17.3.3 Absorption Spectrum

The absorption spectrum is basically the reverse of the emission spectrum. Instead of revealing the energies of the photons emitted by gas atoms, they reveal the energies of the photons absorbed by gas atoms.

To produce an absorption spectrum, we need to pass a beam of white light through a gas. The beam of “white” light contains photons of a continuous spectrum of energies. This light beam passing through the gas is like photons trying to sneak past gas atoms without getting gobbled up by the gas atoms. Since a gas atom has discrete energy levels, most photons will pass through without being absorbed. This is because a gas atom is only able to absorb a photon only if the absorption excites the atom from a lower energy level E_L to a higher one E_H. In other words, for a photon to be absorbed, its photons energy must correspond to

$\displaystyle \frac{{hc}}{\lambda }={{E}_{H}}-{{E}_{L}}$

After being “filtered” by the gas, the light beam is passed through a diffraction grating. On the other side of the grating we observe an absorption spectrum, which is basically a brightly lit continuous spectrum punctured by a few dark lines. Those dark lines are called the absorption lines. From the position of the absorption lines, we can tell which frequencies of photons have been absorbed by the gas. Since each type of gas has its own set of discrete energy levels, each type of gas produces its own characteristic absorption lines.

Re-emission

What happens after the atoms have absorbed those photons? Don’t these excited atoms eventually (if not immediately) return to ground level and inadvertently re-emit the same photons that they just absorbed? That would “spill” the absorbed photons back into the “white” light beam and “fill-up” the dark lines in the continuous spectrum, wouldn’t it? Well, yes and no. You see, the photons are indeed re-emitted, but they are re-emitted randomly and uniformly in all directions. So only a tiny fraction will be re-emitted in the original direction (towards the grating). In that sense, scatter may be a more accurate word to describe this “absorb and then re-emit” process. Furthermore, to form an absorption line, it is not even necessary for to remove all the photons of that absorption line from the white light beam. As long as a small fraction is scattered, the dip in intensity is enough to cast a dark line against the very bright continuous spectrum.

Example

An absorption spectrum is formed by passing white light through a gas consisting of the hypothetical atoms mentioned in worked example 3. Find the wavelength of the absorption lines.

Solution

In general, all the possible down transitions are also the possible up transitions.

The photons energies that the atom is capable of emitting, are also the photons energies that the atom is capable of absorbing.

Transition	E_photon	λ_photon	EM region
1→4	3.70 eV	336 nm	UV
1→3	2.90 eV	429 nm	violet
1→2	1.90 eV	654 nm	red
2→4	1.80 eV	691 nm	red
2→3	1.00 eV	1240 nm	IR
3→4	0.80 eV	1550 nm	IR

This is why the absorption spectrum for a particular gas often “matches” its emission spectrum: the emission lines in the emission spectrum are also the absorption lines in the absorption spectrum.

Cool Gas

This is assuming that all the possible up transitions are taking place. This may not be true if a gas is at such low temperature that there are too few atoms existing at the higher energy levels. For example, suppose that gas in our worked example is so “cool” that at any one time, practically all the atoms are at the ground state. Since there is hardly any atom waiting at E₂, E₃ and E₄, the 2→4, 2→3 and 3→4 absorptions would not be happening. The only absorptions occurring would be those that begin from the ground level, i.e. 1→2, 1→3 and 1→4. In such a scenario, the 691 mm absorption line (2→4) will not show up in the absorption spectrum.