Emission spectra and models of the atom
Photon energy and wavelength, how excited electrons produce line emission spectra, the flame test investigation and its limitations, and the Bohr and Schrödinger models of the atom compared.
When the fireworks go off over Sydney Harbour on New Year's Eve, the colours are chemistry. Strontium salts give the reds, barium the greens, copper the blues and sodium the golden yellows. Each metal's atoms, heated in the burst, give out light of exactly the same colours every time, and nobody has ever found a way to make strontium glow green. The reason those colours are fixed is the reason we know electrons sit in definite energy levels.
Light carries energy in packets
Visible light is part of the electromagnetic spectrum, with wavelengths from about 400 nm (violet) to 700 nm (red). Light behaves as a wave, with $c = f\lambda$, but it also arrives in packets of energy called photons. The energy of one photon is proportional to its frequency:
$$E = hf = \frac{hc}{\lambda}$$
where h = 6.626 × 10⁻³⁴ J s (Planck's constant), c = 3.00 × 10⁸ m s⁻¹ and λ is in metres. Shorter wavelength means higher energy: a violet photon carries almost twice the energy of a red one.
Sodium's yellow light has a wavelength of 589 nm. Calculate the energy of one photon, and of one mole of photons.
$E = \dfrac{hc}{\lambda} = \dfrac{(6.626\times10^{-34})(3.00\times10^{8})}{589\times10^{-9}} = 3.37\times10^{-19}\text{ J}$
Per mole: $3.37\times10^{-19} \times 6.022\times10^{23} = 2.03\times10^{5}$ J mol⁻¹ = 203 kJ mol⁻¹, comparable to the energy of a chemical bond.
How atoms emit light
An atom with all its electrons in the lowest available levels is in its ground state. Heat or an electric discharge can give an electron enough energy to jump to a higher level: the atom is now in an excited state. Excited states are unstable. Within about a hundred-millionth of a second the electron falls back to a lower level and the energy difference is released as a single photon.
Because the energy levels in an atom have fixed values, the gaps between them have fixed values too. So an element emits only certain wavelengths. Passing the light through a prism or diffraction grating spreads it into an emission line spectrum: a few bright coloured lines on a dark background. Every element's line spectrum is different, like a fingerprint. A hot solid such as a light bulb filament gives a continuous spectrum instead, containing every colour.
The reverse also happens. When white light passes through a cool gas, atoms absorb exactly the photons that match their energy gaps, leaving dark lines in a continuous spectrum: an absorption spectrum. In 1868 astronomers saw an unknown yellow line in the Sun's spectrum; the element responsible was named helium after the Greek word for the Sun, 27 years before it was found on Earth.
In a hydrogen atom, the energy difference between levels n = 3 and n = 2 is 3.03 × 10⁻¹⁹ J. Calculate the wavelength of light emitted when an electron falls from n = 3 to n = 2, and state its colour.
$\lambda = \dfrac{hc}{E} = \dfrac{(6.626\times10^{-34})(3.00\times10^{8})}{3.03\times10^{-19}} = 6.56\times10^{-7}\text{ m}$
656 nm is red: the bright red line in the hydrogen spectrum.
Flame tests
A Bunsen flame is hot enough to excite the outer electrons of some metal ions, so each gives a characteristic colour. Not every metal does: magnesium, for example, has energy gaps that correspond to ultraviolet light, so a magnesium salt leaves the flame unchanged to your eye.
| Metal ion | Flame colour |
|---|---|
| Lithium, Li⁺ | crimson red |
| Sodium, Na⁺ | intense yellow-orange |
| Potassium, K⁺ | lilac (pale violet) |
| Calcium, Ca²⁺ | brick red (orange-red) |
| Strontium, Sr²⁺ | scarlet red |
| Barium, Ba²⁺ | pale (apple) green |
| Copper, Cu²⁺ | blue-green |
- In Reference spectra, pick Na and leave the cursor on 589 nm. The photon energy reads 3.375 × 10⁻¹⁹ J (2.11 eV). Now move the cursor to 410 nm on hydrogen: why is the energy higher?
- Open Mystery 1. The flame is yellow. Turn on the cobalt-blue glass and use the reference buttons to identify both metals.
- Solve Mystery 2 and Mystery 3. Lithium and strontium are both red in a flame: how does the spectrum tell them apart?
- Mystery 5 has three metals. What is the minimum number of lines you must match before you can be sure a metal is present?
The Bohr model
In 1913 Niels Bohr proposed that the electron in a hydrogen atom moves in circular orbits of fixed radius and fixed energy, labelled n = 1, 2, 3 ... An electron in an allowed orbit does not radiate energy. It absorbs or emits a photon only when it jumps between orbits, and the photon's energy equals the difference between them. Bohr calculated the energies of hydrogen's orbits and predicted the wavelengths of its spectral lines with remarkable accuracy.
Strengths: it explained why spectra are lines, gave the right wavelengths for hydrogen, and introduced quantised energy levels, which survive in all later models.
Limitations: it only worked for hydrogen and other one-electron ions; it could not explain the spectra of atoms with two or more electrons, why some lines are brighter than others, or why some lines split into closely spaced pairs; and it gave no reason why an orbiting electron should not radiate and spiral into the nucleus.
The Schrödinger model
In the 1920s Louis de Broglie suggested that electrons behave as waves, and in 1926 Erwin Schrödinger wrote an equation describing an electron as a standing wave around the nucleus. Its solutions are orbitals: three-dimensional regions in which there is a high probability of finding the electron. The model does not give the electron an exact path. Heisenberg's uncertainty principle says that is impossible anyway: you cannot know an electron's position and momentum precisely at the same time.
The solutions come out naturally in families that match the s, p, d and f subshells, with exactly the shapes and numbers of orbitals in lesson 1D. That is why the Schrödinger model, now called the quantum mechanical model, explains the spectra of all elements and the structure of the periodic table.
| Bohr (1913) | Schrödinger (1926) | |
|---|---|---|
| Electron described as | a particle in a circular orbit | a wave; position given only as a probability |
| Energy levels | quantised shells n = 1, 2, 3 | quantised shells split into s, p, d, f subshells |
| Location of electron | exact radius | orbital (region of high probability) |
| Explains | hydrogen's line spectrum | spectra of all atoms, the periodic table, bond shapes |
| Main weakness | fails for atoms with more than one electron | mathematically complex; solved exactly only for hydrogen and approximated for others |
Test yourself
Which statement best explains why elements produce line spectra rather than continuous spectra?
Quantised energy levels give fixed gaps and therefore fixed photon energies.
A photon has an energy of 4.85 × 10⁻¹⁹ J. Its wavelength is
λ = hc/E = 1.988 × 10⁻²⁵ ÷ 4.85 × 10⁻¹⁹ = 4.10 × 10⁻⁷ m = 410 nm (violet).
A main limitation of the Bohr model is that it
It worked for hydrogen but failed for multi-electron atoms.
A sample gives a yellow flame. Through cobalt-blue glass, a lilac flame is seen. The sample contains
The glass absorbs sodium's yellow, revealing the lilac of potassium.
In the Schrödinger model, an orbital is
Orbitals describe probability, not paths.
Label the emission diagram.
Which model does each statement describe?
An atom with all electrons in their lowest levels is in its state. When an electron falls to a lower level, a is emitted. The energy of the photon equals the between the two energy levels. Light of shorter carries more energy. The yellow flame of can hide other flame colours.
A student performs flame tests on solutions of lithium chloride and strontium chloride. Both flames appear red. (a) Explain, in terms of electrons and energy levels, how the red light is produced. (b) Explain how the student could reliably distinguish the two metals. (c) Evaluate the flame test as a method of identifying metal ions.
(a) The flame's heat gives outer electrons enough energy to move to higher energy levels (excited state). These states are unstable, and the electrons fall back to lower levels. Each fall releases a photon with energy equal to the gap between the levels, E = hc/λ. In both metals some gaps match red wavelengths.
(b) View each flame through a hand spectroscope. Lithium gives one strong red line at 671 nm (plus weaker lines), while strontium gives several red-orange lines and bands between about 600 and 690 nm and a blue line at 461 nm. Matching the pattern against reference spectra identifies each metal.
(c) The flame test is quick, cheap and needs little equipment, and line spectra are specific to each element. However, flame colours are judged by eye and several metals give similar colours; trace sodium contamination swamps other colours; some metals, such as magnesium, produce no visible colour; and the test only shows whether a metal is present, not how much. It is useful for a rapid preliminary identification, but a spectroscope or instrumental method such as atomic absorption spectroscopy is needed for a reliable or quantitative result.
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