MYP Physics · topic MYP-17.01
Atomic energy levels and spectra
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
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.