in a sample to decay — equivalently, the time for the activity (or count rate) to fall to half its value. Every isotope has its own fixed half-life, from fractions of a second to billions of years.
Deriving the "fraction remaining" rule from first principles. Start with N0 radioactive nuclei.
After 1 half-life, half have decayed, so 21N0 remain.
After 2 half-lives, half of those remain: 21×21N0=41N0.
After 3 half-lives: 21×41N0=81N0.
Each half-life multiplies the amount left by another factor of 21, so after n whole half-lives:
N=(21)nN0⟹fraction remaining=(21)n
The fraction that has decayed is therefore 1−(21)n. The same rule works for the mass of the isotope and for the activity, because all three halve together.
Half-lives elapsed, n
Fraction remaining
As %
0
1
100%
1
1/2
50%
2
1/4
25%
3
1/8
12.5%
4
1/16
6.25%
Reading half-life off a decay curve. A graph of activity (or count rate) against time is an exponential decay curve — it falls steeply at first, then flattens, but never quite reaches zero. To find the half-life: pick any starting value on the curve, find the time for it to fall to half that value, and read off the time interval. Doing it from several starting points and averaging improves accuracy.
Each half-life the activity falls by a further factor of ½ (N₀ → N₀/2 → N₀/4 → N₀/8) over equal time intervals. The curve approaches zero but never reaches it.
Background radiation — the correction you must not forget. We are all bathed in low-level background radiation (from cosmic rays, rocks such as granite, radon gas, food and medical sources). A detector always registers this background on top of the reading from your source. So before finding a half-life you must subtract the background count rate from every measurement:
corrected count rate=measured count rate−background count rate
If you skip this step, your "activity" never falls towards zero (it flattens out at the background level) and every half-life you read off will be too long. Examiners flag this omission year after year.
Half-life t½ = time for half the nuclei/activity to decay; it is fixed for each isotope.
After n whole half-lives: fraction remaining = (½)ⁿ; fraction decayed = 1 − (½)ⁿ.
Always subtract the background count rate before reading a half-life off a decay curve.
The same ionising ability that damages cells makes radiation useful — matched to the right radiation and half-life.
Radiation is chosen for a job according to its penetration, ionising power and half-life — the very properties above. You are expected to discuss these qualitatively.
Everyday and industrial uses
Smoke detectors use a weak alpha source (typically americium-241). Alpha ionises the air in a small gap so a current flows; smoke particles absorb the alpha, the current drops and the alarm sounds. Alpha is ideal because it is easily stopped — it cannot escape the detector to reach you.
Thickness monitoring in manufacturing (paper, foil, sheet metal) uses beta: the amount getting through depends on the sheet's thickness, so the detector reading controls the rollers automatically. Beta is chosen because alpha would be blocked completely and gamma would pass straight through regardless of thickness.
Radioactive tracers in medicine and industry use gamma emitters with a short half-life: gamma penetrates out of the body to be detected, and the short half-life means the activity soon falls to a safe level.
Medicine
Radiotherapy uses focused gamma beams to destroy cancerous cells (gamma penetrates deep to reach internal tumours).
Sterilisation of surgical instruments and some foods uses gamma to kill bacteria without heating.
Dating
Carbon-14 dating exploits the known half-life (~5700 years) of C-14 to estimate the age of once-living material from its residual activity (after background subtraction).
Hazards. All three radiations are ionising: they knock electrons off atoms in living tissue, which can damage or kill cells and alter DNA, potentially causing mutations or cancer.
Outside the body, gamma (and to a lesser extent beta) is the greater danger because it penetrates the skin; alpha is largely stopped by the outer dead layer of skin.
Inside the body (if a source is swallowed or inhaled), alpha is by far the most dangerous, because its intense ionisation is deposited entirely within nearby tissue.
Safety precautions follow directly: keep sources far away (intensity falls with distance), limit exposure time, and use appropriate shielding (lead/concrete for gamma) and remote handling. These trade-offs — matching the radiation and half-life to the task while minimising dose — are exactly what "evaluate/discuss" questions reward.
Alpha: smoke detectors (easily stopped, safe outside the device).
Beta: thickness gauges (transmission depends on sheet thickness).
Gamma: medical tracers, radiotherapy and sterilisation (penetrating); hazards from ionisation of tissue — alpha worst internally, gamma worst externally.
Quick recap
Radioactive decay is the spontaneous and random disintegration of an unstable nucleus, unaffected by temperature, pressure or chemistry.
The strong nuclear force is short-range and attractive between nucleons; stability needs the right N–Z balance (light nuclei N ≈ Z, heavier nuclei N > Z).
Alpha (⁴₂He, +2) is most ionising/least penetrating; gamma (photon, 0) is least ionising/most penetrating; beta (electron/positron) is in between and deflects opposite to alpha in a field.
Balance decay equations by conserving both nucleon number A and proton number Z: α (A−4, Z−2), β⁻ (Z+1), β⁺ (Z−1), γ (unchanged).
The continuous beta energy spectrum is the evidence for a third particle — the antineutrino (β⁻) or neutrino (β⁺) — sharing the energy.
Half-life is the time for half the nuclei/activity to decay; after n half-lives a fraction (½)ⁿ remains — and you must subtract background radiation from count-rate data first.
Memorise this
Verbatim phrases, formulae and definitions IB DP mark schemes credit (key for AO1 knowledge marks on Paper 1).
Radioactivity = spontaneous + random decay of unstable nuclei (unaffected by external conditions).
Strong nuclear force: short-range (~10⁻¹⁵ m), attractive between nucleons, overcomes proton repulsion.
Penetration: α < β < γ. Ionising power: α > β > γ (the two orders are opposite).
Conserve A and Z: α → A−4, Z−2; β⁻ → Z+1; β⁺ → Z−1; γ → A and Z unchanged.
β⁻: n → p + e⁻ + antineutrino; β⁺: p → n + e⁺ + neutrino. Continuous β spectrum ⇒ third particle.
Fraction remaining after n half-lives = (½)ⁿ; fraction decayed = 1 − (½)ⁿ.
Always subtract background count rate before finding a half-life.
How it’s examined
E.3 appears across all three papers. Paper 1A (MCQ, no calculator): identifying a radiation from its charge/penetration or its deflection in a field, completing the changes to A and Z in a decay, and reading a fraction remaining after a whole number of half-lives. Paper 1B (data-based): determining a half-life from a decay curve or a table of count-rate data — where the mark scheme specifically checks that you subtracted the background count rate first, and rewards taking the half-life from more than one starting point. Paper 2: balancing multi-stage decay chains, explaining the continuous beta spectrum as evidence for the (anti)neutrino, and 'discuss/evaluate' questions matching a radiation and half-life to a use (tracer, gauge, smoke detector) with its hazards. Command terms: state, identify, determine, calculate, describe, explain, discuss. Examiner reports repeatedly flag: not subtracting background radiation, conserving only the nucleon number (not the proton number), reversing the ionising/penetrating orders, omitting the antineutrino/neutrino in beta equations, and treating decay as predictable rather than random. Show the halving steps or the (½)ⁿ working explicitly for method marks.
Sources: IB Diploma Programme Physics Guide (first assessment 2025) — Theme E: Nuclear and quantum physics (E.3 Radioactive decay); IB Physics Data Booklet (2025); IB Physics subject reports and specimen papers (2023–2025). Last reviewed 2026-07-21.
Take this whole topic with you
Step-by-step worked examples — Radioactive decay
Step-by-step solutions to past-paper-style questions on radioactive decay, written exactly the way a tutor would explain them at the board.
Radium-226 (88226Ra) decays by alpha emission. Write the balanced nuclear equation and identify the daughter nucleus. (3 marks)
Step-by-step solution
Step 1
An alpha particle is a helium nucleus 24He. Alpha decay removes 4 from the nucleon number and 2 from the proton number.
88226Ra→ZAY+24α
Step 2
Conserve nucleon number (top) and proton number (bottom): A=226−4=222 and Z=88−2=86.
A=222,Z=86
Step 3
Z=86 is radon (Rn) on the periodic table, so the daughter is radon-222.
88226Ra→86222Rn+24α
Answer
88226Ra→86222Rn+24α; the daughter nucleus is radon-222.
Examiner tip
Mark scheme: (1) alpha = ⁴₂He; (2) A and Z both conserved (222 and 86); (3) daughter correctly named radon. Check 226 = 222 + 4 and 88 = 86 + 2 — both must balance.
Strontium-90 (3890Sr) is a beta-minus emitter. Write the balanced nuclear equation, including the antineutrino, and identify the daughter. (3 marks)
Step-by-step solution
Step 1
In β⁻ decay a neutron turns into a proton, emitting an electron −10e and an antineutrino νˉ. The nucleon number is unchanged; the proton number increases by 1.
n→p+−10e+νˉ
Step 2
Conserve numbers: A=90 (unchanged) and Z=38+1=39 (the electron counts as Z=−1, so 38=39+(−1)).
A=90,Z=39
Step 3
Z=39 is yttrium (Y), so the daughter is yttrium-90.
3890Sr→3990Y+−10e+νˉ
Answer
3890Sr→3990Y+−10e+νˉ; the daughter is yttrium-90.
Examiner tip
Mark scheme: (1) β⁻ = ⁰₋₁e with an antineutrino shown; (2) A unchanged, Z increases by 1; (3) daughter named yttrium. Omitting the antineutrino, or forgetting that Z goes UP in β⁻ decay, loses marks.
A radioactive source is tested with absorbers. The count rate (already corrected for background) is unchanged by a sheet of paper, falls to almost zero behind 4 mm of aluminium, and is barely reduced by that aluminium for a second component that also passes through thick lead. Identify the radiation(s) emitted. (3 marks)
Step-by-step solution
Step 1
Paper does not reduce the count → there is no alpha (alpha would be stopped by paper).
Step 2
A few mm of aluminium stops most of the radiation → this component is beta (beta is stopped by a few mm of aluminium).
Step 3
A component that still penetrates thick lead is the most penetrating radiation → gamma. So the source emits beta and gamma, but not alpha.
Answer
The source emits beta and gamma radiation (no alpha, since paper has no effect).
Examiner tip
Mark scheme: (1) no alpha (paper has no effect); (2) beta stopped by ~mm of aluminium; (3) gamma penetrates thick lead. Always reason from the absorber that stops each component — and note the data is already background-corrected.
4Reading a half-life from a decay curve
Building confidenceGraph or diagram• half-life, decay curve, AO2, AO3
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Question
The corrected activity of a sample is 800 Bq at t = 0 and falls to 200 Bq at t = 30 minutes. Determine the half-life of the isotope. (3 marks)
Step-by-step solution
Step 1
Find how many times the activity has halved: 800→400→200 is two halvings, so 2 half-lives have elapsed.
800200=41=(21)2⇒n=2
Step 2
Those 2 half-lives took 30 minutes in total.
2t1/2=30min
Step 3
Solve for the half-life.
t1/2=230=15min
Answer
The half-life is 15 minutes.
Examiner tip
Mark scheme: (1) recognise 800→200 is two halvings (fraction ¼); (2) 2 half-lives = 30 min; (3) t½ = 15 min. Counting the number of halvings is the key skill — the activity must be the background-corrected value.
5Fraction remaining after several half-lives
Building confidenceDirect calculation• half-life, fraction, AO2
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Question
A sample contains 6.4 × 10²⁰ atoms of a radioactive isotope with a half-life of 8.0 days. Calculate (a) the number of these atoms remaining after 32 days and (b) the fraction that has decayed. (4 marks)
Mark scheme: (1) n = 4 half-lives; (2) fraction remaining = 1/16; (3) 4.0 × 10¹⁹ atoms; (4) fraction decayed = 15/16. The classic slip is quoting the fraction remaining when asked for the fraction decayed — read the question.
6Correcting for background before finding half-life
Building confidenceMulti-step problem• half-life, background, AO2, AO3
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Question
A detector reads a background count rate of 20 counts per minute. With a source present it reads 340 counts per minute at t = 0 and 100 counts per minute after 40 minutes. Determine the half-life of the source. (4 marks)
Step-by-step solution
Step 1
Subtract the background from each reading to get the corrected count rate from the source alone.
at t=0:340−20=320;at t=40:100−20=80(cpm)
Step 2
Find how many halvings link 320 to 80.
32080=41=(21)2⇒n=2
Step 3
These 2 half-lives took 40 minutes.
2t1/2=40min
Step 4
Solve for the half-life.
t1/2=240=20min
Answer
The half-life is 20 minutes.
Examiner tip
Mark scheme: (1) subtract background from BOTH readings (320 and 80); (2) 320→80 is two halvings; (3) 2 half-lives = 40 min; (4) t½ = 20 min. Skipping the background subtraction gives 340→100 (not a clean ratio) and the wrong half-life — this is the most penalised error in the topic.
Fluorine-18 (918F), used in PET scans, decays by beta-plus emission. (a) Write the balanced nuclear equation including the neutrino. (b) State what happens to a proton inside the nucleus and how the proton number changes. (4 marks)
Step-by-step solution
Step 1
In β⁺ decay a proton turns into a neutron, emitting a positron +10e and a neutrino ν. Nucleon number is unchanged; proton number decreases by 1.
p→n++10e+ν
Step 2
Conserve numbers: A=18 (unchanged) and Z=9−1=8 (the positron counts as Z=+1, so 9=8+1).
A=18,Z=8
Step 3
Z=8 is oxygen (O), so the daughter is oxygen-18.
918F→818O++10e+ν
Step 4
(b) A proton in the nucleus changes into a neutron, so the proton number falls by one (from 9 to 8).
Answer
(a) 918F→818O++10e+ν; (b) a proton becomes a neutron, so Z decreases by 1 (9 → 8).
Examiner tip
Mark scheme: (1) positron ⁰₊₁e and neutrino shown; (2) A unchanged, Z decreases by 1; (3) daughter named oxygen; (4) proton → neutron stated. In β⁺ the proton number goes DOWN (opposite to β⁻) and it is a neutrino (not antineutrino) that is emitted.
A radioactive source has an initial activity of 2400 Bq and a half-life of 6.0 hours. (a) Calculate the time for the activity to fall to 150 Bq. (b) Calculate the mass of the isotope remaining after this time if the initial mass was 8.0 μg. (4 marks)
Step-by-step solution
Step 1
Find the fraction of the original activity, then the number of halvings needed.
2400150=161=(21)4⇒n=4
Step 2
Multiply the number of half-lives by the half-life.
t=nt1/2=4×6.0=24hours
Step 3
The mass halves with the activity, so after 4 half-lives the fraction remaining is again 161.
m=161×8.0μg=0.50μg
Answer
(a) 24 hours; (b) 0.50 μg of the isotope remains.
Examiner tip
Mark scheme: (1) fraction 1/16 → n = 4; (2) t = 24 hours; (3) mass = 1/16 × 8.0 = 0.50 μg. The activity, the number of undecayed nuclei and the mass of the isotope all halve together — the same (½)ⁿ factor applies to each.
Model Answers — Radioactive decay
High-scoring sample answers for radioactive decay on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.
Question 1
2 marks
Q (2 marks). Explain what is meant by describing radioactive decay as (a) spontaneous and (b) random.
Model answer
(a) Spontaneous means the decay happens by itself, from within the nucleus, and is not affected by external conditions such as temperature, pressure or the atom's chemical state.
(b) Random means it is impossible to predict when any particular nucleus will decay; every undecayed nucleus of the isotope has the same probability of decaying in the next interval of time.
Why this scores
Why this scores 2/2. (1) spontaneous = unaffected by external conditions / arises within the nucleus; (2) random = cannot predict which/when a nucleus decays, only the probability. Vague answers ('it just happens') do not earn the marks.
Question 2
3 marks
Q (3 marks). State the nature of alpha and gamma radiation, and explain why alpha is more strongly ionising but less penetrating than gamma.
Model answer
Alpha is a helium nucleus (24He: 2 protons + 2 neutrons), carrying a charge of +2. Gamma is a high-energy electromagnetic photon with no charge and no mass.
Alpha is more strongly ionising because it is relatively massive, slow-moving and doubly charged, so it interacts strongly with the atoms it passes, knocking off many electrons. Because it loses its energy so rapidly in these many interactions, it travels only a short distance and is less penetrating (stopped by paper or skin). Gamma, being uncharged, interacts only weakly, so it ionises little but penetrates far (needing thick lead to reduce it).
Why this scores
Why this scores 3/3. (1) alpha = He nucleus / +2, gamma = EM photon / uncharged; (2) alpha ionises strongly because it is charged and massive; (3) strong interaction means rapid energy loss → short range (less penetrating). The inverse link (more ionising ⇒ less penetrating) is the assessed idea.
Question 3
3 marks
Q (3 marks). Polonium-210 (84210Po) decays by alpha emission to an isotope of lead (Pb). Write the balanced nuclear equation and show that both nucleon number and proton number are conserved.
Model answer
An alpha particle is 24He, so:
84210Po→82206Pb+24α
Nucleon number:210=206+4 ✓
Proton number:84=82+2 ✓
Both are conserved, and Z=82 correctly identifies the daughter as lead-206.
Why this scores
Why this scores 3/3. (1) correct alpha particle and daughter A = 206; (2) daughter Z = 82 (lead); (3) explicit check that both totals balance. Always show the two conservation sums — 'show that' questions require them written out.
Question 4
4 marks
Q (4 marks). When a nucleus undergoes beta-minus decay, the emitted beta particles have a continuous range of kinetic energies. Explain how this observation provides evidence for the existence of the antineutrino.
Model answer
If the decay produced only the daughter nucleus and the electron (a two-body split), then conservation of energy and momentum would force every emitted electron to carry away the same fixed kinetic energy.
Instead, the beta particles are observed with a continuous spread of energies, from almost zero up to a maximum value Emax.
This spread cannot be explained by a two-body decay, so the fixed energy released must be shared among more than two particles.
Therefore a third particle — the antineutrino — must also be emitted, carrying away the balance of the energy (and momentum). It is uncharged and interacts very weakly, which is why it is not directly detected; the electron gets nearly Emax when the antineutrino takes almost none, and much less when the antineutrino takes more.
Why this scores
Why this scores 4/4. (1) two-body decay would give a single fixed energy; (2) observed spectrum is continuous up to E_max; (3) energy must be shared among 3+ particles; (4) conclude a third particle (antineutrino) carries the balance of energy/momentum. The reasoning from conservation laws is what earns the marks, not just naming the particle.
Question 5
4 marks
Q (4 marks). A radioactive isotope has a half-life of 12 years. A sample initially contains 8.0 × 10¹⁶ undecayed nuclei. Calculate (a) the number of undecayed nuclei remaining after 48 years and (b) the number that have decayed in that time.
Model answer
(a) Number of half-lives in 48 years:
n=1248=4
Fraction remaining =(21)4=161, so:
N=161×8.0×1016=5.0×1015nuclei
(b) Number decayed = initial − remaining:
8.0×1016−5.0×1015=7.5×1016nuclei
Why this scores
Why this scores 4/4. (1) n = 4 half-lives; (2) fraction remaining 1/16; (3) 5.0 × 10¹⁵ remain; (4) 7.5 × 10¹⁶ decayed (initial − remaining). The number decayed is not the fraction remaining — subtract from the initial number.
Question 6
5 marks
Q (5 marks). A student measures the count rate from a source using a Geiger–Müller tube. Before analysing the data, explain why the background count rate must be measured and subtracted, and describe how failing to do so affects a half-life determined from the data. Include how the background could be measured.
Model answer
Background radiation is always present (from cosmic rays, rocks such as granite, radon gas, food and medical sources), so the detector registers this on top of the counts from the source.
The reading from the source alone is the corrected count rate = measured count rate − background count rate. This subtraction must be done for every reading before the data represent the source's true decay.
The background is measured by recording the count rate with the source removed (well away from the detector), ideally averaged over a long time to reduce random fluctuation.
If the background is not subtracted, the count rate never falls towards zero — it levels off at the background value rather than following the true exponential decay.
As a result, each apparent "halving" takes longer, so the half-life obtained is too large (overestimated).
Why this scores
Why this scores 5/5. (1) background is always present from natural/other sources; (2) corrected = measured − background, for every reading; (3) measure background with the source removed; (4) uncorrected data flattens at the background level; (5) consequence — half-life overestimated. This background correction is the single most examined skill in E.3 data questions.
Question 7
6 marks
Q (6 marks). A nucleus 90232Th undergoes a decay series of, in order, one alpha decay followed by two beta-minus decays. (a) Determine the nucleon number and proton number of the nucleus after all three decays. (b) State, with a reason, how the final nucleus compares with 90232Th in terms of the number of neutrons.
Model answer
(a) Track A and Z through each step.
Alpha decay (A−4, Z−2):
90232Th→88228Ra+24α
First beta-minus (A unchanged, Z+1):
88228Ra→89228Ac+−10e+νˉ
Second beta-minus (A unchanged, Z+1):
89228Ac→90228Th+−10e+νˉ
So the final nucleus has nucleon number A=228 and proton number Z=90.
(b) The final nucleus 90228Th has the same proton number (90) as the original but a smaller nucleon number (228 instead of 232). Since neutron number N=A−Z:
Original: N=232−90=142.
Final: N=228−90=138.
So the final nucleus has 4 fewer neutrons. (This makes sense: the alpha removed 2 neutrons and 2 protons, and the two beta-minus decays each converted a neutron into a proton — removing 2 more neutrons — while restoring the 2 protons.)
Why this scores
Why this scores 6/6. (a) 4 marks: correct A and Z after each of the three steps (alpha then two β⁻), ending at A = 228, Z = 90; (b) 2 marks: same Z but lower A, so N = A − Z is 138 vs 142 → 4 fewer neutrons. Tracking both numbers step by step, and using N = A − Z, are the assessed skills.
Question 8
6 marks
Q (6 marks). The table shows the measured count rate from a source. The background count rate is a constant 30 counts per minute.
Time / min
0
10
20
30
40
Measured count rate / cpm
510
270
150
90
60
(a) Correct the readings for background. (b) Determine the half-life of the source, using more than one interval to check your answer.
Model answer
(a) Subtract the 30 cpm background from each measured value:
Time / min
0
10
20
30
40
Corrected count rate / cpm
480
240
120
60
30
(b) Look for halvings in the corrected data:
480→240 takes 10 min (one half-life).
240→120 takes another 10 min.
120→60 takes another 10 min.
60→30 takes another 10 min.
Each halving takes the same 10 minutes, from several independent intervals, so:
t1/2=10minutes
Why this scores
Why this scores 6/6. (a) 2 marks: all five readings background-corrected (480, 240, 120, 60, 30); (b) 4 marks: identify a halving (e.g. 480→240 in 10 min), confirm with at least one further interval, and state t½ = 10 min. Checking the half-life is constant across intervals is exactly what Paper 1B rewards — and it only works once the background has been subtracted.
Question 9
10 marks
Q (10 marks — extended response). A hospital needs (i) a source for a smoke detector, (ii) a source injected as a medical tracer whose gamma emission is detected outside the body, and (iii) a source for a gauge that continuously monitors the thickness of aluminium foil during manufacture.
(a) For each application, state which type of radiation (alpha, beta or gamma) is most suitable and justify your choice in terms of ionising and penetrating power.
(b) Explain why a tracer should have a relatively short half-life.
(c) Discuss the relative hazards of alpha and gamma radiation to the human body, distinguishing between exposure from outside the body and from a source inside the body.
Model answer
(a) Matching the radiation to the task
(i) Smoke detector — alpha. Alpha strongly ionises the air in the detector so a small current flows; smoke particles absorb the alpha and the current drops, triggering the alarm. Alpha is ideal because it is the least penetrating, so it is easily contained inside the device and cannot reach the user, yet it is the most ionising, giving a clear signal over the tiny gap.
(ii) Medical tracer — gamma. The radiation must escape the body to reach an external detector, so it must be highly penetrating: only gamma can pass out through tissue. Its low ionising power also limits the damage done to the patient's cells along the way.
(iii) Thickness gauge — beta. The amount of radiation transmitted must depend on the foil thickness so the reading can control the rollers. Alpha would be stopped completely by the foil (no transmitted signal) and gamma would pass straight through almost unchanged regardless of thickness. Beta is partly absorbed by a few mm of metal, so the transmitted count rate changes measurably with thickness — the property the gauge relies on.
(b) Why a tracer needs a short half-life
A tracer stays inside the patient's body, continuously irradiating tissue for as long as it remains active. A short half-life means the activity falls quickly to a safe level soon after the scan, minimising the total radiation dose the patient receives. (It must not be too short, or it would decay before the measurement is finished.)
(c) Relative hazards of alpha and gamma
From outside the body:gamma is the greater hazard. It is highly penetrating, so it passes through the skin and irradiates internal organs. Alpha is much less dangerous externally because it is stopped by the outer (dead) layer of skin or even a few cm of air, so it never reaches living tissue.
From inside the body (a source swallowed, inhaled or injected): alpha is by far the more dangerous. Because it is the most strongly ionising, it deposits all of its energy in a very short range within nearby cells, causing intense, concentrated damage to DNA. Gamma, being weakly ionising, spreads its (smaller) effect out and much of it may pass out of the body without interacting.
Overall: the same property — ionising power — that makes alpha nearly harmless outside the body makes it the most damaging inside it, while gamma's penetrating power makes it the dominant external hazard. Choosing and shielding a source therefore depends on where the exposure occurs.
Why this scores
Why this scores 10/10. (a) 6 marks — 2 per application: correct radiation AND a justification tied to ionising/penetrating power (alpha contained + ionising; gamma penetrates out of body; beta transmission varies with thickness). (b) 2 marks: short half-life → activity falls quickly → lower patient dose. (c) 2 marks: gamma worse externally (penetrates skin), alpha worse internally (most ionising, energy deposited in nearby tissue). The discriminator in this extended response is linking each choice explicitly to a property, and recognising that alpha's hazard depends on where the source is.
Key Formulae — Radioactive decay
The formulae you need to memorise for radioactive decay on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.
Fraction remaining after n half-lives
▼
N=(21)nN0
N
number of undecayed nuclei (or activity/mass) remaining
N0
initial number of undecayed nuclei (or activity/mass)
n
number of whole half-lives elapsed
When to use
For a whole number of half-lives — the SL method. Count how many times the quantity has halved, then apply the (½)ⁿ factor.
Example
After 3 half-lives: N = (½)³N₀ = ⅛N₀, so 12.5% remains and 87.5% has decayed.
Number of half-lives from elapsed time
▼
n=t1/2t
n
number of half-lives elapsed
t
total time elapsed
t1/2
half-life of the isotope
When to use
To convert an elapsed time into a number of half-lives before applying the (½)ⁿ rule.
Example
In 32 days with a half-life of 8.0 days: n = 32/8.0 = 4 half-lives.
Correcting count rate for background
▼
Rcorrected=Rmeasured−Rbackground
Rcorrected
count rate due to the source alone
Rmeasured
count rate recorded with the source present
Rbackground
count rate with no source present (background)
When to use
Before any half-life analysis of count-rate data — apply it to every reading.
Example
Measured 340 cpm, background 20 cpm → corrected 320 cpm from the source.
Neutron number from A and Z
▼
N=A−Z
N
number of neutrons
A
nucleon (mass) number
Z
proton (atomic) number
When to use
To find how the neutron count changes through a decay, or to place a nucleus on the N–Z curve.
Example
²³²Th (A = 232, Z = 90): N = 232 − 90 = 142 neutrons.
Beta-minus decay (nucleon process)
▼
n→p+−10e+νˉ
n
neutron (converted inside the nucleus)
p
proton (produced; Z increases by 1)
−10e
emitted electron (the beta-minus particle)
νˉ
antineutrino (carries away the balance of energy)
When to use
When writing β⁻ decays — remember A is unchanged and Z increases by 1, and include the antineutrino.
Example
¹⁴C → ¹⁴N + ⁰₋₁e + ν̄ (Z: 6 → 7, A unchanged).
Key Definitions and Keywords — Radioactive decay
Definitions to memorise and the exact keywords mark schemes credit for radioactive decay answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.
Radioactivity (radioactive decay)
Examiner keyword▼
The spontaneous and random disintegration of an unstable nucleus, emitting alpha, beta and/or gamma radiation. Unaffected by temperature, pressure or chemical state.
Spontaneous
Examiner keyword▼
Describes decay that occurs of its own accord within the nucleus, without any external trigger or influence.
Random (decay)
Examiner keyword▼
It is impossible to predict when a given nucleus will decay; each undecayed nucleus has the same fixed probability of decaying in the next interval of time.
Strong nuclear force
Examiner keyword▼
A very short-range (~10⁻¹⁵ m) attractive force between nucleons that binds the nucleus together, overcoming the electrostatic repulsion between protons.
Nucleon
▼
A particle found in the nucleus — a proton or a neutron. The nucleon number A is the total number of protons and neutrons.
Alpha particle (α)
Examiner keyword▼
A helium nucleus (⁴₂He): 2 protons and 2 neutrons, charge +2. The most ionising and least penetrating radiation (stopped by paper or skin).
Beta-minus particle (β⁻)
Examiner keyword▼
A fast-moving electron (⁰₋₁e) emitted when a neutron changes into a proton; charge −1. Moderately ionising, stopped by a few mm of aluminium.
Beta-plus particle (β⁺)
Examiner keyword▼
A positron (⁰₊₁e, the antiparticle of the electron) emitted when a proton changes into a neutron; charge +1.
Gamma radiation (γ)
Examiner keyword▼
A high-energy electromagnetic photon (no charge, no mass) emitted when a nucleus loses energy. The least ionising and most penetrating radiation (reduced by thick lead or concrete).
(Anti)neutrino
Examiner keyword▼
A neutral, very-low-mass, weakly interacting particle emitted in beta decay: an antineutrino (ν̄) in β⁻ decay, a neutrino (ν) in β⁺ decay. It carries away the balance of energy and momentum.
Nucleon number (A)
Examiner keyword▼
The total number of protons and neutrons in a nucleus (also called the mass number). Conserved in every nuclear decay.
Proton number (Z)
Examiner keyword▼
The number of protons in a nucleus (also called the atomic number); it identifies the element. Conserved in every nuclear decay.
Half-life (t½)
Examiner keyword▼
The time taken for half the radioactive nuclei in a sample to decay, equivalently the time for the activity (or count rate) to fall to half its value.
Activity
Examiner keyword▼
The number of nuclei decaying per unit time in a sample, measured in becquerel (Bq), where 1 Bq = 1 decay per second.
Background radiation
Examiner keyword▼
The low level of ionising radiation always present in the environment (from cosmic rays, rocks, radon gas, food and medical sources). Its count rate must be subtracted from measurements before analysis.
Ionising power
Examiner keyword▼
A radiation's ability to knock electrons off atoms it passes, creating ions. Order: alpha > beta > gamma (the reverse of the penetration order).
Common Mistakes and Misconceptions — Radioactive decay
The traps other students keep falling into on radioactive decay questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.
✕Not subtracting background radiation before finding a half-life
IB Physics Theme E subject reports
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Why it happens
Students use the raw measured count rate straight from the detector and forget that background is always added on top.
How to avoid it
Subtract the background count rate from every reading first (corrected = measured − background). Without this the decay curve levels off above zero and the half-life comes out too large. This is the most penalised error in the topic.
✕Balancing only the nucleon number and not the proton number in decay equations
IB Physics Theme E subject reports
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Why it happens
The top numbers are easy to add up, so students stop there and misname the daughter nucleus.
How to avoid it
Conserve both A (top) and Z (bottom). Remember the electron counts as Z = −1 and the positron as Z = +1, so the bottom numbers must still add up on each side.
✕Reversing the ionising and penetrating orders of the three radiations
IB Physics Theme E subject reports
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Why it happens
Students memorise one order and assume the other is the same, but they are opposite.
How to avoid it
Learn them as a pair: penetration α < β < γ but ionising power α > β > γ. Alpha is most ionising yet least penetrating; gamma is least ionising yet most penetrating.
✕Forgetting the antineutrino (or neutrino) in a beta decay equation
IB Physics Theme E subject reports
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Why it happens
It has no charge or nucleon number, so it seems to make no difference to the balancing.
How to avoid it
Always include ν̄ in β⁻ decay and ν in β⁺ decay. Although it does not change A or Z, the continuous beta spectrum is the evidence it must be there, and mark schemes require it.
✕Treating radioactive decay as predictable or as affected by external conditions
IB Physics Theme E subject reports
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Why it happens
Students confuse the reliable statistical behaviour of a large sample (a fixed half-life) with predictability of an individual nucleus.
How to avoid it
State clearly that decay is random and spontaneous: you cannot predict when one nucleus decays, and heating, pressurising or chemically bonding a source does not change its half-life.
✕Mixing up β⁻ and β⁺ and the direction the proton number changes
IB Physics Theme E subject reports
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Why it happens
Both are 'beta', so students assume they do the same thing to Z, or think a negative particle leaving must lower Z.
How to avoid it
β⁻ (electron): a neutron becomes a proton, so Z increases by 1. β⁺ (positron): a proton becomes a neutron, so Z decreases by 1. The nucleon number A is unchanged in both.