Cambridge International A-Level Physics · topic 23
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
Work through the linked stages and explain what each result means.
A half-life hidden by background counts
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.
Balancing a fission channel and controlling a chain reaction
Consider the idealised fission channel ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + x¹₀n. Each fission releases 200 MeV; 1 eV = 1.60×10⁻¹⁹ J. An illustrative reactor's successive-generation neutron population is supplied, with each generation lasting the same time. Supplied data (Generation; Neutrons in relative units): (0, 100), (1, 110), (2, 121), (3, 133.1). (a) Find x and the multiplication factor in the example. Explain the distinct roles of moderator and control rods. [3 marks] (b) Calculate the fission rate for 1.0 GW thermal output and explain the origin of the released energy. [3 marks]
Illustrative neutron multiplication
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Neutrons in relative units
Horizontal axis (x): Generation
Read the exact data as a table
| Generation | Neutrons in relative units |
|---|---|
| 0 | 100 |
| 1 | 110 |
| 2 | 121 |
| 3 | 133.1 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find x and the multiplication factor in the example. Explain the distinct roles of moderator and control rods.
Reveal the worked solution
- Nucleon number gives 235+1 = 141+92+x, hence x = 3. Charge checks as 92 = 56+36; neutrons carry no charge.
- The measured population ratio is 110/100 = 121/110 = 1.10, so this illustrative sequence grows by 10% per generation.
- A moderator slows neutrons in a thermal reactor without being intended to absorb many; control rods absorb neutrons to adjust the effective multiplication. These functions are not interchangeable.
- Energy per fission is 200×10⁶×1.60×10⁻¹⁹ = 3.20×10⁻¹¹ J.
- Rate = power/energy per event = 1.0×10⁹/(3.20×10⁻¹¹) = 3.125×10¹⁹ fissions s⁻¹.
- The products have lower total rest mass and greater total binding energy; the mass difference appears as released energy. Nucleon and charge conservation do not imply equality of the sum of the separate rest masses.
Answer (a) x = 3; effective multiplication factor = 1.10; control must bring it toward 1 for steady operation. (b) Fission rate = 3.125×10¹⁹ s⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A fusion power estimate and a star's fuel lifetime
In an idealised net stellar fusion cycle, four hydrogen nuclei ultimately produce one helium-4 nucleus plus other particles and radiation, releasing 26.7 MeV per completed cycle. Use 1 MeV = 1.60×10⁻¹³ J and c = 3.0×10⁸ m s⁻¹. The supplied graph gives cumulative energy from completed cycles. A star emits 4.0×10²⁶ W and has 2.0×10²⁹ kg of available hydrogen fuel; take four hydrogen nuclei as 6.68×10⁻²⁷ kg. Supplied data (Completed cycles / 10³⁸; Energy / 10²⁶ J): (0, 0), (1, 4.272), (2, 8.544), (3, 12.816). (a) Find energy per cycle, mass defect per cycle and the cycle rate supplying the stated luminosity. [3 marks] (b) Estimate the fuel-consumption rate and idealised lifetime, taking 1 yr = 3.16×10⁷ s. Explain two limits of this estimate. [3 marks]
Fusion energy release
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Energy / 10²⁶ J
Horizontal axis (x): Completed cycles / 10³⁸
Read the exact data as a table
| Completed cycles / 10³⁸ | Energy / 10²⁶ J |
|---|---|
| 0 | 0 |
| 1 | 4.272 |
| 2 | 8.544 |
| 3 | 12.816 |
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 energy per cycle, mass defect per cycle and the cycle rate supplying the stated luminosity.
Reveal the worked solution
- Convert 26.7 MeV to joules: E = 26.7(1.60×10⁻¹³) = 4.272×10⁻¹² J.
- The decrease in total rest mass is Δm = E/c² = 4.272×10⁻¹²/(9.0×10¹⁶)≈4.75×10⁻²⁹ kg per net cycle.
- Cycle rate is L/E = 4.0×10²⁶/(4.272×10⁻¹²)≈9.36×10³⁷ s⁻¹. Nuclear binding-energy differences, rather than chemical combustion, power this process.
- Each cycle uses about 6.68×10⁻²⁷ kg of hydrogen, so consumption is (9.36×10³⁷)(6.68×10⁻²⁷)≈6.25×10¹¹ kg s⁻¹.
- Time is available fuel divided by its consumption rate: 2.0×10²⁹/(6.25×10¹¹)≈3.20×10¹⁷ s≈1.01×10¹⁰ yr.
- Luminosity changes as the star evolves, only part of its hydrogen is accessible to core fusion, and some fusion energy can escape in neutrinos. A full stellar model must account for structure, composition and evolving reactions.
Answer (a) Energy = 4.272×10⁻¹² J; mass defect≈4.75×10⁻²⁹ kg; rate≈9.36×10³⁷ cycles s⁻¹. (b) Fuel consumption≈6.25×10¹¹ kg s⁻¹; idealised lifetime≈1.01×10¹⁰ years.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.