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v0.1.1038
NotesPhysicsTopic 5.3
Unit 5 · Nuclear and quantum physics · Topic 5.3

IB Physics — Radioactive decay

Topic 5.3 of IB Physics covers Radioactive decay, which is part of Unit 5: Nuclear and quantum physics. Students explore key concepts including Types of radiation and their properties, Decay equations and conservation laws, Energy released in radioactive decay, Half-life, activity and background radiation. A strong understanding of radioactive decay is essential for IB Physics exams and builds the foundation for connected topics across the syllabus.

Exam technique guidePractice questions

Key concepts in Radioactive decay

Key Idea: An unstable nucleus calms down by throwing out radiation — and this topic is the four things the IB asks about that. What the radiation is and how it behaves (α, β⁻, γ); how to balance a decay equation so the new nucleus falls out; how much energy the decay releases through E = mc²; and how fast a source weakens, measured by its half-life. It is examined on both papers. Paper 1A is quick multiple-choice — identify a radiation from how it penetrates or deflects, pick the daughter's A and Z, or halve a count rate over a whole number of half-lives. Paper 2 is longer structured work — a show that the released energy is about 5 MeV from a mass defect, a show that the alpha carries about 98% of it, completing a nuclear equation, or correcting a count rate for background.

📋 Key formulas & rules

Only one equation in this topic carries the data-booklet badge — E = mc². Everything else is a rule you memorise: how A and Z change in each decay, and the halving rule for half-life. The exponential decay law A = A₀e⁻λᵗ is not on the SL booklet, so at SL you only ever halve over whole half-lives.

E=mc2E = mc^{2}E=mc2
Mass–energy equivalence (given in the data booklet, Theme E). Here m is the mass defect Δm and E is the energy released. Fast route in MeV: keep Δm in u and multiply by 931.5, since 1 u = 931.5 MeV c⁻².
EEE
energy released in the decay, the disintegration energy Q (J, or MeV)
mmm
mass defect Δm — the mass that disappears in the decay
ccc
speed of light, 3.00 × 10⁸ m s⁻¹ (given constant)
ZAX  ⟶  Z−2A−4Y  +  24α{}^{A}_{Z}X \;\longrightarrow\; {}^{A-4}_{Z-2}Y \;+\; {}^{4}_{2}\alphaZA​X⟶Z−2A−4​Y+24​α
Alpha decay — NOT printed in the booklet; memorise it. The daughter has 4 fewer nucleons and 2 fewer protons. Check: tops (A − 4) + 4 = A; bottoms (Z − 2) + 2 = Z. Balanced.
ZAX{}^{A}_{Z}XZA​X
parent nuclide before decay (A on top = protons + neutrons, Z below = protons)
Z−2A−4Y{}^{A-4}_{Z-2}YZ−2A−4​Y
daughter: nucleon number falls by 4, proton number by 2
24α{}^{4}_{2}\alpha24​α
alpha particle = a helium-4 nucleus (2 protons + 2 neutrons)
ZAX  ⟶  Z+1    AY  +  −1    0e  +  νˉ{}^{A}_{Z}X \;\longrightarrow\; {}^{\;\;A}_{Z+1}Y \;+\; {}^{\;\;0}_{-1}e \;+\; \bar{\nu}ZA​X⟶Z+1A​Y+−10​e+νˉ
Beta-minus decay — NOT in the booklet; memorise it. A neutron becomes a proton, so A is unchanged and Z rises by 1. The electron's −1 charge is what forces Z UP by one to keep the bottoms balanced.
ZAX{}^{A}_{Z}XZA​X
parent nuclide before decay
Z+1    AY{}^{\;\;A}_{Z+1}YZ+1A​Y
daughter: nucleon number unchanged, proton number rises by 1
−1    0e{}^{\;\;0}_{-1}e−10​e
beta-minus particle = an electron (made when a neutron becomes a proton)
νˉ\bar{\nu}νˉ
antineutrino — emitted with the electron (no charge, ≈ no mass)
A=A0(12)nA = A_{0}\left(\tfrac{1}{2}\right)^{n}A=A0​(21​)n
The halving rule for half-life — NOT a booklet equation. After n WHOLE half-lives the activity (or count rate) is the start value halved n times. Work out n = total time ÷ half-life first.
AAA
the activity (or count rate) after the time has passed
A0A_{0}A0​
the starting activity (or count rate)
nnn
the number of WHOLE half-lives that have passed, n = total time ÷ half-life
KEαKEtotal=mdmd+mα\frac{KE_{\alpha}}{KE_{\text{total}}} = \frac{m_{d}}{m_{d} + m_{\alpha}}KEtotal​KEα​​=md​+mα​md​​
Energy-sharing ratio — NOT printed; it comes from the parent being at rest (equal and opposite momentum) and KE = p²/2m. The LIGHT product (the alpha) takes the bigger share, close to but just under 100% for a heavy parent.
KEαKE_{\alpha}KEα​
kinetic energy carried by the alpha (the light product)
KEtotalKE_{\text{total}}KEtotal​
total energy released in the decay
mdm_{d}md​
mass of the daughter (the heavy product)
mαm_{\alpha}mα​
mass of the alpha (the light product)

☢️ The three radiations side by side

Going α → β → γ: penetration goes UP (paper → aluminium → lead) and ionising power goes DOWN (α strongest → γ weakest). The best ioniser travels the shortest distance — α dumps its energy fastest, so it is stopped first. And only γ (neutral) is not bent by a field.

⚖️ How A and Z change in each decay

The top numbers add up the same on both sides (nucleon number A conserved), and the bottom numbers add up the same (proton number Z conserved). That single check finds the daughter every time. For a chain of two decays, apply the changes one at a time and keep a running tally; use N = A − Z if asked for neutrons.

📉 Half-life vs energy — the two calculations


✏️ Worked exam-style questions

IB-style question — identify the radiation

An unknown radiation passes straight through a sheet of paper but is stopped by a 3 mm aluminium plate. When it crosses a magnetic field it is deflected. State which type of radiation it is, giving a reason from each observation.

Solution:

  1. Use the penetration clues to narrow it down — paper does NOT stop it (so it is not the least-penetrating alpha), but a few mm of aluminium does (so it is not the very penetrating gamma, which needs lead):

  2. Medium penetration ⇒ the radiation is beta-minus, OR could still be something charged — check the field clue.

  3. It is DEFLECTED by a field, so it must carry a charge — gamma (neutral) would pass straight through undeflected.

  4. Medium penetration AND charged points to one answer:

Final answer:

Beta-minus (β⁻). Paper does not stop it (not α), a few mm of aluminium does (not the lead-stopping γ), and it deflects in a field, so it is charged (not the neutral γ). Use penetration to narrow to two, then the deflection to decide.

IB-style question — balance a decay chain

Radium-226 (A = 226, Z = 88) emits an alpha particle, and the nucleus it forms then emits a beta-minus particle. Find the nucleon number, proton number AND neutron number of the FINAL nuclide.

Solution:

  1. Alpha step — nucleon number falls by 4, proton number falls by 2:

    A=226−4=222Z=88−2=86A = 226 - 4 = 222 \qquad Z = 88 - 2 = 86A=226−4=222Z=88−2=86
  2. Beta-minus step — nucleon number unchanged, proton number rises by 1:

    A=222Z=86+1=87A = 222 \qquad Z = 86 + 1 = 87A=222Z=86+1=87
  3. Neutron number = nucleon number − proton number:

    N=A−Z=222−87=135N = A - Z = 222 - 87 = 135N=A−Z=222−87=135
Final answer:

The final nuclide has A = 222, Z = 87 and N = 135 (it is francium-222). Apply each decay one at a time, keeping a running tally of A and Z, then use N = A − Z for the neutrons.

IB-style question — energy released, and the alpha's share

A nucleus at rest decays by alpha emission. The masses are: parent = 230.033130 u, daughter = 226.025410 u, alpha = 4.002600 u. (a) Show that the energy released is about 5 MeV. (b) The daughter has mass 226 u and the alpha 4 u — show that the alpha carries about 98% of that energy. (1 u = 931.5 MeV c⁻².)

Solution:

  1. (a) Find the mass defect Δm = parent mass − total mass of the products:

    Δm=230.033130−(226.025410+4.002600)=0.005120 u\Delta m = 230.033130 - (226.025410 + 4.002600) = 0.005120\ \text{u}Δm=230.033130−(226.025410+4.002600)=0.005120 u
  2. (a) Turn the lost mass into energy with the given E = mc²; with Δm in u, the c² is built into 931.5, so just multiply:

    E=0.005120×931.5=4.77 MeV≈5 MeVE = 0.005120 \times 931.5 = 4.77\ \text{MeV} \approx 5\ \text{MeV}E=0.005120×931.5=4.77 MeV≈5 MeV
  3. (b) The parent is at rest, so the two products carry equal and opposite momentum. With KE = p²/2m, the alpha's share is the daughter's mass over the total mass:

    KEαKEtotal=mdmd+mα=226226+4\frac{KE_{\alpha}}{KE_{\text{total}}} = \frac{m_{d}}{m_{d} + m_{\alpha}} = \frac{226}{226 + 4}KEtotal​KEα​​=md​+mα​md​​=226+4226​
  4. (b) Evaluate as a percentage:

    226230=0.983≈98%\frac{226}{230} = 0.983 \approx 98\%230226​=0.983≈98%
Final answer:

(a) Δm = 0.005120 u, so E = 0.005120 × 931.5 ≈ 4.77 MeV ≈ 5 MeV. (b) The alpha's share = 226 ÷ 230 ≈ 0.98 = 98%. Mass defect FIRST, then E = mc²; the light product (the alpha) always gets most of the energy. Keep every decimal place when subtracting the masses.

IB-style question — count rate after two half-lives

A detector near a fresh source reads 124 counts per second. With the source removed the background reads 4 counts per second. The source has a half-life of 15 minutes. Find the count rate the SAME detector reads after 30 minutes.

Solution:

  1. Subtract the background FIRST to get the source's true count rate now:

    124−4=120 counts s−1124 - 4 = 120\ \text{counts s}^{-1}124−4=120 counts s−1
  2. Find the number of whole half-lives, n = total time ÷ half-life:

    n=3015=2n = \frac{30}{15} = 2n=1530​=2
  3. Halve the source rate n = 2 times — multiply by (½)² = ¼:

    120→60→30 counts s−1120 \to 60 \to 30\ \text{counts s}^{-1}120→60→30 counts s−1
  4. Add the background back on — the detector still records it:

    30+4=34 counts s−130 + 4 = 34\ \text{counts s}^{-1}30+4=34 counts s−1
Final answer:

The detector reads 34 counts per second after 30 minutes. Always subtract the background before halving, halve once per half-life (never subtract a fixed amount), then add the background back if the question wants the measured value.


🧠 Quick self-check

Tap each card to reveal the answer.


🎯 Exam tips

Exam Tips

  • α → β → γ: penetration goes UP (paper → aluminium → lead) and ionising power goes DOWN (α strongest → γ weakest). Only γ (neutral) is not deflected by a field. To identify a radiation, use penetration to narrow it down, then deflection to confirm whether it is charged.
  • Most penetrating ≠ most dangerous. From OUTSIDE the body alpha is safe (the skin stops it) but gamma is the bigger hazard; INSIDE the body (breathed in/swallowed) alpha is the most dangerous because of its strong ionising power.
  • Every decay equation: the TOP numbers balance (nucleon number A conserved) and the BOTTOM numbers balance (proton number Z conserved). Alpha: A − 4, Z − 2. Beta-minus: A unchanged, Z + 1 — the electron's −1 charge forces Z UP, so never drop it.
  • For a decay CHAIN, apply each emission one at a time and keep a running tally of A and Z. Use N = A − Z if asked for the neutron number.
  • Energy released: find the mass defect FIRST (parent − total products), then E = mc². In MeV, just multiply Δm(in u) by 931.5. Keep every decimal place when subtracting masses — Δm is a tiny number and early rounding ruins it.
  • Energy sharing: the parent is at rest, so the products have equal and opposite momentum; with KE = p²/2m the LIGHT product (alpha) carries most of the energy. Its share = mdₐᵤgₕₜₑᵣ ÷ (mdₐᵤgₕₜₑᵣ + mₐₗₚₕₐ), close to but just under 100% for a heavy parent.
  • Half-life: ALWAYS subtract the background count rate before halving, then add it back if the question wants the measured value. Work out n = time ÷ half-life and multiply by (½)ⁿ — halve once per half-life, never subtract a fixed amount. Activity is in becquerel (Bq) = decays per second.
  • Two samples with the SAME half-life keep the same RATIO of activities over time, because both fall by the same factor (½)ⁿ.

What you'll learn in Topic 5.3

  • 5.3.1 Types of radiation and their properties
  • 5.3.2 Decay equations and conservation laws
  • 5.3.3 Energy released in radioactive decay
  • 5.3.4 Half-life, activity and background radiation
Suggested study order: Read the notes for each sub-topic below → test yourself with flashcards → attempt practice questions → review exam technique.

Study resources — 5.3 Radioactive decay

5.3.1

Types of radiation and their properties

Notes
5.3.2

Decay equations and conservation laws

Notes
5.3.3

Energy released in radioactive decay

Notes
5.3.4

Half-life, activity and background radiation

Notes

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Topic 5.3 Radioactive decay forms a core part of Unit 5: Nuclear and quantum physics in IB Physics. Mastering these concepts will strengthen your understanding of connected topics across the syllabus and prepare you for exam questions that require analysis, evaluation, and real-world application.

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