People just love those bright and colourful “neon” signs. They are basically tubes containing gases at low pressure. When a voltage (at least a few hundred volts) is applied across the ends of such a discharge tube, some of the gas atoms become ionized, thus enabling a discharge current. The gas gets heated up and gives off light.

All gaseous atoms display a discrete line spectrum. This is easily confirmed by passing the light through diffraction gratings. In fact, each element in its gaseous state has a unique line spectrum. This suggests that atoms of each element has its own discrete energy levels. And the transitions between each element’s unique set of energy levels give rise to each element’s signature spectral lines.

Unfortunately, the energy levels of atoms with two or more electrons cannot be described by a simple formula like the one for the hydrogen atom. To be honest, Bohr’s model of electrons as particles orbiting around the nucleus has long become obsolete[1]. Nevertheless, its main ideas of quantization and electronic transition remain true to all atoms.

Example

A hypothetical atom has four energy levels: the ground level and 1.90 eV, 2.90 eV and 3.70 eV above the ground level respectively. Find the wavelengths of the spectral lines that this atom can emit when excited.

Solution

Firstly, let’s sketch the energy lines for this atom. Note that rightfully, all the energy levels should be negative in value (since the electrons are still bound to the nucleus). But for convenience, we often depict the ground level as the reference level of 0 eV and the excited levels as positive energy levels.

Suppose the atom is excited to energy level *E*_{4}. From here, the atom can return to ground level in one de-excitation, producing one single photon. Or it can make brief stops at each intermediate energy level along the away, returning to ground state in three de-excitations, producing three (less energetic) photons. Or it can return to ground state in two “jumps”, producing two photons.

In general, given *N* energy levels, there are possible down transitions (since we are looking for the number of ways to choose 2 out of *N* energy lines). For this hypothetical atom, the maximum possible number of spectral lines is . They come from the three down transitions ending at *E*_{1}, two ending at *E*_{2}, and one ending at *E*_{3}.[2]

We can calculate the wavelength of the emitted photon for any de-excitation using

Using the de-excitation from *E*_{2} to *E*_{1} as an example, we have

Calculations for all the possible transitions are tabulated below.

Transition | E_{photon} | λ_{photon} | EM region |

4→1 | 3.70 eV | 336 nm | UV |

3→1 | 2.90 eV | 429 nm | violet |

2→1 | 1.90 eV | 654 nm | red |

4→2 | 1.80 eV | 691 nm | red |

3→2 | 1.00 eV | 1240 nm | IR |

4→3 | 0.80 eV | 1550 nm | IR |

It turns out that only three spectral lines (429 nm, 654 nm and 691 nm) fall in the visible spectrum. The rest are in the ultraviolet and infrared regions.

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**Demonstrations**

Emission Spectrum of Discharge Tube and Filament Lamps

**Concept Test**

**Comics**

[1] It’s a long (but truly fascinating) story. But our current understanding is that electrons exist as a wave-like probability cloud around the nucleus.

[2] In reality, momentum and other constraints mean transitions between some energy levels are impossible. So the actual number of spectral lines are much fewer than * ^{N}*C

_{2}.