Coaching Centre
Matter waves and Schrödinger’s atom
De Broglie's proposal that matter has a wavelength, λ = h/mv; how standing electron waves explain Bohr's allowed orbits; the experimental evidence from Davisson and Germer and from G. P. Thomson's electron diffraction; why everyday objects show no wave behaviour; Heisenberg's uncertainty principle; and Schrödinger's contribution to the modern model of the atom with orbitals and quantum numbers.
In 1924 a French prince, Louis de Broglie, submitted a PhD thesis with a radical idea: if light waves can behave like particles, perhaps particles can behave like waves. His examiners weren't sure whether it was genius or nonsense, so they sent it to Einstein, who replied that de Broglie had "lifted a corner of the great veil". Three years later, electrons fired at a crystal made a diffraction pattern.
De Broglie's hypothesis
By the 1920s light was known to behave as a wave (diffraction, interference) and as a stream of particles (the photoelectric effect, where each photon carries $E = hf$ and momentum $p = h/\lambda$). De Broglie proposed that the same relationship works in reverse for matter: any particle with momentum $p = mv$ has a wavelength
$$\lambda = \frac{h}{mv}$$
where $h = 6.626 \times 10^{-34}$ J s is Planck's constant. This is the de Broglie wavelength. The idea is wave–particle duality: matter and light both have wave and particle properties, and which one shows depends on the experiment.
For an electron accelerated from rest through a potential difference $V$, its kinetic energy is $\tfrac12 mv^2 = qV$, so $mv = \sqrt{2mqV}$ and $\lambda = \dfrac{h}{\sqrt{2mqV}}$ (provided the speed is well below the speed of light).
An electron is accelerated from rest through 150 V. Find its de Broglie wavelength.
$\tfrac12 m v^2 = qV$, so $v = \sqrt{\dfrac{2qV}{m}} = \sqrt{\dfrac{2 \times 1.602 \times 10^{-19} \times 150}{9.109 \times 10^{-31}}} = 7.26 \times 10^{6}$ m s−1
$\lambda = \dfrac{h}{mv} = \dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 7.26 \times 10^{6}} = 1.00 \times 10^{-10}$ m $= 0.100$ nm
That is about the spacing of atoms in a crystal, so crystals should diffract electrons just as they diffract X-rays.
Find the de Broglie wavelength of a 160 g cricket ball bowled at 40 m s−1.
$\lambda = \dfrac{h}{mv} = \dfrac{6.626 \times 10^{-34}}{0.160 \times 40} = 1.0 \times 10^{-34}$ m
This is about $10^{19}$ times smaller than a proton. Diffraction is only noticeable when a wave passes through a gap or around an object comparable in size to its wavelength. Nothing that small exists, so the ball's wave nature can never be observed.
Explaining Bohr's orbits
De Broglie's idea gave the first physical reason for Bohr's mysterious rule. If the electron is a wave travelling round the nucleus, a stable orbit is one where the wave fits exactly, so it joins up with itself and forms a standing wave. The circumference must be a whole number of wavelengths:
$$2\pi r = n\lambda = n\frac{h}{mv} \quad\Rightarrow\quad mvr = \frac{nh}{2\pi}$$
This is exactly Bohr's quantisation of angular momentum. In any other orbit the wave would interfere destructively with itself after each loop and could not persist. A standing wave also doesn't transfer energy, which suggests why an electron in an allowed orbit doesn't radiate.
The experimental evidence
Davisson and Germer (1927). At Bell Labs in New York, Clinton Davisson and Lester Germer were studying how electrons scatter off a nickel surface. An accident (air got into the vacuum tube and they heated the nickel to clean it) turned the nickel into a few large crystals. Afterwards the scattered electrons came off strongly at particular angles. With 54 eV electrons they found a sharp peak at 50° from the incident beam.
Treat the rows of surface atoms (spacing $d = 0.215$ nm) as a diffraction grating. Constructive interference occurs where $d\sin\theta = m\lambda$. For the first order at 50°: $\lambda = 0.215 \times \sin 50° = 0.165$ nm. De Broglie's formula predicts
$$\lambda = \frac{h}{\sqrt{2m_e qV}} = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-19} \times 54}} = 0.167\ \text{nm}$$
The agreement (about 1%) confirmed $\lambda = h/mv$. Changing the voltage moved the peak to new angles, exactly as predicted.
G. P. Thomson (1927). At the University of Aberdeen, George Paget Thomson fired fast electrons through very thin metal foils and photographed concentric diffraction rings, like those made by X-rays passing through powdered crystals. The ring sizes matched de Broglie wavelengths. In a nice twist, J. J. Thomson won the 1906 Nobel Prize for showing the electron is a particle; his son shared the 1937 prize with Davisson for showing it is a wave.
Since then, diffraction and interference have been shown for neutrons, atoms and molecules as large as C60 and beyond. Electron microscopes and neutron diffraction (such as at ANSTO's OPAL reactor at Lucas Heights) use matter waves every day to image and study materials.
- In the diffraction tube, set 4.0 kV. Read the wavelength (about 19.4 pm) and the inner ring radius (about 12.3 mm).
- Increase the voltage to 8 kV. What happens to the rings? Use $\lambda = h/\sqrt{2meV}$ to explain why.
- Explain why there are two rings. (They come from two different spacings of planes of atoms in the graphite.)
- Switch to Compare wavelengths. Find a particle whose wavelength is close to atomic spacings and one whose wavelength is smaller than a nucleus.
- Compare the cricket ball and the walking person. Why do neither ever diffract?
Heisenberg's uncertainty principle
A wave is spread out, so a particle that is also a wave can't have a perfectly sharp position. In 1927 Werner Heisenberg showed that the more precisely a particle's position is known, the less precisely its momentum can be known, and vice versa. The product of the uncertainties can never be smaller than about $h/4\pi$. This is not a limit of our instruments but a property of nature. It means an electron in an atom can't follow a definite path like Bohr's circular orbits.
Schrödinger's contribution
In 1926 Erwin Schrödinger wrote an equation for how matter waves behave, the quantum equivalent of Newton's laws. Its solution for a particle is a wave function, $\psi$. Max Born showed that $|\psi|^2$ at a point gives the probability of finding the particle there.
Solving Schrödinger's equation for the hydrogen atom:
- reproduced Bohr's energy levels, $E_n = -13.6/n^2$ eV, but without any assumptions added by hand: quantisation arises naturally, the way a guitar string can only vibrate at certain frequencies.
- replaced orbits with orbitals: three-dimensional regions where the electron is likely to be found, drawn as probability clouds.
- introduced further quantum numbers, giving each orbital its shape and orientation. With electron spin (a fourth quantum number) and Pauli's exclusion principle (no two electrons in an atom can have the same set of four), this explains the arrangement of electrons in shells and subshells, and so the structure of the periodic table.
- allowed the probabilities of transitions to be calculated, explaining why some spectral lines are brighter than others, and could be extended to atoms with many electrons and to molecules.
| Bohr model (1913) | Schrödinger's quantum mechanical model (1926) | |
|---|---|---|
| Electron | A particle on a definite circular orbit | A wave function; only the probability of position is known |
| Quantisation | Assumed (angular momentum = $nh/2\pi$) | Arises from the wave equation and boundary conditions |
| Atoms it works for | Hydrogen and one-electron ions only | All atoms (with approximations for many electrons) |
| Line intensities, fine structure, chemistry | Not explained | Explained (with spin and relativity) |
Schrödinger's model is still the basis of atomic physics and chemistry today, refined by relativistic quantum theory.
Test yourself
An electron's speed doubles. Its de Broglie wavelength
$\lambda = h/mv$: doubling v halves λ.
A proton and an electron have the same speed. Compared with the electron, the proton's de Broglie wavelength is
Same v, so λ ∝ 1/m. The proton is 1836 times more massive.
The Davisson–Germer experiment provided evidence that
Electrons were diffracted by a crystal, a wave behaviour, with λ matching h/mv.
De Broglie's hypothesis explains Bohr's allowed orbits because in them
$2\pi r = n\lambda$ leads directly to $mvr = nh/2\pi$.
Which is a feature of Schrödinger's model but not Bohr's?
Both models have energy levels, transitions and a nucleus; only Schrödinger's replaces definite orbits with probability.
Label the Davisson–Germer experiment.
Which model of the atom has each feature?
De Broglie proposed that the wavelength of a particle is $\lambda = h/$. Electrons were shown to off a nickel crystal by Davisson and Germer. Schrödinger described electrons using a wave , whose square gives the of finding the electron. Regions where electrons are likely to be found are called .
In a Davisson–Germer type experiment, electrons accelerated through 54 V are diffracted by a crystal surface with atom spacing 0.215 nm. Calculate the de Broglie wavelength of the electrons and the angle of the first-order maximum, and explain how the result supports de Broglie's hypothesis.
$v = \sqrt{2qV/m} = \sqrt{2 \times 1.602 \times 10^{-19} \times 54 / 9.109 \times 10^{-31}} = 4.36 \times 10^{6}$ m s⁻¹. $\lambda = h/mv = 6.626 \times 10^{-34}/(9.109 \times 10^{-31} \times 4.36 \times 10^{6}) = 1.67 \times 10^{-10}$ m = 0.167 nm. First order: $\sin\theta = \lambda/d = 0.167/0.215 = 0.777$, so $\theta = 51°$. A strong maximum is observed at about 50°. Diffraction maxima are a wave property, and the measured angle matches the wavelength predicted by $\lambda = h/mv$, so electrons behave as waves with de Broglie's wavelength.
Analyse the contribution of Schrödinger to the current model of the atom.
Building on de Broglie's idea that electrons are waves, Schrödinger (1926) wrote an equation describing how a matter wave behaves in a potential such as that of a nucleus. Its solutions are wave functions, and $|\psi|^2$ gives the probability of finding the electron at each point. This replaced Bohr's definite circular orbits with orbitals, three-dimensional probability clouds, consistent with Heisenberg's uncertainty principle, which rules out precise paths.
For hydrogen, the equation reproduces Bohr's energy levels, $-13.6/n^2$ eV, but quantisation now arises naturally from the wave having to fit its boundary conditions, rather than being assumed. The solutions also give further quantum numbers that describe each orbital's shape and orientation; together with spin and the Pauli exclusion principle these explain the arrangement of electrons in subshells and the periodic table. Unlike Bohr's model, the approach can be applied to atoms with many electrons and to molecules, and it allows transition probabilities, and therefore the relative intensities of spectral lines, to be calculated.
Schrödinger's contribution was therefore central: it gave a single theory from which atomic structure follows, and it remains the foundation of the current quantum mechanical model of the atom.