Question 1Multiple choice · 1 markA* challenge
An electron is accelerated from rest through a potential difference of 100 V.
What is its de Broglie wavelength?
- A 4.1 × 10−17 m
- B 1.2 × 10−10 m
- C 1.7 × 10−10 m
- D 2.5 × 10−10 m
Show answer
Answer: B
10 free IB Physics E Nuclear & Quantum Physics questions written in the style of IB Physics papers: 3 multiple-choice and 7 structured, each with its answer and mark scheme. 4 of them are A* challenge questions, built to catch the mistakes strong students still make.
This topic covers E.1 Structure of the atom, E.2 Quantum physics, E.3 Radioactive decay, E.4 Fission, E.5 Fusion & stars. Work through the questions below, then generate a full E Nuclear & Quantum Physics paper with a fresh set every time.
Generate a full E Nuclear & Quantum Physics paper →An electron is accelerated from rest through a potential difference of 100 V.
What is its de Broglie wavelength?
Answer: B
A radioactive sample has a half-life of 5 days. What fraction of the original nuclei remains undecayed after 25 days?
Answer: A
An electron in a hydrogen atom moves from the −1.51 eV level to the −3.40 eV level.
What is the wavelength of the emitted photon?
Answer: B
A radioactive source contains 2.0 × 1018 nuclei of an isotope with a half-life of 8.0 days.
Calculate the decay constant in s−1.
Calculate the initial activity.
Calculate the number of nuclei remaining after 20 days.
Calculate the time taken for the activity to fall to 1.0 × 1011 Bq.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 1.0 × 10−6 s−1 λ = ln2/T½ ✓ = 0.693/(8.0 × 86 400) = 1.0 × 10−6 s−1 ✓ | 2 |
| (b) | 2.0 × 1012 Bq A = λN = 2.0 × 1012 Bq ✓ | 1 |
| (c) | 3.5 × 1017 20 days = 2.5 half-lives ✓ N = 2.0 × 1018 × 0.52.5 = 3.5 × 1017 ✓ | 2 |
| (d) | 35 days activity falls by a factor of 20 ✓ t = 8.0 × log220 = 35 days ✓ | 2 |
An electron is accelerated from rest through a potential difference of 150 V.
Calculate its momentum.
Calculate its de Broglie wavelength and suggest why such electrons can be diffracted by crystals.
A proton has the same de Broglie wavelength. Calculate its speed.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 6.6 × 10−24 kg m s−1 p = √(2meV) ✓ = 6.6 × 10−24 kg m s−1 ✓ | 2 |
| (b) | 1.0 × 10−10 m λ = h/p = 1.0 × 10−10 m ✓ similar to the spacing of atoms ✓ | 2 |
| (c) | 4.0 × 103 m s−1 same momentum ✓ v = 6.6 × 10−24/1.67 × 10−27 = 4.0 × 103 m s−1 ✓ | 2 |
Ultraviolet light of wavelength 200 nm is incident on a metal surface of work function 4.3 eV.
Calculate the energy of a photon in eV.
Calculate the maximum kinetic energy of the emitted electrons and the stopping potential.
Explain why the photoelectric effect provides evidence for the particle nature of light.
Calculate the de Broglie wavelength of the fastest emitted electrons.
Suggest why these electrons would show measurable diffraction from a crystal.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 6.22 eV E = hcλ ✓ 6.22 eV ✓ | 2 |
| (b) | 1.92 eV; 1.92 V Ek,max = hf − Φ = 1.92 eV ✓ Vs = 1.92 V ✓ | 2 |
| (c) | threshold frequency, instant emission there is a threshold frequency below which no electrons are emitted, whatever the intensity ✓ emission is instantaneous ✓ explained by one photon giving all its energy to one electron ✓ | 3 |
| (d) | 8.87 × 10−10 m p = √2mEk ✓ λ = hp = 8.87 × 10−10 m ✓ | 2 |
| (e) | λ comparable to atomic spacing their wavelength is comparable with the spacing of atoms in a crystal (~10−10 m) ✓ | 1 |
The mass of a helium-4 nucleus is 4.001 506 u. The mass of a proton is 1.007 276 u and the mass of a neutron is 1.008 665 u.
Calculate the binding energy per nucleon of helium-4, in MeV.
Explain, with reference to binding energy per nucleon, why the fusion of light nuclei releases energy.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 7.1 MeV mass of nucleons = 4.031 882 u ✓ mass defect = 0.030 376 u ✓ binding energy = 0.030 376 × 931.5 = 28.3 MeV ✓ per nucleon = 7.07 MeV ✓ | 4 |
| (b) | BE per nucleon increases the product has a higher binding energy per nucleon ✓ so the total binding energy increases and the difference is released ✓ | 2 |
In a nuclear reactor, uranium-235 absorbs a neutron: 23592U + 10n → 14456Ba + 8936Kr + x 10n. Each fission releases about 200 MeV.
Determine x.
Explain, with reference to binding energy per nucleon, why energy is released.
The reactor produces 900 MW of electrical power with an efficiency of 32%. Calculate the number of fissions per second.
Calculate the mass of uranium-235 used in one year.
Outline the roles of the moderator and the control rods.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 3 x = 3 (nucleon numbers balance) ✓ | 1 |
| (b) | products more tightly bound the products have a greater binding energy per nucleon ✓ so the total binding energy increases and the mass decreases ✓ | 2 |
| (c) | 8.8 × 1019 s−1 thermal power = 2813 MW ✓ 200 MeV = 3.20 × 10−11 J ✓ 8.8 × 1019 s−1 ✓ | 3 |
| (d) | 1082 kg moles per year = fissions ÷ NA ✓ 1082 kg ✓ | 2 |
| (e) | slow neutrons; absorb neutrons moderator slows neutrons so they are more likely to cause fission ✓ control rods absorb neutrons to control the reaction rate ✓ | 2 |
This question is about the operation of a nuclear reactor.
Explain the role of the control rods.
State what is meant by critical mass.
Explain why spent fuel rods must be stored for a long time.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | absorb neutrons they absorb neutrons ✓ so that on average one neutron per fission causes a further fission / the rate is controlled ✓ | 2 |
| (b) | minimum mass for sustained chain reaction the minimum mass of fuel for a self-sustaining chain reaction ✓ | 1 |
| (c) | long half-life products fission products are highly radioactive ✓ some have very long half-lives ✓ | 2 |
A U-235 nucleus absorbs a neutron and splits into Ba-141, Kr-92 and three neutrons, releasing 173 MeV.
Explain why energy is released, with reference to binding energy per nucleon.
Calculate the mass defect in kg.
Explain the role of the moderator in a nuclear reactor.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | products more tightly bound products have higher binding energy per nucleon ✓ total binding energy increases, mass decreases ✓ | 2 |
| (b) | 3.1 × 10−28 kg 173 MeV = 2.77 × 10−11 J ✓ Δm = E/c2 = 3.1 × 10−28 kg ✓ | 2 |
| (c) | slows neutrons by collisions neutrons lose kinetic energy in collisions with moderator nuclei ✓ slow (thermal) neutrons are more likely to cause fission ✓ | 2 |
In the SL & HL syllabus, E Nuclear & Quantum Physics covers E.1 Structure of the atom, E.2 Quantum physics, E.3 Radioactive decay, E.4 Fission, E.5 Fusion & stars.
This page has 10 E Nuclear & Quantum Physics questions with answers. For more, the free generator on mentorriyaz.in builds a new E Nuclear & Quantum Physics paper with a full mark scheme each time you click.
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Meet Riyaz →Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in