IGCSE Physics Radioactive Decay Questions and Worked Solutions (5.2.3) | 0625

IGCSE Physics 0625 · Topic 5.2.3

IGCSE Physics Radioactive Decay Questions and Worked Solutions

50 topic-wise practice questions on Radioactive Decay (Unit 5: Nuclear Physics), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions11 Core29 Extended10 Challenge131 marksTextbook §5.2 · pp. 260–274
Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Radioactive Decay

Radioactive decay is a change in an unstable nucleus. Alpha and beta decay produce a different element; gamma emission removes energy only. (Extended) Decay equations show the changes in proton and nucleon numbers.

Key formulas

α decay: A → A − 4, Z → Z − 2alpha emissionExtended
β decay: A unchanged, Z → Z + 1beta emission (n → p + e)Extended
γ emission: A and Z unchangedgamma emissionExtended

Prerequisite knowledge

  • Nuclide notation (5.1.2)
  • Types of emission (5.2.2)

Exam tips & common mistakes

  • Check that A and Z balance on both sides
  • The product of β decay has the same A
  • Gamma does not change the element

Key facts from the syllabus (5.2.3)

  • Decay is spontaneous and random
  • During α or β decay the nucleus changes to a different element
  • Extended isotopes can be radioactive because they have too many neutrons and/or are too heavy
  • Extended decay increases stability and reduces the number of excess neutrons
  • Extended in β-decay a neutron changes into a proton and an electron
  • Extended write decay equations in nuclide notation for α, β and γ

Textbook: §5.2 Radioactivity, pp. 260–274

1
Core[1 mark]

Define radioactive decay.

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Method: Unstable nucleus change.

A change in an unstable nucleus that results in the emission of alpha, beta and/or gamma radiation
2
Core[1 mark]

State two words describing the nature of radioactive decay.

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Method: Spontaneous and random.

Spontaneous and random
3
Core[2 marks]

State what happens to the element when a nucleus emits an alpha or beta particle.

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Method: New element.

It changes into a different element
4
Core[1 mark]

State whether gamma emission changes the element.

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Method: No change in Z.

No
5
Core[2 marks]

Explain what is meant by 'random' in radioactive decay.

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Method: Unpredictable.

It is impossible to predict which nucleus will decay next or when
6
Core[1 mark]

State whether temperature or pressure affect the rate of radioactive decay.

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Method: Not affected.

No, decay is not affected by temperature or pressure
7
Core[2 marks]

State what is meant by 'spontaneous' in radioactive decay.

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Method: No outside cause.

It happens without any external trigger
8
Core[2 marks]

Explain why it is impossible to predict when a particular nucleus will decay.

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Method: Random.

Decay is a random process
9
Core[2 marks]

State which emissions change the proton number of a nucleus.

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Method: α and β.

Alpha and beta
10
Core[2 marks]

State the type of decay that produces a nucleus of the same element.

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Method: Gamma.

Gamma emission
11
Core[2 marks]

A sample contains 1000 unstable nuclei. Explain why you cannot say exactly how many will decay in the next minute.

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Method: Random.

Decay is random; only the probability of decay is fixed, so the actual number varies
12
Extended[2 marks]

State two reasons why isotopes may be radioactive.

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Method: Excess neutrons, too heavy.

Too many neutrons; nucleus too heavy
13
Extended[2 marks]

Describe the change in the nucleus during beta emission.

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Method: n → p + e.

A neutron changes into a proton and an electron; the electron is emitted
14
Extended[2 marks]

State the effect of alpha decay on the proton number and nucleon number.

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Method: Z − 2, A − 4.

Z decreases by 2; A decreases by 4
15
Extended[2 marks]

State the effect of beta decay on the proton number and nucleon number.

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Method: Z + 1, A same.

Z increases by 1; A unchanged
16
Extended[2 marks]

State the effect of gamma emission on the nucleus.

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Method: Energy only.

Z and A unchanged; the nucleus loses energy
17
Extended[3 marks]

Write the decay equation for ²²⁶₈₈Ra emitting an alpha particle (product Rn).

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Method: A − 4, Z − 2.

²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He
18
Extended[3 marks]

Write the decay equation for ¹⁴₆C emitting a beta particle (product N).

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Method: Z + 1.

¹⁴₆C → ¹⁴₇N + ⁰₋₁e
19
Extended[3 marks]

Write the decay equation for ⁹⁰₃₈Sr beta decay (product Y).

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Method: Z + 1.

⁹⁰₃₈Sr → ⁹⁰₃₉Y + ⁰₋₁e
20
Extended[3 marks]

²³⁸₉₂U decays by alpha emission. State the proton and nucleon numbers of the product.

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Method: Z − 2, A − 4.

Z = 90, A = 234 (thorium-234)
21
Extended[3 marks]

A nucleus ²³⁴₉₀Th emits a beta particle. State the product.

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Method: Z + 1.

²³⁴₉₁Pa
22
Extended[3 marks]

A nucleus emits two alpha particles and one beta particle starting from ²³⁸₉₂U. State the final Z and A.

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Method: 2α: Z − 4, A − 8; β: Z + 1.

Z = 89, A = 230
23
Extended[2 marks]

Explain why alpha or beta decay increases the stability of a nucleus.

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Method: Reduces excess.

It reduces the excess of neutrons or the size of a heavy nucleus
24
Extended[2 marks]

Write the equation for the gamma emission of an excited ⁶⁰₂₈Ni nucleus.

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Method: Unchanged Z, A.

⁶⁰₂₈Ni* → ⁶⁰₂₈Ni + ⁰₀γ
25
Extended[2 marks]

Explain why beta emission reduces the number of excess neutrons.

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Method: n → p.

A neutron turns into a proton, so there is one fewer neutron and one more proton
26
Extended[2 marks]

State why a nucleus with too many neutrons is likely to undergo beta decay.

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Method: n → p.

Beta decay changes a neutron into a proton, reducing the excess of neutrons
27
Extended[2 marks]

State why very heavy nuclei are likely to undergo alpha decay.

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Method: Reduce size.

Alpha decay reduces the mass and proton number, making the nucleus less heavy and more stable
28
Extended[3 marks]

Write the decay equation for ²¹⁰₈₄Po emitting an alpha particle (product Pb).

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Method: A − 4, Z − 2.

²¹⁰₈₄Po → ²⁰⁶₈₂Pb + ⁴₂He
29
Extended[3 marks]

Write the decay equation for ⁴⁰₁₉K emitting a beta particle (product Ca).

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Method: Z + 1.

⁴⁰₁₉K → ⁴⁰₂₀Ca + ⁰₋₁e
30
Extended[3 marks]

Write the decay equation for ¹³¹₅₃I beta decay (product Xe).

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Method: Z + 1.

¹³¹₅₃I → ¹³¹₅₄Xe + ⁰₋₁e
31
Extended[3 marks]

Write the decay equation for ²⁴¹₉₅Am alpha decay (product Np).

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Method: A − 4, Z − 2.

²⁴¹₉₅Am → ²³⁷₉₃Np + ⁴₂He
32
Extended[3 marks]

²²²₈₆Rn decays to ²¹⁸₈₄Po. State the type of decay and explain.

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Method: A − 4, Z − 2.

Alpha decay: nucleon number falls by 4 and proton number by 2
33
Extended[3 marks]

²¹⁴₈₂Pb decays to ²¹⁴₈₃Bi. State the type of decay and explain.

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Method: Z + 1.

Beta decay: proton number rises by 1, nucleon number unchanged
34
Extended[3 marks]

A nucleus undergoes alpha decay then two beta decays. State the overall change in Z and A.

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Method: α: −2, −4; 2β: +2.

Z unchanged; A decreases by 4 (it becomes an isotope of the original element)
35
Extended[2 marks]

State what particle is emitted from the nucleus in beta decay and where it comes from.

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Method: Electron from neutron.

An electron, produced when a neutron changes into a proton
36
Extended[2 marks]

Explain why gamma emission often follows alpha or beta decay.

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Method: Excess energy.

The new nucleus is left with excess energy, which it releases as gamma radiation
37
Extended[3 marks]

²³⁸₉₂U decays to ²⁰⁶₈₂Pb by a series of alpha and beta decays. Calculate the number of alpha decays.

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Method: ΔA = 32.

Number of α = 32 ÷ 4 = 8
38
Extended[3 marks]

For the decay series in the previous question, calculate the number of beta decays.

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Method: 8α gives Z − 16; need Z − 10.

Number of β = 16 − 10 = 6
39
Extended[2 marks]

State the symbol used for a beta particle in a decay equation.

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Method: ⁰₋₁e.

⁰₋₁e (or ⁰₋₁β)
40
Extended[2 marks]

State the symbol used for an alpha particle in a decay equation.

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Method: ⁴₂He.

⁴₂He (or ⁴₂α)

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

41
ChallengeCore[3 marks]

Explain what is meant by radioactive decay being spontaneous and random.

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Method: No trigger; unpredictable.

It happens without any outside cause, and it is impossible to predict which nucleus will decay or when
42
ChallengeExtended[4 marks]

Write decay equations for ²³⁸₉₂U → Th by alpha decay, then the Th → Pa by beta decay.

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Method: A − 4, Z − 2; Z + 1.

²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He; ²³⁴₉₀Th → ²³⁴₉₁Pa + ⁰₋₁e
43
ChallengeExtended[4 marks]

⁶⁰₂₇Co decays by beta emission followed by gamma emission. Write both equations (product Ni).

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Method: Beta then gamma.

⁶⁰₂₇Co → ⁶⁰₂₈Ni* + ⁰₋₁e; ⁶⁰₂₈Ni* → ⁶⁰₂₈Ni + γ
44
ChallengeExtended[5 marks]

²³²₉₀Th decays to ²⁰⁸₈₂Pb by a series of α and β decays. Calculate the number of each.

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Method: ΔA/4 for α; then β from Z.

α = 24 ÷ 4 = 6; Z after 6α = 78, so β = 82 − 78 = 4
45
ChallengeExtended[4 marks]

A nucleus emits an alpha particle then two beta particles. Show that the final nucleus is an isotope of the original.

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Method: Net change.

α: Z − 2, A − 4; 2β: Z + 2. Net: Z unchanged, A − 4 → same element, different mass number
46
ChallengeExtended[4 marks]

Explain why carbon-14 (6 protons, 8 neutrons) decays by beta emission, and state the product.

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Method: Neutron excess.

It has too many neutrons; a neutron changes into a proton and an electron; the product is ¹⁴₇N
47
ChallengeExtended[4 marks]

Explain why very heavy nuclei such as radium tend to decay by alpha emission.

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Method: Too heavy.

The nucleus is too heavy to be stable; emitting an alpha particle reduces its mass and charge significantly
48
ChallengeExtended[4 marks]

Explain why gamma emission does not change the element, but often follows alpha or beta decay.

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Method: Excess energy.

Gamma carries away energy only (Z and A unchanged); after α or β decay the nucleus is often left excited
49
ChallengeExtended[4 marks]

An unknown nucleus ᴬ_Z X decays by beta emission to ⁹⁰₃₉Y. Find A and Z and identify X if Z corresponds to strontium (38).

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Method: Reverse beta.

A = 90, Z = 38: ⁹⁰₃₈Sr
50
ChallengeExtended[4 marks]

A nucleus ²¹⁸₈₄Po decays to ²¹⁴₈₂Pb then to ²¹⁴₈₃Bi. Identify each decay and write the overall change.

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Method: Compare A and Z.

First α (A − 4, Z − 2), then β (Z + 1); overall A − 4, Z − 1

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