IB Diploma Physics · topic E.1
Structure of the atom
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.
Inferring discrete levels from a line spectrum
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.
The rare deflections that changed the atomic model
A narrow alpha-particle beam strikes a very thin gold foil. Detectors covering equal solid angles record the supplied count rates; values are illustrative rather than a full scattering-law fit. Most alpha particles pass close to the forward direction. Gold-197 has atomic number 79, and an alpha particle is a helium-4 nucleus of charge +2e. Supplied data (Scattering angle / °; Counts per minute): (10, 100000), (30, 10000), (60, 1000), (90, 100), (150, 10). (a) Explain what the common small deflections and rare large deflections imply, and compare them with a model of diffuse positive charge. [3 marks] (b) Determine proton, neutron and electron numbers in neutral gold-197 and in a gold-197 ion of charge +3e. Describe an isotope and balance alpha emission from a parent with A = 201, Z = 81. [3 marks]
Illustrative alpha-scattering observations
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Counts per minute
Horizontal axis (x): Scattering angle / °
Read the exact data as a table
| Scattering angle / ° | Counts per minute |
|---|---|
| 10 | 100000 |
| 30 | 10000 |
| 60 | 1000 |
| 90 | 100 |
| 150 | 10 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Explain what the common small deflections and rare large deflections imply, and compare them with a model of diffuse positive charge.
Reveal the worked solution
- Most projectiles cross the foil with little change in direction, so atoms are not solid dense obstacles across their full dimensions.
- A rare close encounter with concentrated positive nuclear charge can strongly repel an alpha particle. Very large deflections require a strong local interaction and a sufficiently massive scattering centre.
- A diffuse positive-charge model would spread the repulsion over the atomic volume and struggles to explain the observed rare strong deflections. The evidence supports a compact nucleus, while this count table alone does not determine an exact nuclear radius.
- Atomic number gives 79 protons. Neutrons are A−Z = 197−79 = 118; neutrality requires 79 electrons.
- A +3e ion has lost three electrons and has 76; its nucleus is unchanged. An isotope has the same proton number but a different neutron number.
- Alpha emission removes A = 4 and Z = 2: ²⁰¹₈₁X → ¹⁹⁷₇₉Au + ⁴₂He. Both nucleon and charge numbers balance.
Answer (a) Most atomic volume is sparsely occupied; positive charge and most mass are concentrated in a small nucleus. (b) Neutral Au-197: 79 p, 118 n, 79 e; Au³⁺ has 76 e. Alpha daughter: A = 197, Z = 79.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.