IB DP Physics: Nuclear structure, binding energy and types of radioactive decay (DP Physics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover the first part of sub-topic E.3, Radioactive decay, from the IB Physics guide (first assessment 2025): nuclear structure, binding energy and types of decay. Questions apply to SL and HL except those marked “HL only”. Learn the content in the study guide and the revision notes. Related pages: the course hub, the printable checklist and the free 10-minute diagnostics.

Data where needed: 1 u = 1.661 × 10⁻²⁷ kg = 931.5 MeV/c²; 1 MeV = 1.602 × 10⁻¹³ J; e = 1.602 × 10⁻¹⁹ C; k = 8.99 × 10⁹ N m² C⁻²; G = 6.67 × 10⁻¹¹ N m² kg⁻². Mass of hydrogen-1 atom 1.007825 u; neutron 1.008665 u; proton 1.673 × 10⁻²⁷ kg.

Questions

1. Chlorine has proton number 17. Chlorine-35 and chlorine-37 are isotopes. (a) State the number of neutrons in each nuclide. [2] (b) Explain why the two isotopes have the same chemical properties. [1]

2. A nuclear reaction has a mass defect of 0.0123 u. Express this mass in kg and in MeV/c². [3]

3. Nitrogen-14 has proton number 7 and atomic mass 14.003074 u. Calculate its binding energy and its binding energy per nucleon, in MeV. [4]

4. On a graph of binding energy per nucleon plotted against A, the line climbs to a broad maximum near A = 60 and then declines gradually. Explain, using the curve, why energy is released when two light nuclei join and also when a very heavy nucleus splits. [4]

5. Complete each equation, giving the nucleon number and proton number of the daughter nuclide and any extra particle. (a) Radon-222 (Z = 86) emits an alpha particle to form polonium. [2] (b) Cobalt-60 (Z = 27) undergoes beta-minus decay to nickel. [2] (c) Sodium-22 (Z = 11) undergoes beta-plus decay to neon. [2]

6. Radon-222 decays by alpha emission to polonium-218. Atomic masses: Rn-222 222.017578 u, Po-218 218.008973 u, He-4 4.002603 u. The radon nucleus is at rest. Calculate the energy released and the kinetic energy of the alpha particle. [5]

7. A technician must choose a radioactive source for three jobs. (a) State which of alpha, beta and gamma radiation is the most ionizing. [1] (b) To find a leak in a buried pipe, a tracer is added to the water. Identify the type of radiation to use and justify it. [2] (c) Aluminium foil of changing thickness is made on a rolling line, and a source and detector monitor it. Identify the type of radiation to use and justify it. [2]

8. Two protons are 2.0 × 10⁻¹⁵ m apart inside a nucleus. (a) Calculate the electric force between them. [2] (b) Calculate the gravitational force between them. [2] (c) State two properties of the strong nuclear force that allow the nucleus to hold together. [2]

9. Describe what is meant by the statement that radioactive decay is random and spontaneous. [3]

10. (HL only) Iron-52 (Z = 26) and strontium-90 (Z = 38) are both unstable. Stable iron is mostly iron-56. (a) Explain why stable heavy nuclides have more neutrons than protons. [2] (b) Predict the decay type of each of iron-52 and strontium-90, with a reason. [3]

11. (HL only) (a) Beta particles from a sample are found with energies spread from zero up to a maximum. Explain what this suggests. [2] (b) A nuclide emits alpha particles of two energies, 5.80 MeV and 5.54 MeV, and the sample also emits gamma photons of about 0.26 MeV. Explain how this supports discrete nuclear energy levels. [2]

12. Cobalt-60 decays by beta-minus emission to nickel-60. Atomic masses: Co-60 59.933817 u, Ni-60 59.930786 u. (a) Calculate the total energy released in the decay. [3] (b) (HL only) The daughter nucleus emits two gamma photons of energy 1.17 MeV and 1.33 MeV in turn. State what this shows about the nickel-60 nucleus formed and find the energy of its excited level above the ground state. [2] (c) (HL only) Ignoring the nuclear recoil, find the maximum kinetic energy of the electron when the nickel-60 nucleus is formed in this excited level. [2]

Answers

1. (a) N = A − Z [1]. Cl-35 has 18 neutrons and Cl-37 has 20 neutrons [1]. (b) Same number of protons, so the same number of electrons and electron arrangement, which fixes the chemistry [1]. Examiner insight: Linking chemistry to electron arrangement, not just “same atomic number”, is what turns a restatement into an explanation.

2. Mass in kg = 0.0123 × 1.661 × 10⁻²⁷ [1] = 2.04 × 10⁻²⁹ kg [1]. Mass in MeV/c² = 0.0123 × 931.5 = 11.5 MeV/c² [1]. Examiner insight: Carry the unit symbol through each line; a bare number for a mass in MeV/c² is easy to misread as an energy.

3. Separated nucleons (using the hydrogen atom mass) = 7 × 1.007825 + 7 × 1.008665 = 7.054775 + 7.060655 = 14.115430 u [1]. Mass defect = 14.115430 − 14.003074 = 0.112356 u [1]. Binding energy = 0.112356 × 931.5 = 104.7 MeV [1]. Per nucleon = 104.7 ÷ 14 = 7.48 MeV [1]. Examiner insight: Show the separated mass, the subtraction and the multiplication on separate lines; clear working lets a reader see where a slip happened and follow your reasoning.

4. Binding energy per nucleon is lower for light nuclei than at the peak [1], so two light nuclei joining form a nucleus nearer the peak, with higher binding energy per nucleon [1]. A very heavy nucleus is on the falling part of the curve, so its fragments, nearer the peak, have higher binding energy per nucleon [1]. Either way the products are more tightly bound overall and have less total mass, and the missing mass is carried off as kinetic energy and radiation [1]. Examiner insight: Refer to the curve explicitly (which side of the peak) and say total mass falls; “it is more stable” alone does not link to the energy released.

5. (a) ²²²₈₆Rn → ²¹⁸₈₄Po + ⁴₂He: A = 218 [1], Z = 84 [1]. (b) ⁶⁰₂₇Co → ⁶⁰₂₈Ni + ⁰₋₁e + antineutrino: A = 60, Z = 28 [1], with electron and antineutrino [1]. (c) ²²₁₁Na → ²²₁₀Ne + ⁰₊₁e + neutrino: A = 22, Z = 10 [1], with positron and neutrino [1]. Examiner insight: Check that both top and bottom numbers balance before moving on, and keep neutrino and antineutrino apart: beta-minus has the antineutrino.

6. Mass lost = 222.017578 − 218.008973 − 4.002603 [1] = 0.006002 u [1]. Energy = 0.006002 × 931.5 = 5.59 MeV [1]. Equal and opposite momenta, so the energy is shared in inverse proportion to mass: alpha share = 5.59 × 218/222 [1] = 5.49 MeV [1]. Examiner insight: Explain the sharing by momentum conservation; writing that the alpha particle gets all the energy is the usual lost mark.

7. (a) Alpha [1]. (b) Gamma [1] because it is the most penetrating and can pass out through the pipe wall and soil to reach a detector [1]. (c) Beta [1] because it is partly absorbed by thin aluminium, so the detected count changes with thickness; alpha would not get through and gamma would barely change [1]. Examiner insight: A justification must name a property (penetration or absorption) and link it to the job; naming only the radiation earns nothing extra.

8. (a) F = kq²/r² = 8.99 × 10⁹ × (1.602 × 10⁻¹⁹)² ÷ (2.0 × 10⁻¹⁵)² [1] = 58 N (57.7 N) [1]. (b) F = Gm²/r² = 6.67 × 10⁻¹¹ × (1.673 × 10⁻²⁷)² ÷ (2.0 × 10⁻¹⁵)² [1] = 4.7 × 10⁻³⁵ N [1]. (c) It is attractive [1] and stronger than the electric repulsion at nuclear distances, and it acts over a very short range [1]. Examiner insight: Square the separation as well as the charge or mass, and write the answer with its unit; the ratio of the two answers makes the point that gravity cannot be the binding force.

9. Spontaneous: decay needs no external trigger and is unaffected by conditions such as temperature or pressure [1]. Random: it cannot be predicted when a particular nucleus will decay [1]. Only the probability of decay is fixed, so a large sample follows a predictable pattern while individual decays do not [1]. Examiner insight: Keep the two words apart: conditions belong to spontaneous, timing belongs to random.

10. (a) Protons repel each other electrically, and more protons means more repulsion [1]. Extra neutrons add strong-force attraction without adding repulsion, so heavier stable nuclides need N greater than Z [1]. (b) Iron-52 has N = 26 and Z = 26, fewer neutrons than stable iron-56, so it is proton-rich and decays by beta-plus [1]. Strontium-90 has N = 52 and Z = 38, a high neutron-to-proton ratio, so it decays by beta-minus [1]. Each decay moves the ratio towards the stable value [1]. Examiner insight: Compare the neutron number with the stable nuclides of the same element or nearby mass, and state the direction of change in Z.

11. (a) If only an electron and a daughter nucleus were emitted, the electron would have a single fixed energy [1]. A spread of energies, with energy and momentum seemingly missing, suggests a third particle, the neutrino, shares the energy [1]. (b) The two alpha energies differ by 0.26 MeV, which matches the gamma photon energy [1], so the daughter nucleus is formed in either of two discrete levels, and the upper one drops to the lower by emitting a photon of the energy gap [1]. Examiner insight: Match the numbers: the difference in alpha energies should be compared with the photon energy, and the word “discrete” or “quantized” should appear.

12. (a) Mass lost = 59.933817 − 59.930786 [1] = 0.003031 u [1]. Energy = 0.003031 × 931.5 = 2.82 MeV [1]. (b) Nickel-60 is formed in an excited nuclear energy level [1]. Level above ground = 1.17 + 1.33 = 2.50 MeV [1]. (c) Energy left for the electron and antineutrino = 2.82 − 2.50 [1]. Maximum electron kinetic energy (antineutrino carries almost none) = 0.32 MeV [1]. Examiner insight: State the assumption that the antineutrino and recoil take negligible energy at the maximum; an unexplained subtraction is hard to credit.

Where marks are usually lost

• Using the proton mass instead of the hydrogen atom mass with atomic masses, or the other way round.
• Reversing the subtraction so the mass defect is negative.
• Stopping at total binding energy when the question asks per nucleon.
• Mixing u, kg and MeV/c² in one line.
• Giving the alpha particle all of the released energy.
• Writing a beta equation with no neutrino or antineutrino, or the wrong one.
• Describing penetration without linking it to the job.
• Mixing random with spontaneous.
• Giving a 2-digit answer where three significant figures are expected.

Next steps

• Re-read the revision notes and the study guide.
• Visit the IB DP Physics hub and tick off topics on the printable checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Physics guide, first assessment 2025, International Baccalaureate Organization: sub-topic E.3, Radioactive decay (nuclear structure, binding energy and decay types, including additional higher level understandings).