The quantum mechanical model
The Bohr model explained the line emission spectrum of hydrogen, but it pictured the electron as a tiny ball circling the nucleus on a fixed path, like a planet around the Sun. Two ideas from twentieth-century physics ruled out that picture: matter can behave like a wave, and there is a fundamental limit to how precisely we can know a particle's motion. If an electron has no path, how can we say where it is?
Wave–particle duality: de Broglie's hypothesis
In 1927, Davisson and Germer fired electrons at a nickel crystal and saw a diffraction pattern just like the one produced by light. How can a particle with mass and charge behave like a wave?
Diffraction patterns, one electron at a time
If electrons are sent through a thin crystal foil one at a time, each one lands at a single point on the detector. The first few impacts look random. After many thousands of electrons, the impacts build up a pattern of bright rings and dark gaps, exactly like light interference. Each electron arrives at a single point, not spread out. The wave tells us where it is likely to land: $|\psi|^2$ gives the probability of arriving at each point. A single impact cannot be predicted, but when many impacts are added together the probabilities build up the overall pattern.
For those who want to go further (this is not required at Ordinary Level), the rings follow the Bragg condition for constructive diffraction from planes a distance $d$ apart, $n\lambda = 2d\sin\theta$. Here $n$ is the order of the ring and $\theta$ is the angle between the beam and the planes. Once you know $\lambda = h/mv$ and $d$, you can predict the angle of each ring. The key idea for the exam is simpler: the rings show that electrons behave like waves.
A beam of electrons travels at a constant speed of $2.20 \times 10^{6}\,\mathrm{m\times s^{-1}}$. Data: mass of the electron…
A beam of electrons travels at a constant speed of $2.20 \times 10^{6}\,\mathrm{m\times s^{-1}}$. Data: mass of the electron $m_{\mathrm{e}} = 9,109 \times 10^{-31}\,\mathrm{kg}$; Planck constant $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$.
- a) Calculate the de Broglie wavelength associated with each electron in the beam.
- b) Justify whether this wavelength is comparable to typical interatomic distances in a solid (of the order of $10^{-10}\,\mathrm{m}$) and what physical consequence this has for particle diffraction.
Consider two particles with the same speed of $15.0\,\mathrm{m\times s^{-1}}$: a very slow free neutron and a microscopic dust particle with a mass of…
Consider two particles with the same speed of $15.0\,\mathrm{m\times s^{-1}}$: a very slow free neutron and a microscopic dust particle with a mass of $2.00 \times 10^{-10}\,\mathrm{g}$. Data: mass of the neutron $m_{\mathrm{n}} = 1,675 \times 10^{-27}\,\mathrm{kg}$; Planck constant $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$.
- a) Calculate the de Broglie wavelength associated with each of the two particles.
- b) Compare the values obtained with observable dimensions (such as visible light or atomic spacing) and explain why wave behaviour is relevant for the neutron but completely negligible for the dust particle.
A particle impact detector records individual electrons transmitted through a thin polycrystalline metal foil with interplanar spacing $d = 0.125\,\mathrm{nm}$. At the…
A particle impact detector records individual electrons transmitted through a thin polycrystalline metal foil with interplanar spacing $d = 0.125\,\mathrm{nm}$. At the start, when only 20 electrons have arrived, the screen records apparently random point impacts. After the continued passage of $10^{6}$ electrons, the impacts accumulate to form a sharp pattern of concentric rings identical to optical interference patterns. Data: $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$; $m_{\mathrm{e}} = 9,109 \times 10^{-31}\,\mathrm{kg}$.
- a) Given that the first-order constructive diffraction rings follow the geometric condition $\lambda = 2d\sin\theta$ and that the first ring appears at an angle $\theta = 30.0^\circ$, determine the speed of the incident electrons.
- b) Analyse this experiment in light of wave-particle duality and the quantum mechanical concept of probability: clarify whether each electron behaves as a particle or as a wave during its propagation and detection.
Heisenberg's uncertainty principle
The speed of an electron is determined experimentally as $v = 4.50 \times 10^{5}\,\mathrm{m\times s^{-1}}$ with an uncertainty of $0.10\,\%$. Data:…
The speed of an electron is determined experimentally as $v = 4.50 \times 10^{5}\,\mathrm{m\times s^{-1}}$ with an uncertainty of $0.10\,\%$. Data: $m_{\mathrm{e}} = 9,109 \times 10^{-31}\,\mathrm{kg}$; $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$; $\pi = 3.1416$.
- a) Determine the minimum uncertainty in the measurement of its position ($\Delta x$).
- b) Using the result, explain why the classical concept of a defined trajectory is not applicable to an electron inside the atom.
The uncertainty in the measurement of motion is analysed for two bodies: an electron confined in a theoretical nuclear region of…
The uncertainty in the measurement of motion is analysed for two bodies: an electron confined in a theoretical nuclear region of $\Delta x = 1.00 \times 10^{-14}\,\mathrm{m}$ and a tennis ball of mass $58.0\,\mathrm{g}$ located with a precision of $\Delta x = 1.00 \times 10^{-6}\,\mathrm{m}$. Data: $m_{\mathrm{e}} = 9,109 \times 10^{-31}\,\mathrm{kg}$; $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$; $c = 3.00 \times 10^{8}\,\mathrm{m\times s^{-1}}$; $\pi = 3.1416$.
- a) First calculate $\Delta p_{\min}=h/(4\pi\Delta x)$ and the value that would result for $\Delta v$ if the non-relativistic approximation $p=mv$ were applied in both cases.
- b) Explain why the formal value obtained for the electron shows that the non-relativistic approximation is no longer valid, and does not show that an electron can exceed the speed of light. Compare this case with the ball.
In an electron gun, electrons start from rest and are accelerated by an electric potential difference $\Delta V = 150\,\mathrm{V}$. The beam is then passed through a…
In an electron gun, electrons start from rest and are accelerated by an electric potential difference $\Delta V = 150\,\mathrm{V}$. The beam is then passed through a collimator that restricts its transverse coordinate with an uncertainty $\Delta x = 0.200\,\mathrm{nm}$. Data: elementary charge $e = 1,602 \times 10^{-19}\,\mathrm{C}$; mass of the electron $m_{\mathrm{e}} = 9,109 \times 10^{-31}\,\mathrm{kg}$; Planck constant $h = 6,626 \times 10^{-34}\,\mathrm{J\times s}$; $\pi = 3.1416$.
- a) Calculate the speed reached by the electrons and determine their associated de Broglie wavelength.
- b) Calculate the minimum uncertainty introduced in the transverse component of their velocity ($\Delta v_x$) and determine the approximate minimum angular spread of the beam $\theta \approx \Delta v_x / v$. Conclude whether it is acceptable to describe the beam using classical rays with straight-line trajectories and no divergence.
Atomic orbitals: the wave function and probability
If we cannot measure an electron's path, how can we describe where it is? Quantum mechanics answers with probabilities, not paths.
Electrons in an atom are arranged in main energy levels (n = 1, 2, 3, ...). Each main level is divided into sublevels (s, p, d), and each sublevel contains one or more orbitals. For example, $1s$ and $2s$ are s orbitals in the first and second main levels, and $2p$ is a set of p orbitals in the second main level. Each orbital holds a maximum of two electrons.
The s orbitals are spherical. A p orbital has two lobes on opposite sides of the nucleus, with a nodal plane through the nucleus where the probability of finding the electron is zero. The three 2p orbitals ($2p_x$, $2p_y$, $2p_z$) point along the three axes.
For a $2p_z$ orbital the nodal plane is the $xy$ plane. The probability of finding the electron anywhere in that plane is zero, even though the electron can be found in both lobes.
Within quantum mechanics, the concept of orbit in the Bohr model is replaced by that of atomic orbital. a) Define precisely the term "atomic orbital", stating which…
Within quantum mechanics, the concept of orbit in the Bohr model is replaced by that of atomic orbital.
- a) Define precisely the term "atomic orbital", stating which physical quantity the square of the wave function ($|\psi|^2$) represents.
- b) State whether knowing the orbital allows the position of an electron to be predicted with certainty before it is measured, and justify your answer by citing the relevant physical principle.
The following illustration shows a two-dimensional cross-section of the dot density corresponding to an atomic orbital of type $s$, highlighting an inner region of high…
The following illustration shows a two-dimensional cross-section of the dot density corresponding to an atomic orbital of type $s$, highlighting an inner region of high relative density (A) and a peripheral region of lower density (B), bounded by a contour surface. Observe the diagram
and answer the following questions:
- a) Explain the physical meaning of the dot density in this diagram and justify why a precise circular or elliptical orbit is not drawn in the quantum model.
- b) Describe what the outer contour line usually shown in textbooks for this type of orbital represents, and explain whether the algebraic signs ($+$ or $-$) that are sometimes assigned to wave functions represent positive or negative electric charge.
You can do it on paper and send a photo with your answer.
Back to the start: the electron has no path, so we describe it with an orbital, a region where it is most likely to be found.