On Diwali night the sky fills with colour. The red of a firework comes from a salt of strontium, the green from barium, the golden yellow from sodium and the blue from copper. The makers do not mix paints. They choose the metal, and the metal decides the colour. The same salt gives the same colour in every firework, every year.
The ladder of levels
The energies of the Bohr orbits are easier to use as a picture than as a list. Draw a horizontal line for each allowed energy, lower energies lower down. The result is an energy level diagram, a ladder. For hydrogen the rungs are at \(-13.6\), \(-3.40\), \(-1.51\), \(-0.85\) and \(-0.54\,\text{eV}\), and so on without end.
The lowest rung, \(n = 1\), is the ground state. An atom left alone settles there and stays for ever. Every higher rung is an excited state. An atom lifted to one does not stay long; it usually falls back within about \(10^{-8}\,\text{s}\). At the top, \(E = 0\) stands for an electron at rest, just free of the atom. It is labelled \(n = \infty\).
The rungs are not evenly spaced. The first gap is \(10.2\,\text{eV}\), the next \(1.89\,\text{eV}\), then \(0.66\,\text{eV}\), and the gaps keep shrinking. The upper rungs crowd together just below zero, because \(E_n\) goes as \(1/n^2\). This crowding will show itself in the spectrum, as lines crowding together.
For a hydrogen-like ion every rung is \(Z^2\) times lower, since \(E_n = -13.6\,Z^2/n^2\) eV. The ground state of \(\text{He}^+\) is at \(-54.4\,\text{eV}\), and that of \(\text{Li}^{2+}\) at \(-122.4\,\text{eV}\). The whole ladder stretches by \(Z^2\), as Fig 4.1 shows, so every gap stretches by \(Z^2\) too.
As \(n\) increases, what happens to the energy gap between neighbouring levels of hydrogen?
Lifting the electron
The energy needed to lift the electron from the ground state to a higher level is the excitation energy of that level, \(\Delta E = E_n - E_1\). For hydrogen, the first excitation energy, to \(n = 2\), is \(-3.40 - (-13.6) = 10.2\,\text{eV}\). The second, to \(n = 3\), is \(-1.51 + 13.6 = 12.09\,\text{eV}\). The accelerating voltage that gives an electron this energy is the excitation potential: \(10.2\,\text{V}\) for the first.
Freeing the electron
To remove the electron completely from the ground state, it must be lifted to \(E = 0\). The energy needed is the ionisation energy, \(0 - E_1 = 13.6\,Z^2\,\text{eV}\). For hydrogen it is \(13.6\,\text{eV}\), for \(\text{He}^+\) \(54.4\,\text{eV}\), and for \(\text{Li}^{2+}\) \(122.4\,\text{eV}\). The ionisation potential is the voltage that gives an electron this energy, \(13.6\,\text{V}\) for hydrogen.
An electron already in an excited state is easier to free. From level \(n\) it needs only its binding energy \(|E_n| = 13.6\,Z^2/n^2\) eV, which is \(3.4\,\text{eV}\) from \(n = 2\) of hydrogen. If more energy than this is supplied, the electron leaves with the surplus as kinetic energy.
Find the ionisation energy and the first excitation energy of \(\text{He}^+\) (\(Z = 2\)) and of \(\text{Li}^{2+}\) (\(Z = 3\)). What is the ionisation potential of hydrogen?
Both are hydrogen-like, so every energy is \(Z^2\) times hydrogen's. Hydrogen's values are \(13.6\,\text{eV}\) and \(10.2\,\text{eV}\).
\(E_{\text{ion}} = 13.6\,Z^2\) and \(\Delta E_{1 \to 2} = 13.6\,Z^2 \times \tfrac34 = 10.2\,Z^2\) eV.
\(\text{He}^+\): \(54.4\,\text{eV}\) and \(40.8\,\text{eV}\). \(\text{Li}^{2+}\): \(122.4\,\text{eV}\) and \(91.8\,\text{eV}\).
The ionisation potential of hydrogen is the voltage that gives an electron \(13.6\,\text{eV}\): \(13.6\,\text{V}\). Values used: ground state of hydrogen \(-13.6\,\text{eV}\), \(1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}\).
Find the second excitation energy of hydrogen, from \(n = 1\) to \(n = 3\), in eV.
A hydrogen atom in its ground state absorbs a photon of \(15\,\text{eV}\). What happens?
How much energy, in eV, is needed to free the electron of a hydrogen atom that is already in the state \(n = 2\)?
Light out, light in
When an electron falls from a higher level to a lower one, the atom gives out one photon of energy \(h\nu = E_i - E_f\). With \(hc = 1240\,\text{eV nm}\), the wavelength follows at once. A gap \(\Delta E\) in eV gives a photon of wavelength \(\lambda = 1240/\Delta E\) nm. The fall from \(n = 3\) to \(n = 2\) releases \(1.89\,\text{eV}\), which is red light of \(656.5\,\text{nm}\).
The reverse happens too. A photon whose energy exactly matches a gap from the electron's level can be absorbed, and the electron jumps up. A photon that matches no gap cannot be absorbed at all, and it passes on. The photon is taken whole or not at all.
After an absorption the atom soon gives the energy back. Fig 4.2 shows two ways. The electron may fall from \(n = 3\) straight to \(n = 1\), giving one ultraviolet photon of \(12.09\,\text{eV}\), \(102.6\,\text{nm}\). Or it may stop at \(n = 2\) on the way, giving a red photon of \(1.89\,\text{eV}\) and then an ultraviolet one of \(10.2\,\text{eV}\), \(121.6\,\text{nm}\). Both routes return the same \(12.09\,\text{eV}\).
Emission and absorption spectra
A hot gas, or a gas in an electric discharge, has many atoms in excited states. Their falls give light at a set of sharp wavelengths, the emission spectrum of the element: bright lines on a dark background. Now shine white light, which has every wavelength, through a cool gas. The atoms take out the wavelengths that match their gaps. The result is the absorption spectrum: dark lines on a bright rainbow.
In cool hydrogen almost every atom is in the ground state. So absorption can start only from \(n = 1\). Cool hydrogen therefore takes out only the lines of jumps from \(n = 1\): \(121.6\), \(102.6\) and \(97.3\,\text{nm}\), and so on, all ultraviolet. Hot hydrogen emits many more lines, since its electrons fall from every level to every lower one.
White light passes through cool hydrogen gas. Why is the red line of \(656.5\,\text{nm}\) (the jump between \(n = 2\) and \(n = 3\)) not absorbed?
A hydrogen atom in \(n = 3\) gives out a photon of \(1.89\,\text{eV}\), and then a second photon that takes it to the ground state. What is the energy of the second photon, in eV?
A hydrogen atom falls from \(n = 3\) to \(n = 1\). Find the wavelength of the photon, in nm. Take \(hc = 1240\,\text{eV nm}\).
Now the colours of Diwali. Each element has its own nucleus and its own electrons, so each has its own ladder of levels. In the heat of the burning powder, its atoms are lifted up the ladder and fall back. Each fall gives a photon whose energy is a gap of that ladder, and no other. Sodium's strongest gap is about \(2.1\,\text{eV}\), which gives yellow light of \(589\,\text{nm}\).
In the school chemistry laboratory, a platinum wire dipped in a salt and held in a Bunsen flame names the metal by its colour. Sodium gives golden yellow, potassium lilac, calcium brick red and copper blue-green. Astronomers use the same idea with a spectrograph. The dark lines in sunlight told them which elements are in the Sun, and helium was found there before it was found on the Earth.
- Thinking a photon with more energy than a gap is still absorbed by a bound jump.Between bound levels the match must be exact. Only above the ionisation energy is any surplus allowed, as kinetic energy of the freed electron.
- Using \(\Delta E = 13.6(1 - 1/n^2)\) from a level other than \(n = 1\).In general \(\Delta E = E_i - E_f\). The formula with \(1 - 1/n^2\) is only for jumps from or to the ground state.
- Forgetting that energies of ions go as \(Z^2\).\(\text{He}^+\) needs \(4\) times and \(\text{Li}^{2+}\) \(9\) times hydrogen's energy for the same jump.