IB DP Physics: Atomic structure, scattering, spectra and Bohr model (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 sub-topic E.1 (Structure of the atom) of the IB Physics guide, first assessment 2025. Questions 1 to 5 are SL and HL. Questions 6 to 12 are HL only (additional higher level). Use these values: c = 3.00 × 10⁸ m s⁻¹, h = 6.63 × 10⁻³⁴ J s, electron charge magnitude e = 1.60 × 10⁻¹⁹ C, k = 8.99 × 10⁹ N m² C⁻², u = 1.66 × 10⁻²⁷ kg and electron mass m = 9.11 × 10⁻³¹ kg; and take R₀ = 1.2 fm (1 fm = 10⁻¹⁵ m) and ke² = 1.44 MeV fm. Give numerical answers to three significant figures.

Related: study guide, revision notes, course hub and printable checklist.

Questions

1. A nucleus contains 47 protons and 61 neutrons. State its nucleon number and write it in nuclear notation, using X for the chemical symbol. [2]

2. Describe the Geiger-Marsden-Rutherford experiment and explain what two of its observations showed about the structure of the atom. [5]

3. White light is passed through cool hydrogen gas and then through a prism. Explain why the spectrum shows dark lines, and why they appear at the same wavelengths as the bright lines in the emission spectrum of hydrogen. [3]

4. An atom has energy levels at −9.0 eV (ground state), −4.0 eV, −2.0 eV and 0 (ionised).

(a) Atoms are excited to the −2.0 eV level. State the number of different photon energies that can be emitted as they return to the ground state. [1] (b) Calculate the shortest wavelength emitted in part (a). [3] (c) A beam of 6.0 eV photons is directed at atoms in the ground state. Explain whether they are absorbed. [2]

5. The light from a star shows a continuous spectrum crossed by dark lines. Several of the lines match wavelengths emitted by sodium in the laboratory. Explain what this suggests and how the dark lines arise. [3]

6. (HL only) A nucleus has nucleon number 125.

(a) Calculate its radius. [2] (b) Another nucleus has radius 3.6 fm. Calculate its nucleon number. [2] (c) State how the density of these two nuclei compare, and why. [1]

7. (HL only) A head-on alpha particle of kinetic energy 9.0 MeV is fired at an aluminium nucleus (Z = 13, A = 27), which you should treat as fixed.

(a) Calculate the distance of closest approach. [3] (b) Compare it with the nuclear radius and state what this tells you about the scattering. [2]

8. (HL only) Alpha particles of increasing energy are fired at a thin aluminium foil (Z = 13, A = 27). Above a certain energy, the number scattered through large angles falls below the prediction of Rutherford scattering.

(a) Explain this deviation. [3] (b) Estimate the alpha energy, in MeV, at which the deviation should begin, treating the alpha particle as a point charge. [3]

9. (HL only) A hydrogen electron drops from the fifth Bohr level to the third. Calculate the wavelength of the photon released and say which part of the spectrum it lies in. [4]

10. (HL only) An electron in a hydrogen atom moves at 1.09 × 10⁶ m s⁻¹ in a circular orbit of radius 2.12 × 10⁻¹⁰ m. Show that it is consistent with the quantisation of angular momentum for a whole number n, state n, and state its energy in eV. [4]

11. (HL only) Hydrogen atoms are in the ground state and are illuminated by photons of energy 10.20 eV, 11.00 eV, 12.09 eV and 12.75 eV.

(a) Determine which photons can be absorbed. [3] (b) An atom is excited to n = 3. List the photon energies it can emit and calculate the longest wavelength emitted. [5]

12. (HL only) An alpha particle of kinetic energy 6.8 MeV approaches a lead nucleus head-on (Z = 82, A = 208).

(a) Calculate the nuclear radius. [2] (b) Calculate the distance of closest approach. [3] (c) Comment on how the two values relate. [1] (d) The radius of a lead atom is about 1.0 × 10⁻¹⁰ m. Calculate the ratio of the volume of the atom to the volume of the nucleus, and state what this suggests about the density of the nucleus relative to the atom. [2]

Answers

1. Nucleon number = 47 + 61 = 108 [1]. Notation: ¹⁰⁸₄₇X, with 108 at the top and 47 at the bottom [1]. [2] Examiner insight: the top number counts protons and neutrons, the bottom number protons only; swapping them loses both marks.

2. A narrow beam of alpha particles is aimed at a very thin gold foil in a vacuum and detected at different angles [1]. Most alpha particles passed through undeflected or with small deflections, so most of the atom is empty space [1][1]. A very small number were scattered through large angles, almost straight back [1], so the positive charge and mass are concentrated in a tiny, dense nucleus [1]. [5] Examiner insight: pair every observation with what it shows about the atom; observations alone gain little.

3. Atoms in the gas absorb only photons that match a gap between a pair of their energy levels [1], removing those wavelengths from the white light, which shows as dark lines [1]. The same gaps are the ones crossed in emission, so the absorbed and emitted wavelengths are the same [1]. [3] Examiner insight: name the mechanism, photon energy matching a level gap. “The gas absorbs some colours” does not explain the sharp lines.

4. (a) Three photon energies: −2.0 to −4.0, −4.0 to −9.0 and −2.0 to −9.0 eV [1]. [1] (b) The shortest wavelength comes from the largest gap, −2.0 to −9.0 eV = 7.0 eV [1]. 7.0 × 1.60 × 10⁻¹⁹ = 1.12 × 10⁻¹⁸ J [1]. λ = hc/ΔE = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 1.12 × 10⁻¹⁸ = 1.78 × 10⁻⁷ m [1]. [3] (c) The possible gaps from the ground state are 5.0 eV, 7.0 eV and 9.0 eV [1]. 6.0 eV matches none of them, so the photons are not absorbed [1]. [2] Examiner insight: convert eV to joules before using h; mixing eV and J gives a wrong answer that is hard to credit.

5. The star’s hot interior emits a continuous spectrum [1]. Cooler gas in the outer atmosphere absorbs photons at wavelengths that match its atoms’ level gaps, leaving dark lines [1]. Lines matching sodium suggest sodium is present in the star’s atmosphere [1]. [3] Examiner insight: state where the absorbing gas is and why the pattern identifies an element.

6. (a) R = 1.2 × 125^(1/3) [1] = 1.2 × 5 = 6.0 fm [1]. [2] (b) A = (R/R₀)³ = (3.6/1.2)³ [1] = 3³ = 27 [1]. [2] (c) Both have the same density, since density is 3u/(4πR₀³), independent of A [1]. [1] Examiner insight: explain why the densities match (mass and volume are both proportional to A); stating “the same” without a reason is incomplete.

7. (a) Kinetic energy = 9.0 × 10⁶ × 1.60 × 10⁻¹⁹ = 1.44 × 10⁻¹² J [1]. At closest approach, E_k = k(2e)(13e)/d, so d = 2 × 8.99 × 10⁹ × 13 × (1.60 × 10⁻¹⁹)² / 1.44 × 10⁻¹² [1] = 4.16 × 10⁻¹⁵ m [1]. [3] (b) R = 1.2 × 27^(1/3) = 3.6 fm [1]. d (4.16 fm) is greater than R, so the alpha particle does not reach the nucleus and the electric force alone describes the scattering [1]. [2] Examiner insight: use the alpha charge 2e and the nuclear charge 13e; using a single elementary charge for either particle gives the wrong factor.

8. (a) At high energy the alpha particle approaches closer to the nucleus [1]. When the distance is about the nuclear radius, the strong nuclear force (attractive, between nucleons) acts on the alpha particle [1]. Rutherford’s prediction includes only electric repulsion, so measured counts differ from it [1]. [3] (b) Set the closest approach equal to R = 1.2 × 27^(1/3) = 3.6 fm [1]. E_k = 2kZe²/R = 2 × 13 × 1.44 MeV fm / 3.6 fm [1] = 10.4 MeV [1]. [3] Examiner insight: name the force responsible for the deviation; “the model breaks down” is too vague for credit.

9. E₅ = −13.6/25 = −0.544 eV and E₃ = −13.6/9 = −1.51 eV [1]. ΔE = 0.967 eV = 1.55 × 10⁻¹⁹ J [1]. λ = hc/ΔE = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 1.55 × 10⁻¹⁹ = 1.29 × 10⁻⁶ m [1]. This is longer than visible wavelengths, so it is infrared [1]. [4] Examiner insight: subtract upper minus lower so the photon energy is positive.

10. mvr = 9.11 × 10⁻³¹ × 1.09 × 10⁶ × 2.12 × 10⁻¹⁰ = 2.105 × 10⁻³⁴ J s [1]. n = mvr × 2π / h = 2.105 × 10⁻³⁴ × 2π / 6.63 × 10⁻³⁴ = 1.995 [1], so n = 2 [1]. E = −13.6/4 = −3.40 eV [1]. [4] Examiner insight: give the unrounded value before stating the integer n.

11. (a) Gaps from n = 1: to n = 2, 10.20 eV; to n = 3, 12.09 eV; to n = 4, 12.75 eV [1]. 10.20, 12.09 and 12.75 eV photons are absorbed [1]. 11.00 eV is not absorbed because it matches no gap [1]. [3] (b) The possible transitions are 3 to 1, 3 to 2 and 2 to 1 [1]. Photon energies 12.09 eV, 1.89 eV and 10.20 eV [1][1]. The longest wavelength comes from the smallest energy, 1.89 eV = 3.02 × 10⁻¹⁹ J [1]. λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 3.02 × 10⁻¹⁹ = 6.58 × 10⁻⁷ m [1]. [5] Examiner insight: show the gaps you checked against each photon energy.

12. (a) R = 1.2 × 208^(1/3) [1] = 7.11 fm [1]. [2] (b) E_k = 6.8 × 10⁶ × 1.60 × 10⁻¹⁹ = 1.09 × 10⁻¹² J [1]. d = 2kZe²/E_k = 2 × 8.99 × 10⁹ × 82 × (1.60 × 10⁻¹⁹)² / 1.09 × 10⁻¹² [1] = 3.47 × 10⁻¹⁴ m [1]. [3] (c) d (34.7 fm) is about five times R, so the alpha particle stops well outside the nucleus [1]. [1] (d) Radius ratio = 1.0 × 10⁻¹⁰ / 7.11 × 10⁻¹⁵ = 1.41 × 10⁴, so volume ratio = (1.41 × 10⁴)³ = 2.8 × 10¹² [1]. Almost all the mass is in the nucleus, so its density is far greater than the atom’s average density [1]. [2] Examiner insight: the volume ratio is the cube of the radius ratio, not the square.

Where marks are usually lost

• Writing the photon energy as a level energy rather than the difference between two levels.
• Substituting eV straight into E = hf without converting to joules.
• Describing the scattering experiment without the conclusion each observation supports.
• Using charge e instead of 2e for the alpha particle in a closest-approach calculation.
• Stating that deviations occur at high energy without naming the strong nuclear force or the approach to the nuclear radius.
• Using a positive sign for Bohr energy levels.

Next steps

• Revise from the revision notes and re-read the study guide.
• Go to the IB Physics course hub and tick off the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Physics guide, first assessment 2025, published by the International Baccalaureate Organization: sub-topic E.1 Structure of the atom.