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Why do atoms emit and absorb light at only specific wavelengths?

Topic 15.2 The Bohr Model and Atomic Spectra: relate quantised energy levels to the emission and absorption spectra of atoms.

A focused answer to AP Physics 2 Topics 15.2 and 15.3, covering the Bohr model of quantised electron energy levels, the emission of a photon when an electron drops between levels, the relation E = hf for the photon energy, and the line emission and absorption spectra that result, with full worked examples.

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  1. What this topic is asking
  2. The Bohr model: quantised energy levels
  3. Photons from transitions
  4. Emission and absorption spectra
  5. Try this

What this topic is asking

The College Board (Topics 15.2 and 15.3) want you to relate the quantised energy levels of the Bohr model to the emission and absorption spectra of atoms, using E=hfE = hf for the photon released or absorbed in a transition.

The Bohr model: quantised energy levels

The Bohr model's key idea is quantisation: an electron in an atom cannot have just any energy, only the specific values of the allowed levels. The levels are usually negative (the electron is bound), with the lowest level (the ground state) the most negative. An electron sits in a level without radiating, and only changes energy by jumping to another allowed level, never to an energy in between. This discreteness is what makes atomic spectra sharp.

Photons from transitions

This is conservation of energy applied to the atom: the photon carries away (or supplies) exactly the energy gap between two levels. Because the levels are fixed and discrete, the gaps are fixed and discrete, so the photon energies, and hence the frequencies f=E/hf = E/h and wavelengths, can take only specific values. A larger jump emits a higher-energy (higher-frequency, shorter-wavelength) photon. This single rule, photon energy equals the level difference, generates the entire spectrum.

Emission and absorption spectra

Because only certain transitions are allowed, an atom emits light at only specific wavelengths, producing an emission spectrum of sharp bright lines on a dark background. Conversely, when white light passes through a cool gas, the atoms absorb exactly those same wavelengths (lifting electrons up levels), leaving an absorption spectrum of dark lines at the identical positions. Each element has its own unique pattern of lines, a spectral fingerprint, which is how astronomers determine what distant stars are made of. The strategic role of these topics is that they apply the photon idea of Topic 15.1 (E=hfE = hf) to the atom, showing that line spectra are direct evidence of quantised energy levels, the same quantisation that the wave nature of electrons (de Broglie) explains. This quantum picture of the atom is the foundation for understanding the nucleus and radioactivity in the unit's final topics.

Try this

Q1. State what an electron does to emit a photon, in terms of energy levels. [1 point]

  • Cue. It drops from a higher energy level to a lower one, emitting a photon equal to the energy difference.

Q2. State why each element produces a unique line spectrum. [1 point]

  • Cue. Each element has its own set of quantised energy levels, so its own set of allowed transition energies and wavelengths.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)7 marksSection II (long FRQ). An electron in an atom drops from an energy level of −2.4×10−19-2.4 \times 10^{-19} J to a lower level of −5.4×10−19-5.4 \times 10^{-19} J. Take h=6.63×10−34h = 6.63 \times 10^{-34} J s and c=3.0×108c = 3.0 \times 10^8 m/s. (a) Calculate the energy of the emitted photon. (b) Calculate its frequency. (c) Explain why atoms emit light at only certain discrete wavelengths.
Show worked answer →

A 7-point FRQ on the Bohr model and spectra.

(a) Photon energy (2 points): the photon carries the energy lost by the electron, E=Ehigh−Elow=(−2.4×10−19)−(−5.4×10−19)=3.0×10−19E = E_{high} - E_{low} = (-2.4 \times 10^{-19}) - (-5.4 \times 10^{-19}) = 3.0 \times 10^{-19} J.
(b) Frequency (3 points): f=Eh=3.0×10−196.63×10−34=4.5×1014f = \dfrac{E}{h} = \dfrac{3.0 \times 10^{-19}}{6.63 \times 10^{-34}} = 4.5 \times 10^{14} Hz.
(c) Discrete wavelengths (2 points): the electron can occupy only certain quantised energy levels, so the energy difference between levels, and thus the photon energy and wavelength, can take only specific values, giving a line spectrum.

Markers reward the photon energy as the level difference, the frequency from E=hfE = hf, and the quantised-levels explanation of discrete spectra.

AP 2023 (style)1 marksSection I (multiple choice). An atom's emission spectrum consists of a few sharp colored lines rather than a continuous band. This is best explained by (A) the atom being very hot (B) electrons occupying only discrete energy levels (C) the photons having no energy (D) total internal reflection. Justify your reasoning.
Show worked answer →

A 1-point MCQ on line spectra. The answer is (B).

Sharp spectral lines arise because electrons can occupy only discrete (quantised) energy levels, so transitions release photons of only specific energies and wavelengths. A continuous range of levels would give a continuous spectrum. The trap is (A): temperature affects which lines appear and their brightness, but the discreteness comes from quantised levels.

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