Why do atoms emit and absorb light only at specific wavelengths, and how does the Bohr model explain this?
Explain atomic emission and absorption spectra using the Bohr model and the relationship between photon energy and energy-level differences.
How discrete atomic energy levels in the Bohr model produce line emission and absorption spectra, and the link between photon energy and the difference between energy levels.
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What this dot point is asking
You need to explain emission and absorption line spectra using the Bohr model and the relationship between photon energy and the difference between atomic energy levels.
The Bohr model
Bohr proposed that electrons can only exist in certain allowed orbits, each with a fixed energy - the atom's energy levels. The lowest level is the ground state; higher levels are excited states. An electron cannot have an energy between the allowed levels.
A photon whose energy does not match a level difference is simply not absorbed.
Why spectra are lines, not continuous
Because the energy levels are discrete, only certain transitions - and therefore only certain photon energies and wavelengths - are possible. Each element has its own unique set of levels, so it produces a unique pattern of spectral lines, like a fingerprint.
This is how astronomers identify the elements in stars: dark absorption lines in starlight reveal which elements are present in the star's atmosphere.
Reading energy-level diagrams
Energy levels are usually drawn as horizontal lines, with the ground state at the bottom (most negative energy) and the ionisation level at eV. The energy is negative because the electron is bound; raising it to eV frees it (ionisation). Larger downward jumps emit higher-energy, shorter-wavelength photons.
Exam-style practice questions
Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2023 SACE Stage 22 marksExplain how the presence of discrete frequencies in the line emission spectrum of hydrogen provides evidence for the existence of states with discrete electron energy-levels in hydrogen.Show worked answer →
A photon is emitted when an electron drops from a higher energy level to a lower one, with photon energy equal to the difference between the levels: E photon = E high - E low, and frequency given by E = h f.
If the electron energy levels were continuous, transitions of every possible energy would occur and the spectrum would be a continuous band of all frequencies.
Instead, hydrogen emits only certain discrete frequencies (sharp coloured lines). Each line corresponds to one specific energy gap, so the energies that can be released are restricted to fixed values. This is only possible if the electron can occupy only certain discrete energy levels.
1 mark for linking each emitted frequency to a transition between two specific levels via E = h f, 1 mark for concluding that discrete frequencies imply discrete (quantised) energy levels.
2024 SACE Stage 24 marksA hydrogen atom has electron energy levels n = 1 at -13.6 eV, n = 2 at -3.40 eV, n = 3 at -1.51 eV and n = 4 at -0.85 eV. An electron in the n = 4 state transitions down to the n = 1 state. Calculate the frequency of the photon emitted as a result of this transition.Show worked answer →
The photon energy equals the difference between the two energy levels: E = E4 - E1 = (-0.85) - (-13.6) = 12.75 eV.
Convert to joules using 1 eV = 1.60 x 10^-19 J: E = 12.75 x 1.60 x 10^-19 = 2.04 x 10^-18 J.
Use E = h f to find the frequency: f = E / h = (2.04 x 10^-18) / (6.63 x 10^-34) = 3.08 x 10^15 Hz.
1 mark for the energy difference in eV, 1 mark for the sign handling, 1 mark for the conversion to joules, 1 mark for using E = h f to reach about 3.08 x 10^15 Hz (in the ultraviolet region).
2023 SACE Stage 24 marksAn electron raised to the n = 3 energy-level transitions from n = 3 to n = 2 emitting a photon of frequency 8.45 x 10^14 Hz. The n = 3 level is at -8.0 eV. Determine the energy-level of the n = 2 state. Give your answer in eV.Show worked answer →
The energy of the emitted photon is E = h f = (6.63 x 10^-34)(8.45 x 10^14) = 5.60 x 10^-19 J.
Convert to eV: E = (5.60 x 10^-19) / (1.60 x 10^-19) = 3.50 eV.
This photon energy equals the gap between the levels: E photon = E3 - E2, so E2 = E3 - E photon.
E2 = (-8.0) - 3.50 = -11.5 eV.
1 mark for the photon energy in joules, 1 mark for converting to eV, 1 mark for E2 = E3 - E photon, 1 mark for the answer of -11.5 eV. The n = 2 level is lower (more negative) than n = 3, as expected.