MYP Physics · unit 17
Atomic and 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.
Inferring discrete levels from a line spectrum
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. An idealised atom has the supplied energy levels relative to ionisation at 0 eV. Use h = 4.14×10⁻¹⁵ eV s and c = 3.0×10⁸ m s⁻¹. The atom initially occupies the −1.5 eV level; allowed transitions between all listed levels may emit photons. Supplied data (Level index; Energy / eV): (1, -6), (2, -3), (3, -1.5). (a) List the possible emitted photon energies in cascades and find the shortest emitted wavelength. [3 marks] (b) From the −6.0 eV ground level, can isolated 2.0 eV and 6.5 eV photons be absorbed? Explain what line spectra establish about the model. [3 marks]
Idealised energy levels
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Energy / eV
Horizontal axis (x): Level index
Read the exact data as a table
| Level index | Energy / eV |
|---|---|
| 1 | -6 |
| 2 | -3 |
| 3 | -1.5 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): List the possible emitted photon energies in cascades and find the shortest emitted wavelength.
Reveal the worked solution
- Energy drops are (−1.5)−(−3.0)=1.5 eV, (−3.0)−(−6.0)=3.0 eV and (−1.5)−(−6.0)=4.5 eV.
- A direct transition emits one 4.5 eV photon; a two-step cascade emits separate 1.5 and 3.0 eV photons, not one of each plus 4.5 eV in the same decay.
- The greatest photon energy gives shortest wavelength: λ = hc/E = (4.14×10⁻¹⁵)(3.0×10⁸)/4.5≈2.76×10⁻⁷ m.
- A bound–bound absorption must match an allowed energy difference; the listed ground-state excitation gaps are 3.0 and 4.5 eV, not 2.0 eV.
- Ground-state ionisation needs 6.0 eV. Absorbing 6.5 eV can free the electron with 0.5 eV excess kinetic energy, neglecting recoil.
- Discrete lines support quantised energy changes. They do not show electrons following literal tiny planetary trajectories, and a real atom may have selection rules absent from this idealised model.
Answer (a) Possible photons: 1.5, 3.0 and 4.5 eV; shortest wavelength≈276 nm. (b) A 2.0 eV photon matches no listed excitation; a 6.5 eV photon can ionise the atom, leaving 0.5 eV kinetic energy.
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.
Balancing a fission channel and controlling a chain reaction
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. 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 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.