Atomic structure

How is an atom built, and why does each element glow only in its own colours? The answers came from Rutherford’s experiment, Bohr’s quantum postulates and the theory of the hydrogen atom. The topic continues quantum physics: an atom emits and absorbs energy only in definite portions, as photons. Spectral analysis, lasers, fluorescent lamps and even our understanding of chemical bonds rest on this knowledge.

§1. From Thomson’s model to Rutherford’s experiment

After Thomson discovered the electron in 1897, it was clear that atoms contain negative particles. Since an atom is neutral overall, it must also contain positive charge. In Thomson’s model the atom is a positively charged “sphere” with electrons embedded in it like raisins in a pudding.

To test this model, Rutherford (1911) fired α-particles (helium nuclei) at thin gold foil and counted the scattered particles by the flashes they made on a screen:

• most particles passed almost straight through, so the atom is mostly empty space;
• a very small fraction were deflected through large angles and a few even bounced back; this needs a very strong field, so the positive charge and almost all the mass are concentrated in a tiny volume, the nucleus.

The nuclear (planetary) model: in the centre is a positive nucleus about 10⁻¹⁵ m across; electrons move around it at about 10⁻¹⁰ m. The atom is roughly a hundred thousand times larger than its nucleus. In a neutral atom the number of electrons equals the nuclear charge Z (the atomic number in the periodic table).

The problem: according to classical electrodynamics an orbiting (accelerating) electron should radiate electromagnetic waves, lose energy and spiral into the nucleus within a tiny fraction of a second. Such an atom would also give a continuous spectrum. In reality atoms are stable and their spectra consist of lines.

§2. Bohr’s postulates

Bohr (1913) solved the problem with new rules that contradict classical physics but agree with experiment:

1. An electron can exist only in certain stationary orbits, where it does not radiate. Each orbit has a definite energy E_n.
2. When the atom moves from one stationary state to another it emits or absorbs a photon whose energy equals the difference between the levels:

3. The condition for stationary orbits is quantisation of angular momentum:

A jump from a higher to a lower level emits a photon; a jump upwards absorbs one. An atom can absorb only a photon whose energy exactly matches a level difference; other photons pass by. Franck and Hertz (1914) showed directly that levels are discrete by colliding electrons with mercury vapour.

§3. The hydrogen atom

Bohr’s theory gives exact results for the one-electron atom, hydrogen:

• n = 1 is the ground state (lowest energy); n > 1 are excited states. An excited atom usually returns to the ground state in about 10⁻⁸ s.
• The energy is negative because the electron is bound to the nucleus; E = 0 means the electron has been removed.
• The ionisation energy from level n is 13.6/n² eV; from the ground state it is 13.6 eV.
• Spectral series: transitions to n = 1 form the Lyman series (ultraviolet), to n = 2 the Balmer series (visible), to n = 3 the Paschen series (infrared).
• Atoms excited up to level n can produce n(n − 1)/2 different lines.

A handy formula: λ (nm) ≈ 1240/E (eV), since hc ≈ 1240 eV·nm.

Worked example

An electron drops from level 4 to level 2. Find the photon energy and wavelength.

1. E = 13.6·(1/4 − 1/16) = 13.6·3/16 = 2.55 eV.
2. In joules: E = 2.55·1.6·10⁻¹⁹ ≈ 4.08·10⁻¹⁹ J.
3. λ = hc/E = 1.989·10⁻²⁵/(4.08·10⁻¹⁹) ≈ 4.87·10⁻⁷ m ≈ 487 nm, a blue-green Balmer line.

Worked example

The ionisation energy of a hydrogen atom is 3.4 eV. What does it become if the electron’s angular momentum increases by ħ?

1. 13.6/n² = 3.4 → n² = 4, n = 2.
2. Since L = nħ, increasing it by ħ gives n = 3.
3. New ionisation energy: 13.6/9 ≈ 1.51 eV.

Worked example

Find the wavelength of the first Lyman line (2 → 1).

1. E = 13.6·(1 − 1/4) = 10.2 eV = 10.2·1.6·10⁻¹⁹ ≈ 1.63·10⁻¹⁸ J.
2. λ = 1.989·10⁻²⁵/(1.63·10⁻¹⁸) ≈ 1.22·10⁻⁷ m = 122 nm, ultraviolet and invisible to the eye.

§4. De Broglie waves and the quantum-mechanical picture

Why exactly mvr = nħ? De Broglie (1924) proposed that every particle has a wave of wavelength λ = h/(mv). For an orbit to be stationary, a whole number of wavelengths must fit around it:

So Bohr’s third postulate follows from the wave nature of the electron.

In modern quantum mechanics the electron does not follow a definite orbit: we know only the probability of finding it at a given point, and the region where it is likely to be is called an orbital (an “electron cloud”). The state of an electron is described by four quantum numbers: principal (n), orbital (l), magnetic (m) and spin (s). By the Pauli exclusion principle, no two electrons in an atom can have all four quantum numbers the same. This principle explains how electron shells fill and why the periodic table looks the way it does. Bohr’s model gives the hydrogen levels correctly but cannot explain atoms with many electrons.

Worked example

Find the speed of the electron in the first hydrogen orbit and its de Broglie wavelength (m = 9.1·10⁻³¹ kg, ħ ≈ 1.055·10⁻³⁴ J·s).

1. From mvr₁ = ħ: v = ħ/(m·r₁) = 1.055·10⁻³⁴/(9.1·10⁻³¹·0.53·10⁻¹⁰) ≈ 2.19·10⁶ m/s.
2. For n = 1: λ = 2πr₁ = 2·3.14·0.53·10⁻¹⁰ ≈ 3.33·10⁻¹⁰ m.
3. Answer: v ≈ 2.2·10⁶ m/s (more than 100 times slower than light), λ ≈ 0.33 nm.

§5. Spectra and spectral analysis

• A line emission spectrum comes from rarefied atomic gases; each element’s spectrum is unique, like a fingerprint.
• A band spectrum comes from molecules: groups of very closely spaced lines.
• A continuous spectrum comes from hot solids, liquids and dense gases.
• An absorption spectrum shows dark lines on a continuous background: a gas absorbs exactly the wavelengths it emits (Kirchhoff’s law).

Spectral analysis finds the chemical composition of a substance from its spectrum. It is very sensitive and detects tiny amounts of matter. This is how the composition of the Sun and stars was found; helium was first discovered in the solar spectrum (1868) and found on Earth later.

§6. Lasers

An atom can emit a photon by itself (spontaneous emission) or when triggered by another photon (stimulated emission). In stimulated emission the new photon has the same frequency, phase, polarisation and direction as the triggering photon, so one photon becomes two. Normally the lower level is more populated and absorption dominates. In a laser an external energy source (pumping) makes the upper level more populated than the lower one, a population inversion. Between two mirrors (the resonator) photons pass back and forth many times and are amplified in an avalanche; one mirror is partly transparent and lets the beam out.

Properties of laser light: coherence, monochromaticity, very small divergence and high intensity. Uses: fibre-optic communication, eye surgery, cutting and welding metals, barcode scanners, distance measurement.

Check yourself

1. What is the energy of hydrogen’s level 3 (in eV)?

2. Find the photon energy and wavelength for a transition from level 3 to level 2.

3. Radius of hydrogen’s second orbit?

4. Smallest energy needed to excite a hydrogen atom from the ground state?

5. How many spectral lines can atoms excited up to level 5 produce?

6. A ground-state atom absorbs a 12.75 eV photon. Which level does the electron reach?

7. What does it show that most α-particles in Rutherford’s experiment passed straight through?

8. What state must be created in a laser?

Common mistakes

• Mixing up signs: level energies are negative, but the emitted photon energy is positive.
• Using n instead of n²: E₃ = −13.6/9, not −13.6/3.
• Always taking the ionisation energy as 13.6 eV: that holds only for the ground state.
• Putting eV into λ = hc/E without converting to joules.
• Concluding from Rutherford’s experiment that the atom is dense: it is mostly empty.
• Thinking an atom can absorb a photon of any energy: only one matching a level difference.