Cambridge IGCSE Physics · unit 5
Nuclear physics
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
The rare deflections that changed the atomic model
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A narrow alpha-particle beam strikes a very thin gold foil. Detectors covering equal solid angles record the supplied count rates; values are illustrative rather than a full scattering-law fit. Most alpha particles pass close to the forward direction. Gold-197 has atomic number 79, and an alpha particle is a helium-4 nucleus of charge +2e. Supplied data (Scattering angle / °; Counts per minute): (10, 100000), (30, 10000), (60, 1000), (90, 100), (150, 10). (a) Explain what the common small deflections and rare large deflections imply, and compare them with a model of diffuse positive charge. [3 marks] (b) Determine proton, neutron and electron numbers in neutral gold-197 and in a gold-197 ion of charge +3e. Describe an isotope and balance alpha emission from a parent with A = 201, Z = 81. [3 marks]
Illustrative alpha-scattering observations
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Counts per minute
Horizontal axis (x): Scattering angle / °
Read the exact data as a table
| Scattering angle / ° | Counts per minute |
|---|---|
| 10 | 100000 |
| 30 | 10000 |
| 60 | 1000 |
| 90 | 100 |
| 150 | 10 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Explain what the common small deflections and rare large deflections imply, and compare them with a model of diffuse positive charge.
Reveal the worked solution
- Most projectiles cross the foil with little change in direction, so atoms are not solid dense obstacles across their full dimensions.
- A rare close encounter with concentrated positive nuclear charge can strongly repel an alpha particle. Very large deflections require a strong local interaction and a sufficiently massive scattering centre.
- A diffuse positive-charge model would spread the repulsion over the atomic volume and struggles to explain the observed rare strong deflections. The evidence supports a compact nucleus, while this count table alone does not determine an exact nuclear radius.
- Atomic number gives 79 protons. Neutrons are A−Z = 197−79 = 118; neutrality requires 79 electrons.
- A +3e ion has lost three electrons and has 76; its nucleus is unchanged. An isotope has the same proton number but a different neutron number.
- Alpha emission removes A = 4 and Z = 2: ²⁰¹₈₁X → ¹⁹⁷₇₉Au + ⁴₂He. Both nucleon and charge numbers balance.
Answer (a) Most atomic volume is sparsely occupied; positive charge and most mass are concentrated in a small nucleus. (b) Neutral Au-197: 79 p, 118 n, 79 e; Au³⁺ has 76 e. Alpha daughter: A = 197, Z = 79.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A half-life hidden by background counts
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A detector counts the same sealed radioactive source in consecutive equal-duration measurement windows at the supplied times. Background measured without the source is 40 counts per window. Geometry and detector efficiency are fixed. Assume one radionuclide dominates and counting windows are short compared with its half-life. Supplied data (Time / min; Counts per window): (0, 360), (5, 200), (10, 120), (15, 80), (20, 60). (a) Correct for background, infer half-life and predict the raw count after 30 min. [3 marks] (b) Evaluate the inference that the source is harmless when raw counts approach background, and explain how to improve confidence in a weak signal. [3 marks]
Raw detector counts
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Counts per window
Horizontal axis (x): Time / min
Read the exact data as a table
| Time / min | Counts per window |
|---|---|
| 0 | 360 |
| 5 | 200 |
| 10 | 120 |
| 15 | 80 |
| 20 | 60 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Correct for background, infer half-life and predict the raw count after 30 min.
Reveal the worked solution
- Subtract 40 from each raw reading before analysing source decay. The corrected signal halves every 5 min.
- At 30 min six half-lives have elapsed, giving expected net counts 320/2⁶ = 5.
- The detector still records background, so expected raw counts are 5+40 = 45. A real individual reading will fluctuate randomly about this expectation.
- Activity describes decays per second, while counts depend on detector efficiency, radiation type, shielding and geometry. The source can remain active when this setup cannot clearly distinguish it from background.
- Longer matched source and background counting periods reduce relative random uncertainty, while repeated readings reveal scatter. Keep distance and shielding fixed for a fair decay comparison.
- Risk also depends on exposure time, energy, biological uptake and contamination pathways. Use the relevant handling controls rather than inferring safety from one low count.
Answer (a) Net counts are 320, 160, 80, 40 and 20; half-life = 5 min; expected raw count at 30 min = 45. (b) Near-background readings limit detectability; they do not by themselves establish absence of activity or a safe exposure.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A shielding decision needs more than a half-life
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. Two sealed waste samples each have initial activity 800 Bq. Sample A has half-life 2 days and sample B 20 years. Their activity fractions after successive half-lives are supplied. For a simplified external gamma-shielding estimate, each 1 cm of a material halves transmitted intensity. This model ignores buildup and other radiation types. Supplied data (Elapsed half-lives; Activity / initial activity): (0, 1), (1, 0.5), (2, 0.25), (3, 0.125), (4, 0.0625). (a) Find each sample's activity after 8 days or 80 years respectively and the ideal shielding thickness for a 1/32 intensity factor. [3 marks] (b) Evaluate the claim that equal activity means equal risk, and propose a defensible waste-management comparison. [3 marks]
Remaining activity fraction
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Activity / initial activity
Horizontal axis (x): Elapsed half-lives
Read the exact data as a table
| Elapsed half-lives | Activity / initial activity |
|---|---|
| 0 | 1 |
| 1 | 0.5 |
| 2 | 0.25 |
| 3 | 0.125 |
| 4 | 0.0625 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find each sample's activity after 8 days or 80 years respectively and the ideal shielding thickness for a 1/32 intensity factor.
Reveal the worked solution
- Eight days is four half-lives for A; 80 years is four for B. Each activity becomes 800/2⁴ = 50 Bq, despite very different calendar times.
- A transmission factor 1/32 is (1/2)⁵, so five 1 cm layers are needed in the simplified shielding model.
- Decay storage and shielding reduce different aspects of exposure: elapsed time reduces activity, while shielding reduces radiation reaching a location without changing the decay rate.
- A becquerel is one decay per second, not a unit of absorbed or biologically weighted dose. Decay energy, radiation type and uptake can differ.
- A long-lived sample may need containment for far longer, while a short-lived sample can have intense early emission. Daughter products can complicate both histories.
- Compare measured dose rates under specified geometry, containment integrity, transport and storage requirements, and credible contamination pathways. The simple half-value model alone cannot certify a storage design.
Answer (a) Both specified intervals give 50 Bq; five half-value layers require 5 cm. (b) Equal Bq does not imply equal dose or risk; assess radiation, geometry, duration and pathways as well as activity.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.